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New conformal map for the trapezoidal formula for infinite integrals of unilateral rapidly decreasing functions111This work was partially supported by JSPS Grant-in-Aid for Young Scientists (B) JP17K14147.

Tomoaki Okayama Graduate School of Information Sciences, Hiroshima City University, 3-4-1, Ozuka-higashi, Asaminami-ku, Hiroshima 731-3194, Japan okayama@hiroshima-cu.ac.jp Tomoki Nomura Hitachi Information Engineering, Ltd., Hiroshima K building 7F, 6-13, Nakamachi, Naka-ku, Hiroshima 730-0037, Japan Saki Tsuruta Hiroshima Municipal Funairi High School, 1-4-4, Funairi Minami, Naka-ku, Hiroshima 730-0847, Japan
Abstract

While the trapezoidal formula can attain exponential convergence when applied to infinite integrals of bilateral rapidly decreasing functions, it is not capable of this in the case of unilateral rapidly decreasing functions. To address this issue, Stenger proposed the application of a conformal map to the integrand such that it transforms into bilateral rapidly decreasing functions. Okayama and Hanada modified the conformal map and provided a rigorous error bound for the modified formula. This paper proposes a further improved conformal map, with two rigorous error bounds provided for the improved formula. Numerical examples comparing the proposed and existing formulas are also given.

keywords:
trapezoidal formula, Conformal map, Computable error bound
MSC:
[2010] 65D30 , 65D32 , 65G20
††journal: Elsevier

1 Introduction and summary

In this paper, we are concerned with the trapezoidal formula for the infinite integral, expressed as

∫−∞∞f​(x)​dx≈h​∑k=−∞∞f​(k​h),\int_{-\infty}^{\infty}f(x)\mathrm{d}x\approx h\sum_{k=-\infty}^{\infty}f(kh),

where hh is a mesh size. This approximation formula is fairly accurate if the integrand f​(x)f(x) is analytic, which has been known since several decades ago [6, 7]. For example, the approximation

∫−∞∞e−x2⁡d​x≈h​∑k=−∞∞e−(k​h)2\int_{-\infty}^{\infty}\operatorname{\mathrm{e}}^{-x^{2}}\mathrm{d}x\approx h\sum_{k=-\infty}^{\infty}\operatorname{\mathrm{e}}^{-(kh)^{2}}

gives the correct answer in double-precision with h=1/2h=1/2, and the approximation

∫−∞∞14+x2​dx≈h​∑k=−∞∞14+(k​h)2\int_{-\infty}^{\infty}\frac{1}{4+x^{2}}\mathrm{d}x\approx h\sum_{k=-\infty}^{\infty}\frac{1}{4+(kh)^{2}}

gives the correct answer in double-precision with h=1/3h=1/3. In general, however, the infinite sum on the right-hand side cannot be calculated, and thus, the sum has to be truncated at some MM and NN as

∫−∞∞f​(x)​dx≈h​∑k=−MNf​(k​h).\int_{-\infty}^{\infty}f(x)\mathrm{d}x\approx h\sum_{k=-M}^{N}f(kh).

In the case where f​(x)=e−x2f(x)=\operatorname{\mathrm{e}}^{-x^{2}}, this approximation requires h=1/2h=1/2 and M=N=12M=N=12 to obtain the correct answer in double-precision. On the other hand, in the case where f​(x)=1/(4+x2)f(x)=1/(4+x^{2}), this approximation requires h=1/3h=1/3 and M=N=1016M=N=10^{16} to obtain the correct answer in double-precision. This is because f​(x)=e−x2f(x)=\operatorname{\mathrm{e}}^{-x^{2}} is a rapidly decreasing function, i.e., ff decays exponentially as x→±∞x\to\pm\infty, whereas f​(x)=1/(4+x2)f(x)=1/(4+x^{2}) is not.

In the case where the integrand f​(x)f(x) is not a rapidly decreasing function, a useful solution is the application of an appropriate conformal map before applying the (truncated) trapezoidal formula. When f​(x)f(x) decays algebraically as x→±∞x\to\pm\infty like f​(x)=1/(4+x2)f(x)=1/(4+x^{2}), by applying a conformal map x=sinh⁡tx=\sinh t, a new integral is obtained:

∫−∞∞f​(x)​dx=∫−∞∞f​(sinh⁡t)​cosh⁡t​d​t,\int_{-\infty}^{\infty}f(x)\mathrm{d}x=\int_{-\infty}^{\infty}f(\sinh t)\cosh t\,\mathrm{d}t,

where the transformed integrand f​(sinh⁡t)​cosh⁡tf(\sinh t)\cosh t decays exponentially as t→±∞t\to\pm\infty. Therefore, the (truncated) trapezoidal formula should yield an accurate result when applied to the new integral. Appropriate conformal maps for certain typical cases have been usefully summarized by Stenger [8, 9].

One of the cases listed in the summary is rather convoluted: the integrand f​(x)f(x) decays exponentially as x→∞x\to\infty, but decays algebraically as x→−∞x\to-\infty, like f​(x)=1/{(4+x2)​(1+ex)}f(x)=1/\{(4+x^{2})(1+\operatorname{\mathrm{e}}^{x})\}. We refer to such a function as a unilateral rapidly decreasing function. In such a case, Stenger [9] proposed the employment of a conformal map

x=ψ​(t)=sinh⁡(log⁡(arcsinh⁡(et))),\displaystyle x=\psi(t)=\sinh(\log(\operatorname{\mathrm{arcsinh}}(\operatorname{\mathrm{e}}^{t}))),

and applied the trapezoidal formula as

∫−∞∞f​(x)​dx=∫−∞∞f​(ψ​(t))​ψ′​(t)​dt≈h​∑k=−MNf​(ψ​(k​h))​ψ′​(k​h).\int_{-\infty}^{\infty}f(x)\mathrm{d}x=\int_{-\infty}^{\infty}f(\psi(t))\psi^{\prime}(t)\mathrm{d}t\approx h\sum_{k=-M}^{N}f(\psi(kh))\psi^{\prime}(kh). (1)

Furthermore, by appropriately setting hh, MM, and NN depending on the given positive integer nn, he theoretically analyzed the error as O⁡(e−2​π​d​μ′​n)\operatorname{\mathrm{O}}(\operatorname{\mathrm{e}}^{-\sqrt{2\uppi d\mu^{\prime}n}}), where μ′\mu^{\prime} indicates the decay rate of the transformed integrand, and dd indicates the width of the domain in which the transformed integrand is analytic (described in detail further on). Okayama and Hanada [2] slightly modified the conformal map as follows:

x=ψ~​(t)=2​sinh⁡(log⁡(arcsinh⁡(et))),x=\tilde{\psi}(t)=2\sinh(\log(\operatorname{\mathrm{arcsinh}}(\operatorname{\mathrm{e}}^{t}))),

and derived a new approximation formula:

∫−∞∞f​(x)​dx=∫−∞∞f​(ψ~​(t))​ψ~′​(t)​dt≈h​∑k=−MNf​(ψ~​(k​h))​ψ~′​(k​h).\int_{-\infty}^{\infty}f(x)\mathrm{d}x=\int_{-\infty}^{\infty}f(\tilde{\psi}(t))\tilde{\psi}^{\prime}(t)\mathrm{d}t\approx h\sum_{k=-M}^{N}f(\tilde{\psi}(kh))\tilde{\psi}^{\prime}(kh). (2)

Furthermore, they theoretically showed that the error of the modified formula, say EnE_{n}, is bounded by

|En|≤C​e−2​π​d​μ​n,|E_{n}|\leq C\operatorname{\mathrm{e}}^{-\sqrt{2\uppi d\mu n}}, (3)

where μ≥μ′\mu\geq\mu^{\prime}, and CC is explicitly given in a computable form. This inequality not only shows that the modified formula (2) can attain faster convergence than (1), but it also indicates that the error can be rigorously estimated by the right-hand side. This is useful for verified numerical integration.

The present work improves upon their results. Rather than the conformal map x=ψ​(t)x=\psi(t) or x=ψ~​(t)x=\tilde{\psi}(t), we propose a new conformal map

x=ϕ​(t)=2​sinh⁡(log⁡(log⁡(1+et))).x=\phi(t)=2\sinh(\log(\log(1+\operatorname{\mathrm{e}}^{t}))).

The principle of this conformal map is derived from the fact that the convergence rate is improved by replacing arcsinh⁡(et)\operatorname{\mathrm{arcsinh}}(\operatorname{\mathrm{e}}^{t}) with log⁡(1+et)\log(1+\operatorname{\mathrm{e}}^{t}) in some fields [1, 3, 5]. Consequently, the following approximation formula is derived:

∫−∞∞f​(x)​dx=∫−∞∞f​(ϕ​(t))​ϕ′​(t)​dt≈h​∑k=−MNf​(ϕ​(k​h))​ϕ′​(k​h).\int_{-\infty}^{\infty}f(x)\mathrm{d}x=\int_{-\infty}^{\infty}f(\phi(t))\phi^{\prime}(t)\mathrm{d}t\approx h\sum_{k=-M}^{N}f(\phi(kh))\phi^{\prime}(kh). (4)

Furthermore, as the main contribution of this work, we provide two (general and special) theoretical error bounds in the same form as (3), where μ\mu does not change, but a larger dd can be taken as compared to that in the previous studies. This indicates that the improved formula (4) can attain faster convergence than (1) and (2).

The remainder of this paper is organized as follows. First, existing and new theorems are summarized in Section 2. Then, numerical examples are provided in Section 3. Finally, proofs of the new theorems are given in Sections 4 and 5.

2 Summary of existing and new results

Sections 2.1 and 2.2 describe the existing results, and Sections 2.3 and 2.4 describe the new results. First, the relevant notations are introduced. Let 𝒟d\mathscr{D}_{d} be a strip domain defined by 𝒟d={ζ∈ℂ:|Im⁡ζ|<d}\mathscr{D}_{d}=\{\zeta\in\mathbb{C}:|\operatorname{Im}\zeta|<d\} for d>0d>0. Furthermore, let 𝒟d−={ζ∈𝒟d:Re⁡ζ<0}\mathscr{D}_{d}^{-}=\{\zeta\in\mathscr{D}_{d}:\operatorname{Re}\zeta<0\} and 𝒟d+={ζ∈𝒟d:Re⁡ζ≥0}\mathscr{D}_{d}^{+}=\{\zeta\in\mathscr{D}_{d}:\operatorname{Re}\zeta\geq 0\}.

2.1 Error analysis of Stenger’s formula

An error analysis for Stenger’s formula (1) can be expressed as the following theorem, which is a restatement of an existing theorem [9, Theorem 1.5.16].

Theorem 2.1 (Okayama–Hanada [2, Theorem 2.1])

Assume that ff is analytic in ψ​(𝒟d)\psi(\mathscr{D}_{d}) with 0<d<π/20<d<\uppi/2, and that there exist positive constants KK, α\alpha, and β\beta such that

|f​(z)|\displaystyle|f(z)| ≤K​|e−z|2​β\displaystyle\leq K|\operatorname{\mathrm{e}}^{-z}|^{2\beta} (5)
holds for all z∈ψ​(𝒟d+)z\in\psi(\mathscr{D}_{d}^{+}), and
|f​(z)|\displaystyle|f(z)| ≤K​1|z|α+1\displaystyle\leq K\frac{1}{|z|^{\alpha+1}} (6)

holds for all z∈ψ​(𝒟d−)z\in\psi(\mathscr{D}_{d}^{-}). Let μ=min⁡{α,β}\mu=\min\{\alpha,\beta\}, let MM and NN be defined as

{M=n,N=⌈α​n/β⌉(if​μ=α),N=n,M=⌈β​n/α⌉(if​μ=β),\begin{cases}M=n,\quad N=\lceil\alpha n/\beta\rceil&\,\,\,(\text{if}\,\,\,\mu=\alpha),\\ N=n,\quad M=\lceil\beta n/\alpha\rceil&\,\,\,(\text{if}\,\,\,\mu=\beta),\end{cases} (7)

and let hh be defined as

h=2​π​dμ​n.h=\sqrt{\frac{2\uppi d}{\mu n}}. (8)

Then, there exists a constant CC independent of nn, such that

|∫−∞∞f​(x)​dx−h​∑k=−MNf​(ψ​(k​h))​ψ′​(k​h)|≤C​e−2​π​d​μ​n.\left|\int_{-\infty}^{\infty}f(x)\mathrm{d}x-h\sum_{k=-M}^{N}f(\psi(kh))\psi^{\prime}(kh)\right|\leq C\operatorname{\mathrm{e}}^{-\sqrt{2\uppi d\mu n}}.

2.2 Error bound for the formula by Okayama and Hanada

Okayama and Hanada [2] proposed the replacement of ψ\psi with ψ~\tilde{\psi} in Stenger’s formula (1). They also provided the following theoretical error bound for the modified formula (2).

Theorem 2.2 (Okayama–Hanada [2, Theorem 2.2])

Assume that ff is analytic in ψ~​(𝒟d)\tilde{\psi}(\mathscr{D}_{d}) with 0<d<π/20<d<\uppi/2, and that there exist positive constants KK, α\alpha, and β\beta such that

|f​(z)|\displaystyle|f(z)| ≤K​|e−z|β\displaystyle\leq K|\operatorname{\mathrm{e}}^{-z}|^{\beta} (9)

holds for all z∈ψ~​(𝒟d+)z\in\tilde{\psi}(\mathscr{D}_{d}^{+}), and

|f​(z)|\displaystyle|f(z)| ≤K​1|4+z2|(α+1)/2\displaystyle\leq K\frac{1}{|4+z^{2}|^{(\alpha+1)/2}} (10)

holds for all z∈ψ~​(𝒟d−)z\in\tilde{\psi}(\mathscr{D}_{d}^{-}). Let μ=min⁡{α,β}\mu=\min\{\alpha,\beta\}, let MM and NN be defined as (7), and let hh be defined as (8). Then, it holds that

|∫−∞∞f​(x)​dx−h​∑k=−MNf​(ψ~​(k​h))​ψ~′​(k​h)|≤K​(2​C11−e−2​π​d​μ+C2)​e−2​π​d​μ​n,\left|\int_{-\infty}^{\infty}f(x)\mathrm{d}x-h\sum_{k=-M}^{N}f(\tilde{\psi}(kh))\tilde{\psi}^{\prime}(kh)\right|\leq K\left(\frac{2C_{1}}{1-\operatorname{\mathrm{e}}^{-\sqrt{2\uppi d\mu}}}+C_{2}\right)\operatorname{\mathrm{e}}^{-\sqrt{2\uppi d\mu n}},

where C1C_{1} and C2C_{2} are constants defined as

C1\displaystyle C_{1} =γdα​arctan⁡(γd)​{γd2​(1+1sin2⁡1)}α+(1+σ2)​γdβ​{2​eσcos⁡(d/2)}β,\displaystyle=\frac{\gamma_{d}}{\alpha\arctan(\gamma_{d})}\left\{\frac{\gamma_{d}}{2}\left(1+\frac{1}{\sin^{2}1}\right)\right\}^{\alpha}+\frac{(1+\sigma^{2})\sqrt{\gamma_{d}}}{\beta}\left\{\frac{\sqrt{2}\operatorname{\mathrm{e}}^{\sigma}}{\cos(d/2)}\right\}^{\beta},
C2\displaystyle C_{2} =1α​{12​(1+1sin2⁡1)}α+1+σ2β​(eσ2)β,\displaystyle=\frac{1}{\alpha}\left\{\frac{1}{2}\left(1+\frac{1}{\sin^{2}1}\right)\right\}^{\alpha}+\frac{1+\sigma^{2}}{\beta}\left(\frac{\operatorname{\mathrm{e}}^{\sigma}}{2}\right)^{\beta},

where γd=1/cos⁡(d)\gamma_{d}=1/\cos(d) and σ=1/arcsinh⁡(1)\sigma=1/\operatorname{\mathrm{arcsinh}}(1).

In Theorem 2.2, the condition (5) is modified to (9), and the condition (6) is modified to (10). The former constitutes the most significant difference, because β\beta in Theorem 2.2 can be two times greater than that in Theorem 2.1, while α\alpha remains unchanged. Owing to the difference, μ\mu in Theorem 2.2 may be greater than that in Theorem 2.1, which affects the convergence rate O⁡(e−2​π​d​μ​n)\operatorname{\mathrm{O}}(\operatorname{\mathrm{e}}^{-\sqrt{2\uppi d\mu n}}).

Another difference between Theorems 2.1 and 2.2 lies in the constants on the right-hand side of the inequalities. All the constants in Theorem 2.2 are explicitly revealed, and the right-hand side can be computed to provide an error bound. This paper provides two error bounds for the improved formula (4) in the same manner as Theorem 2.2.

2.3 General error bound for the proposed formula

As a general case, we present the following error bound for the improved formula (4). The proof is given in Section 4.

Theorem 2.3

Assume that ff is analytic in ϕ​(𝒟d)\phi(\mathscr{D}_{d}) with 0<d<π0<d<\uppi, and that there exist positive constants KK, α\alpha, and β\beta such that (9) holds for all z∈ϕ​(𝒟d+)z\in\phi(\mathscr{D}_{d}^{+}), and (6) holds for all z∈ϕ​(𝒟d−)z\in\phi(\mathscr{D}_{d}^{-}). Let μ=min⁡{α,β}\mu=\min\{\alpha,\beta\}, let MM and NN be defined as (7), and let hh be defined as (8). Then, it holds that

|∫−∞∞f​(x)​dx−h​∑k=−MNf​(ϕ​(k​h))​ϕ′​(k​h)|≤K​(2​C31−e−2​π​d​μ+C4)​e−2​π​d​μ​n,\left|\int_{-\infty}^{\infty}f(x)\mathrm{d}x-h\sum_{k=-M}^{N}f(\phi(kh))\phi^{\prime}(kh)\right|\leq K\left(\frac{2C_{3}}{1-\operatorname{\mathrm{e}}^{-\sqrt{2\uppi d\mu}}}+C_{4}\right)\operatorname{\mathrm{e}}^{-\sqrt{2\uppi d\mu n}},

where C3C_{3} and C4C_{4} are constants defined as

C3\displaystyle C_{3} =(1α+1+1α)​{e⁡cd(1−log⁡2)​(e−1)}α+1​1+{log⁡(2+cd)}2{log⁡(2+cd)}2​(1+cd)2+(1+λ2)​cdβ​(eλ⁡cd)β,\displaystyle=\left(\frac{1}{\alpha+1}+\frac{1}{\alpha}\right)\left\{\frac{\operatorname{\mathrm{e}}c_{d}}{(1-\log 2)(\operatorname{\mathrm{e}}-1)}\right\}^{\alpha+1}\frac{1+\{\log(2+c_{d})\}^{2}}{\{\log(2+c_{d})\}^{2}}(1+c_{d})^{2}+\frac{(1+\lambda^{2})c_{d}}{\beta}\left(\operatorname{\mathrm{e}}^{\lambda}c_{d}\right)^{\beta}, (11)
C4\displaystyle C_{4} =e1/π3α​(1−log⁡2)α+1+1+λ2β​(eλ)β,\displaystyle=\frac{\operatorname{\mathrm{e}}^{1/\uppi^{3}}}{\alpha(1-\log 2)^{\alpha+1}}+\frac{1+\lambda^{2}}{\beta}\left(\operatorname{\mathrm{e}}^{\lambda}\right)^{\beta}, (12)

where cd=1/cos⁡(d/2)c_{d}=1/\cos(d/2) and λ=1/log⁡2\lambda=1/\log 2.

The crucial difference between Theorems 2.2 and 2.3 is the upper bound of dd; d<π/2d<\uppi/2 in Theorem 2.2, whereas d<πd<\uppi in Theorem 2.3. This implies that in the new approximation (4), dd may be greater than dd in the previous approximation (2). In this case, the convergence rate O⁡(e−2​π​d​μ​n)\operatorname{\mathrm{O}}(\operatorname{\mathrm{e}}^{-\sqrt{2\uppi d\mu n}}) is improved (note that μ\mu is not changed between the two theorems).

This difference in the range of dd originates from the conformal maps ψ~\tilde{\psi} and ϕ\phi. By observing the derivatives of the functions

ψ~′​(ζ)=1+arcsinh2⁡(eζ)1+e−2​ζ​arcsinh2⁡(eζ),ϕ′​(ζ)=1+{log⁡(1+eζ)}2(1+e−ζ)​{log⁡(1+eζ)}2,\tilde{\psi}^{\prime}(\zeta)=\frac{1+\operatorname{\mathrm{arcsinh}}^{2}(\operatorname{\mathrm{e}}^{\zeta})}{\sqrt{1+\operatorname{\mathrm{e}}^{-2\zeta}}\operatorname{\mathrm{arcsinh}}^{2}(\operatorname{\mathrm{e}}^{\zeta})},\quad\phi^{\prime}(\zeta)=\frac{1+\{\log(1+\operatorname{\mathrm{e}}^{\zeta})\}^{2}}{(1+\operatorname{\mathrm{e}}^{-\zeta})\{\log(1+\operatorname{\mathrm{e}}^{\zeta})\}^{2}},

we see that ψ~′​(ζ)\tilde{\psi}^{\prime}(\zeta) is not analytic at ζ=±i⁡(π/2)\zeta=\pm\operatorname{\mathrm{i}}(\uppi/2), and ϕ′​(ζ)\phi^{\prime}(\zeta) is not analytic at ζ=±i⁡π\zeta=\pm\operatorname{\mathrm{i}}\uppi. Accordingly, f​(ψ~​(ζ))​ψ~′​(ζ)f(\tilde{\psi}(\zeta))\tilde{\psi}^{\prime}(\zeta) is analytic at most 𝒟π/2\mathscr{D}_{\uppi/2}, and f​(ϕ​(ζ))​ϕ′​(ζ)f(\phi(\zeta))\phi^{\prime}(\zeta) is analytic at most 𝒟π\mathscr{D}_{\uppi}. Therefore, the range of dd is 0<d<π/20<d<\uppi/2 in Theorem 2.2 and 0<d<π0<d<\uppi in Theorem 2.3.

2.4 Special error bound for the proposed formula

As a special case, restricting the range of dd to 0<d<(1+π)/20<d<(1+\uppi)/2, we present the following error bound for the improved formula (4). The proof is given in Section 5.

Theorem 2.4

Assume that ff is analytic in ϕ​(𝒟d)\phi(\mathscr{D}_{d}) with 0<d<(1+π)/20<d<(1+\uppi)/2, and that there exist positive constants KK, α\alpha, and β\beta such that (9) holds for all z∈ϕ​(𝒟d+)z\in\phi(\mathscr{D}_{d}^{+}), and

|f​(z)|≤K​1|4+z2|1/2​|z|α|f(z)|\leq K\frac{1}{|4+z^{2}|^{1/2}|z|^{\alpha}} (13)

holds for all z∈ϕ​(𝒟d−)z\in\phi(\mathscr{D}_{d}^{-}). Let μ=min⁡{α,β}\mu=\min\{\alpha,\beta\}, let MM and NN be defined as (7), and let hh be defined as (8). Then, it holds that

|∫−∞∞f​(x)​dx−h​∑k=−MNf​(ϕ​(k​h))​ϕ′​(k​h)|≤K​(2​C51−e−2​π​d​μ+C6)​e−2​π​d​μ​n,\left|\int_{-\infty}^{\infty}f(x)\mathrm{d}x-h\sum_{k=-M}^{N}f(\phi(kh))\phi^{\prime}(kh)\right|\leq K\left(\frac{2C_{5}}{1-\operatorname{\mathrm{e}}^{-\sqrt{2\uppi d\mu}}}+C_{6}\right)\operatorname{\mathrm{e}}^{-\sqrt{2\uppi d\mu n}},

where C5C_{5} and C6C_{6} are constants defined as

C5\displaystyle C_{5} =1α​{e⁡cd(1−log⁡2)​(e−1)}α​1+cdlog⁡(2+cd)+(1+λ2)​cdβ​(eλ⁡cd)β,\displaystyle=\frac{1}{\alpha}\left\{\frac{\operatorname{\mathrm{e}}c_{d}}{(1-\log 2)(\operatorname{\mathrm{e}}-1)}\right\}^{\alpha}\frac{1+c_{d}}{\log(2+c_{d})}+\frac{(1+\lambda^{2})c_{d}}{\beta}\left(\operatorname{\mathrm{e}}^{\lambda}c_{d}\right)^{\beta}, (14)
C6\displaystyle C_{6} =1α​(1−log⁡2)α+1+λ2β​(eλ)β,\displaystyle=\frac{1}{\alpha(1-\log 2)^{\alpha}}+\frac{1+\lambda^{2}}{\beta}\left(\operatorname{\mathrm{e}}^{\lambda}\right)^{\beta}, (15)

where cd=1/cos⁡(d/2)c_{d}=1/\cos(d/2) and λ=1/log⁡2\lambda=1/\log 2.

In this theorem, the upper bound of dd is (1+π)/2(1+\uppi)/2, which is smaller than that in Theorem 2.3 (π\uppi). This is because the condition (6) is changed to (13), where 4+{ϕ​(ζ)}24+\{\phi(\zeta)\}^{2} (put z=ϕ​(ζ)z=\phi(\zeta)) has zero points at ζ=log⁡(2​sin⁡(1/2))±i⁡(1+π)/2\zeta=\log(2\sin(1/2))\pm\operatorname{\mathrm{i}}(1+\uppi)/2. However, the constants C5C_{5} and C6C_{6} are considerably smaller than C3C_{3} and C4C_{4}, respectively (comparing the first term). Therefore, Theorem 2.4 is useful for attaining a sharp error bound rather than a large upper bound of dd. It must be noted here that (1+π)/2(1+\uppi)/2 is still greater than π/2\uppi/2 in Theorems 2.1 and 2.2.

3 Numerical examples

This section presents the numerical results obtained in this study. All the programs were written in C language with double-precision floating-point arithmetic. The following three integrals are considered:

∫−∞∞{11+(x/2)2+1−(x/2)}2​exp⁡(−x2−1+(x2)2)​dx\displaystyle\int_{-\infty}^{\infty}\left\{\frac{1}{\sqrt{1+(x/2)^{2}}+1-(x/2)}\right\}^{2}\exp\left(-\frac{x}{2}-\sqrt{1+\left(\frac{x}{2}\right)^{2}}\right)\mathrm{d}x =3−4​e⁡E1​(1),\displaystyle=3-4\operatorname{\mathrm{e}}E_{1}(1), (16)
∫−∞∞14+x2​exp⁡(−x2−1+(x2)2)​dx\displaystyle\int_{-\infty}^{\infty}\frac{1}{4+x^{2}}\exp\left(-\frac{x}{2}-\sqrt{1+\left(\frac{x}{2}\right)^{2}}\right)\mathrm{d}x =Ci⁡(1)​sin⁡1−si⁡(1)​cos⁡1,\displaystyle=\operatorname{\mathrm{Ci}}(1)\sin 1-\operatorname{\mathrm{si}}(1)\cos 1, (17)
∫−∞∞12​(1+x4+x2)​11+e(π/2)​x​dx\displaystyle\int_{-\infty}^{\infty}\frac{1}{2}\left(1+\frac{x}{\sqrt{4+x^{2}}}\right)\frac{1}{1+\operatorname{\mathrm{e}}^{(\uppi/2)x}}\mathrm{d}x =1.136877446810281077257​⋯,\displaystyle=1.136877446810281077257\cdots, (18)

where E1​(x)E_{1}(x) is the exponential integral defined by E1​(x)=∫1∞(e−t​x/t)​dtE_{1}(x)=\int_{1}^{\infty}(\operatorname{\mathrm{e}}^{-tx}/t)\mathrm{d}t, Ci⁡(x)\operatorname{\mathrm{Ci}}(x) is the cosine integral defined by Ci⁡(x)=−∫x∞(cos⁡t/t)​dt\operatorname{\mathrm{Ci}}(x)=-\int_{x}^{\infty}(\cos t/t)\mathrm{d}t, and si⁡(x)\operatorname{\mathrm{si}}(x) is the sine integral defined by si⁡(x)=−∫x∞(sin⁡t/t)​dt\operatorname{\mathrm{si}}(x)=-\int_{x}^{\infty}(\sin t/t)\mathrm{d}t. The third integral (18) is taken from the previous study [2].

Table 1: Parameters for the integral (16).
α\alpha β\beta dd KK
Theorem 2.1 11 1/21/2 3/23/2
Theorem 2.2 11 11 3/23/2 11
Theorem 2.3 11 11 33 7878
Theorem 2.4 11 11 22 6/56/5
Table 2: Parameters for the integral (17).
α\alpha β\beta dd KK
Theorem 2.1 11 1/21/2 3/23/2
Theorem 2.2 11 11 3/23/2 16/916/9
Theorem 2.3 11 11 22 215215
Theorem 2.4 11 11 22 3939
Table 3: Parameters for the integral (18).
α\alpha β\beta dd KK
Theorem 2.1 11 π/4\uppi/4 3/23/2
Theorem 2.2 11 π/2\uppi/2 3/23/2 1212
Theorem 2.3 11 π/2\uppi/2 3/23/2 99
Theorem 2.4 11 π/2\uppi/2 3/23/2 9/29/2

The integrand in (16) satisfies the assumptions of Theorems 2.1, 2.2, 2.3, and 2.4 with the parameters shown in Table 1. In Theorem 2.1, KK is not investigated since KK is not used for computation. In Theorems 2.1 and 2.2, dd is taken as d=3/2d=3/2 since d<π/2d<\uppi/2. In Theorem 2.3, dd is taken as d=3d=3 since d<πd<\uppi. In Theorem 2.4, dd is taken as d=2d=2 since d<(1+π)/2d<(1+\uppi)/2. The results are shown in Figure 1. As seen in the graph, the proposed formula with d=3d=3 shows the fastest convergence as compared to the others. However, the corresponding error bound by Theorem 2.3 is relatively large, because the constant C3C_{3} in (11) is large. In contrast, Theorem 2.4 produces a sharp error for the proposed formula with d=2d=2, although the convergence rate is slightly worse than that from Theorem 2.3.

The integrand in (17) satisfies the assumptions of Theorems 2.1, 2.2, 2.3, and 2.4 with the parameters shown in Table 2. In this case, dd must satisfy d<(1+π)/2d<(1+\uppi)/2 in Theorem 2.3, due to the singular points of 1/(4+x2)1/(4+x^{2}). The results are shown in Figure 2. As seen in the graph, the proposed formula with d=2d=2 shows the fastest convergence as compared to the others. Note that in this example, Theorem 2.3 and Theorem 2.4 have the same dd values, and thus, their approximation formulas are exactly the same. As for the error bound, Theorem 2.4 produces a sharper error than Theorem 2.3 in this case as well.

The integrand in (18) satisfies the assumptions of Theorems 2.1, 2.2, 2.3, and 2.4 with the parameters shown in Table 3. In this case, dd must satisfy d<π/2d<\uppi/2 in both Theorems 2.3 and 2.4, due to the singular points of 1/(1+e(π/2)​x)1/(1+\operatorname{\mathrm{e}}^{(\uppi/2)x}). The results are shown in Figure 3. As seen in the graph, all formulas show a similar convergence rate, mainly because all formulas use the same value of dd. Approximation formulas of Theorem 2.3 and Theorem 2.4 are exactly the same, but Theorem 2.4 produces a sharper error than Theorem 2.3 in this case as well.

1e-151e-101e-05120406080100120140ErrornnStenger’s formulaOkayama and Hanada’s formulaNew formula with d=3d=3New formula with d=2d=2Error bound by Theorem 2.2Error bound by Theorem 2.3Error bound by Theorem 2.4
Figure 1: Numerical results for (16).
1e-151e-101e-05120406080100120140ErrornnStenger’s formulaOkayama and Hanada’s formulaNew formula with d=2d=2New formula with d=2d=2Error bound by Theorem 2.2Error bound by Theorem 2.3Error bound by Theorem 2.4
Figure 2: Numerical results for (17).
1e-151e-101e-05120406080100120140ErrornnStenger’s formulaOkayama and Hanada’s formulaNew formula with d=3/2d=3/2New formula with d=3/2d=3/2Error bound by Theorem 2.2Error bound by Theorem 2.3Error bound by Theorem 2.4
Figure 3: Numerical results for (18).

4 Proofs for Theorem 2.3

This section presents the proof of Theorem 2.3. It is organized as follows. In Section 4.1, the task is decomposed into two lemmas: Lemmas 4.2 and 4.3. To prove these lemmas, useful inequalities are presented in Sections 4.2, 4.3, 4.4, and 4.5. Following this, Lemma 4.2 is proved in Section 4.6, and Lemma 4.3 is proved in Section 4.7.

4.1 Sketch of the proof

Let F​(t)=f​(ϕ​(t))​ϕ′​(t)F(t)=f(\phi(t))\phi^{\prime}(t). The main strategy in the proof of Theorem 2.3 is to split the error into two terms as follows:

|∫−∞∞f​(x)​dx−h​∑k=−MNf​(ϕ​(k​h))​ϕ′​(k​h)|\displaystyle\left|\int_{-\infty}^{\infty}f(x)\mathrm{d}x-h\sum_{k=-M}^{N}f(\phi(kh))\phi^{\prime}(kh)\right| =|∫−∞∞F​(t)​dt−h​∑k=−MNF​(k​h)|\displaystyle=\left|\int_{-\infty}^{\infty}F(t)\mathrm{d}t-h\sum_{k=-M}^{N}F(kh)\right|
≤|∫−∞∞F​(t)​dt−h​∑k=−∞∞F​(k​h)|+|h​∑k=−∞−M−1F​(k​h)+h​∑k=N+1∞F​(k​h)|.\displaystyle\leq\left|\int_{-\infty}^{\infty}F(t)\mathrm{d}t-h\sum_{k=-\infty}^{\infty}F(kh)\right|+\left|h\sum_{k=-\infty}^{-M-1}F(kh)+h\sum_{k=N+1}^{\infty}F(kh)\right|. (19)

The first and second terms are called the discretization error and truncation error, respectively. The following function space is important for bounding the discretization error.

Definition 4.1

Let 𝒟d​(ϵ)\mathscr{D}_{d}(\epsilon) be a rectangular domain defined for 0<ϵ<10<\epsilon<1 by

𝒟d​(ϵ)={ζ∈ℂ:|Re⁡ζ|<1/ϵ,|Im⁡ζ|<d​(1−ϵ)}.\mathscr{D}_{d}(\epsilon)=\{\zeta\in\mathbb{C}:|\operatorname{Re}\zeta|<1/\epsilon,\,|\operatorname{Im}\zeta|<d(1-\epsilon)\}.

Then, 𝐇1​(𝒟d)\mathbf{H}^{1}(\mathscr{D}_{d}) denotes the family of all functions FF that are analytic in 𝒟d\mathscr{D}_{d} such that the norm 𝒩1​(F,d)\mathcal{N}_{1}(F,d) is finite, where

𝒩1​(F,d)=limϵ→0∮∂𝒟d​(ϵ)|F​(ζ)|​|d​ζ|.\mathcal{N}_{1}(F,d)=\lim_{\epsilon\to 0}\oint_{\partial\mathscr{D}_{d}(\epsilon)}|F(\zeta)||\mathrm{d}\zeta|.

For functions belonging to this function space, the discretization error is estimated as follows.

Theorem 4.1 (Stenger [8, Theorem 3.2.1])

Let F∈𝐇1​(𝒟d)F\in\mathbf{H}^{1}(\mathscr{D}_{d}). Then,

|∫−∞∞F​(x)​dx−h​∑k=−∞∞F​(k​h)|≤𝒩1​(F,d)1−e−2​π​d/h​e−2​π​d/h.\left|\int_{-\infty}^{\infty}F(x)\mathrm{d}x-h\sum_{k=-\infty}^{\infty}F(kh)\right|\leq\frac{\mathcal{N}_{1}(F,d)}{1-\operatorname{\mathrm{e}}^{-2\uppi d/h}}\operatorname{\mathrm{e}}^{-2\uppi d/h}.

In this paper, we show the following lemma, which completes estimation of the discretization error. The proof is given in Section 4.6.

Lemma 4.2

Let the assumptions made in Theorem 2.3 be fulfilled. Then, the function F​(ζ)=f​(ϕ​(ζ))​ϕ′​(ζ)F(\zeta)=f(\phi(\zeta))\phi^{\prime}(\zeta) belongs to 𝐇1​(𝒟d)\mathbf{H}^{1}(\mathscr{D}_{d}), and 𝒩1​(F,d)\mathcal{N}_{1}(F,d) is bounded as

𝒩1​(F,d)≤2​K​C3,\mathcal{N}_{1}(F,d)\leq 2KC_{3},

where C3C_{3} is a constant defined as (11).

In addition, we bound the truncation error as follows. The proof is given in Section 4.7.

Lemma 4.3

Let the assumptions made in Theorem 2.3 be fulfilled. Then, setting F​(ζ)=f​(ϕ​(ζ))​ϕ′​(ζ)F(\zeta)=f(\phi(\zeta))\phi^{\prime}(\zeta), we have

|h​∑k=−∞−M−1F​(k​h)+h​∑k=N+1∞F​(k​h)|≤K​C4​e−μ​n​h,\displaystyle\left|h\sum_{k=-\infty}^{-M-1}F(kh)+h\sum_{k=N+1}^{\infty}F(kh)\right|\leq KC_{4}\operatorname{\mathrm{e}}^{-\mu nh},

where C4C_{4} is a constant defined as (12).

Setting hh as (8), the above estimates (Theorem 4.1, Lemmas 4.2, and 4.3) yield the desired result as

|∫−∞∞f​(x)​dx−h​∑k=−MNf​(ϕ​(k​h))​ϕ′​(k​h)|\displaystyle\left|\int_{-\infty}^{\infty}f(x)\mathrm{d}x-h\sum_{k=-M}^{N}f(\phi(kh))\phi^{\prime}(kh)\right| ≤2​K​C31−e−2​π​d/h​e−2​π​d/h+K​C4​e−μ​n​h\displaystyle\leq\frac{2KC_{3}}{1-\operatorname{\mathrm{e}}^{-2\uppi d/h}}\operatorname{\mathrm{e}}^{-2\uppi d/h}+KC_{4}\operatorname{\mathrm{e}}^{-\mu nh}
=K​(2​C31−e−2​π​d​μ​n+C4)​e−2​π​d​μ​n\displaystyle=K\left(\frac{2C_{3}}{1-\operatorname{\mathrm{e}}^{-\sqrt{2\uppi d\mu n}}}+C_{4}\right)\operatorname{\mathrm{e}}^{-\sqrt{2\uppi d\mu n}}
≤K​(2​C31−e−2​π​d​μ+C4)​e−2​π​d​μ​n.\displaystyle\leq K\left(\frac{2C_{3}}{1-\operatorname{\mathrm{e}}^{-\sqrt{2\uppi d\mu}}}+C_{4}\right)\operatorname{\mathrm{e}}^{-\sqrt{2\uppi d\mu n}}.

This completes the proof of Theorem 2.3.

4.2 Useful inequalities on ℝ\mathbb{R}

We prepare two lemmas here.

Lemma 4.4 (Okayama et al. [5, Lemma 4.7])

We have

|log⁡(1+ex)1+log⁡(1+ex)⋅1+exex|\displaystyle\left|\frac{\log(1+\operatorname{\mathrm{e}}^{x})}{1+\log(1+\operatorname{\mathrm{e}}^{x})}\cdot\frac{1+\operatorname{\mathrm{e}}^{x}}{\operatorname{\mathrm{e}}^{x}}\right| ≤1(x∈ℝ).\displaystyle\leq 1\quad(x\in\mathbb{R}). (20)
Lemma 4.5

We have

arccos⁡(t2)≥2−t(0≤t≤2).\arccos\left(\frac{t}{2}\right)\geq\sqrt{2-t}\quad(0\leq t\leq 2). (21)
Proof 1

Integrating both sides of the obvious inequality

2−2​cos⁡w≥0(w≥0),2-2\cos w\geq 0\quad(w\geq 0),

we have

∫0v2​(1−cos⁡w)​dw=2​v−2​sin⁡v≥0(v≥0).\int_{0}^{v}2(1-\cos w)\mathrm{d}w=2v-2\sin v\geq 0\quad(v\geq 0).

In the same manner, integrating both sides of the above inequality, we have

∫0u2​(v−sin⁡v)​dv=u2+2​cos⁡u−2≥0(u≥0).\int_{0}^{u}2(v-\sin v)\mathrm{d}v=u^{2}+2\cos u-2\geq 0\quad(u\geq 0).

Here, putting u=arccos⁡(t/2)u=\arccos(t/2), we rewrite the inequality as

arccos2⁡(t2)≥2−t(0≤t≤2),\arccos^{2}\left(\frac{t}{2}\right)\geq 2-t\quad(0\leq t\leq 2),

which is equivalent to the desired inequality (21).

4.3 Useful inequalities on 𝒟d+\mathscr{D}_{d}^{+}

We prepare three lemmas here. Note that 𝒟¯\overline{\mathscr{D}} denotes the closure of 𝒟\mathscr{D}.

Lemma 4.6

It holds for all ζ∈𝒟π+¯\zeta\in\overline{\mathscr{D}_{\uppi}^{+}} that

|1log⁡(1+eζ)|≤1log⁡2.\left|\frac{1}{\log(1+\operatorname{\mathrm{e}}^{\zeta})}\right|\leq\frac{1}{\log 2}. (22)
Proof 2

Let ζ=x+i⁡y\zeta=x+\operatorname{\mathrm{i}}y where xx and yy are real numbers with x≥0x\geq 0 and |y|≤π|y|\leq\uppi. By the definition of log⁡z\log z, it holds that

|1log⁡(1+eζ)|2=|1log⁡|1+eζ|+i⁡arg⁡(1+eζ)|2=1{log⁡|1+ex+i⁡y|}2+{arg⁡(1+ex+i⁡y)}2.\left|\frac{1}{\log(1+\operatorname{\mathrm{e}}^{\zeta})}\right|^{2}=\left|\frac{1}{\log|1+\operatorname{\mathrm{e}}^{\zeta}|+\operatorname{\mathrm{i}}\arg(1+\operatorname{\mathrm{e}}^{\zeta})}\right|^{2}=\frac{1}{\left\{\log|1+\operatorname{\mathrm{e}}^{x+\operatorname{\mathrm{i}}y}|\right\}^{2}+\left\{\arg(1+\operatorname{\mathrm{e}}^{x+\operatorname{\mathrm{i}}y})\right\}^{2}}.

Since |1+ex+i⁡y||1+\operatorname{\mathrm{e}}^{x+\operatorname{\mathrm{i}}y}| and |arg⁡(1+ex+i⁡y)||\arg(1+\operatorname{\mathrm{e}}^{x+\operatorname{\mathrm{i}}y})| monotonically increase with respect to xx, we have

1{log⁡|1+ex+i⁡y|}2+{arg⁡(1+ex+i⁡y)}2≤1{log⁡|1+e0+i⁡y|}2+{arg⁡(1+e0+i⁡y)}2.\frac{1}{\left\{\log|1+\operatorname{\mathrm{e}}^{x+\operatorname{\mathrm{i}}y}|\right\}^{2}+\left\{\arg(1+\operatorname{\mathrm{e}}^{x+\operatorname{\mathrm{i}}y})\right\}^{2}}\leq\frac{1}{\left\{\log|1+\operatorname{\mathrm{e}}^{0+\operatorname{\mathrm{i}}y}|\right\}^{2}+\left\{\arg(1+\operatorname{\mathrm{e}}^{0+\operatorname{\mathrm{i}}y})\right\}^{2}}.

Furthermore, using

log⁡|1+ei⁡y|\displaystyle\log|1+\operatorname{\mathrm{e}}^{\operatorname{\mathrm{i}}y}| =log⁡(1+cos⁡y)2+sin2⁡y=log⁡4​cos2⁡(y2)=log⁡(2​cos⁡y2),\displaystyle=\log\sqrt{(1+\cos y)^{2}+\sin^{2}y}=\log\sqrt{4\cos^{2}\left(\frac{y}{2}\right)}=\log\left(2\cos\frac{y}{2}\right),
arg⁡(1+ei⁡y)\displaystyle\arg(1+\operatorname{\mathrm{e}}^{\operatorname{\mathrm{i}}y}) =arctan⁡(sin⁡y1+cos⁡y)=arctan⁡(tan⁡y2)=y2,\displaystyle=\arctan\left(\frac{\sin y}{1+\cos y}\right)=\arctan\left(\tan\frac{y}{2}\right)=\frac{y}{2},

and putting t=2​cos⁡(y/2)t=2\cos(y/2), we have

1{log⁡|1+ei⁡y|}2+{arg⁡(1+ei⁡y)}2=1{log⁡(2​cos⁡(y/2))}2+(y/2)2=1(log⁡t)2+arccos2⁡(t/2).\frac{1}{\left\{\log|1+\operatorname{\mathrm{e}}^{\operatorname{\mathrm{i}}y}|\right\}^{2}+\left\{\arg(1+\operatorname{\mathrm{e}}^{\operatorname{\mathrm{i}}y})\right\}^{2}}=\frac{1}{\left\{\log\left(2\cos(y/2)\right)\right\}^{2}+(y/2)^{2}}=\frac{1}{\left(\log t\right)^{2}+\arccos^{2}(t/2)}.

From 0≤t≤20\leq t\leq 2 and (21), we have

1(log⁡t)2+arccos2⁡(t/2)≤1(log⁡t)2+{2−t}2=q​(t),\frac{1}{\left(\log t\right)^{2}+\arccos^{2}(t/2)}\leq\frac{1}{(\log t)^{2}+\{\sqrt{2-t}\}^{2}}=q(t),

where

q​(t)=1(log⁡t)2+2−t.q(t)=\frac{1}{(\log t)^{2}+2-t}.

Since log⁡t≤t−1\log t\leq t-1, it holds that

q′​(t)=t−2​log⁡tt​{(log⁡t)2+2−t}2≥t−2​(t−1)t​{(log⁡t)2+2−t}2=2−tt​{(log⁡t)2+2−t}2≥0.q^{\prime}(t)=\frac{t-2\log t}{t\{(\log t)^{2}+2-t\}^{2}}\geq\frac{t-2(t-1)}{t\{(\log t)^{2}+2-t\}^{2}}=\frac{2-t}{t\{(\log t)^{2}+2-t\}^{2}}\geq 0.

Therefore, q​(t)q(t) monotonically increases, from which we have q​(t)≤q​(2)=1/(log⁡2)2q(t)\leq q(2)=1/(\log 2)^{2}. This completes the proof.

Lemma 4.7

It holds for all ζ∈𝒟π+¯\zeta\in\overline{\mathscr{D}_{\uppi}^{+}} that

|e1/log⁡(1+eζ)|≤e1/log⁡2.\left|\operatorname{\mathrm{e}}^{1/\log(1+\operatorname{\mathrm{e}}^{\zeta})}\right|\leq\operatorname{\mathrm{e}}^{1/\log 2}. (23)
Proof 3

Using (22), we have

|e1/log⁡(1+eζ)|≤e|1/log⁡(1+eζ)|≤e1/log⁡2.\left|\operatorname{\mathrm{e}}^{1/\log(1+\operatorname{\mathrm{e}}^{\zeta})}\right|\leq\operatorname{\mathrm{e}}^{\left|1/\log(1+\operatorname{\mathrm{e}}^{\zeta})\right|}\leq\operatorname{\mathrm{e}}^{1/\log 2}.

This completes the proof.

Lemma 4.8

It holds for all ζ∈𝒟π+¯\zeta\in\overline{\mathscr{D}_{\uppi}^{+}} that

|1+{log⁡(1+eζ)}2{log⁡(1+eζ)}2|≤1+1(log⁡2)2.\left|\frac{1+\{\log(1+\operatorname{\mathrm{e}}^{\zeta})\}^{2}}{\{\log(1+\operatorname{\mathrm{e}}^{\zeta})\}^{2}}\right|\leq 1+\frac{1}{(\log 2)^{2}}. (24)
Proof 4

Using (22), we have

|1+{log⁡(1+eζ)}2{log⁡(1+eζ)}2|=|1+1{log⁡(1+eζ)}2|≤1+|1{log⁡(1+eζ)}2|≤1+1(log⁡2)2.\left|\frac{1+\{\log(1+\operatorname{\mathrm{e}}^{\zeta})\}^{2}}{\{\log(1+\operatorname{\mathrm{e}}^{\zeta})\}^{2}}\right|=\left|1+\frac{1}{\{\log(1+\operatorname{\mathrm{e}}^{\zeta})\}^{2}}\right|\leq 1+\left|\frac{1}{\{\log(1+\operatorname{\mathrm{e}}^{\zeta})\}^{2}}\right|\leq 1+\frac{1}{(\log 2)^{2}}.

This completes the proof.

4.4 Useful inequality on 𝒟d−\mathscr{D}_{d}^{-}

We prepare the following lemma here.

Lemma 4.9

It holds for all ζ∈𝒟π−¯\zeta\in\overline{\mathscr{D}_{\uppi}^{-}} that

1|−1+log⁡(1+eζ)|≤11−log⁡2.\frac{1}{|-1+\log(1+\operatorname{\mathrm{e}}^{\zeta})|}\leq\frac{1}{1-\log 2}. (25)
Proof 5

By the definition of log⁡z\log z, it holds that

1|−1+log⁡(1+eζ)|=1|−1+log|​1+eζ⁡|+i⁡arg⁡(1+eζ)|≤1|−1+log|​1+eζ⁡|+0|.\frac{1}{|-1+\log(1+\operatorname{\mathrm{e}}^{\zeta})|}=\frac{1}{|-1+\log|1+\operatorname{\mathrm{e}}^{\zeta}|+\operatorname{\mathrm{i}}\arg(1+\operatorname{\mathrm{e}}^{\zeta})|}\leq\frac{1}{|-1+\log|1+\operatorname{\mathrm{e}}^{\zeta}|+0|}.

Let ζ=x+i⁡y\zeta=x+\operatorname{\mathrm{i}}y where xx and yy are real numbers with x<0x<0 and |y|≤π|y|\leq\uppi. Then, we have

|1+eζ|≤1+|eζ|=1+|ex+i⁡y|=1+ex<1+e0<e,\left|1+\operatorname{\mathrm{e}}^{\zeta}\right|\leq 1+\left|\operatorname{\mathrm{e}}^{\zeta}\right|=1+\left|\operatorname{\mathrm{e}}^{x+\operatorname{\mathrm{i}}y}\right|=1+\operatorname{\mathrm{e}}^{x}<1+\operatorname{\mathrm{e}}^{0}<\operatorname{\mathrm{e}},

from which we have log⁡|1+eζ|<1\log|1+\operatorname{\mathrm{e}}^{\zeta}|<1. Therefore, it holds that

1|−1+log⁡|1+eζ||=11−log⁡|1+eζ|,\frac{1}{|-1+\log|1+\operatorname{\mathrm{e}}^{\zeta}||}=\frac{1}{1-\log|1+\operatorname{\mathrm{e}}^{\zeta}|},

which is further bounded as

11−log⁡|1+eζ|≤11−log⁡(1+|eζ|)=11−log⁡(1+ex)≤11−log⁡(1+e0)=11−log⁡2.\displaystyle\frac{1}{1-\log|1+\operatorname{\mathrm{e}}^{\zeta}|}\leq\frac{1}{1-\log(1+|\operatorname{\mathrm{e}}^{\zeta}|)}=\frac{1}{1-\log(1+\operatorname{\mathrm{e}}^{x})}\leq\frac{1}{1-\log(1+\operatorname{\mathrm{e}}^{0})}=\frac{1}{1-\log 2}.

This completes the proof.

4.5 Useful inequalities on 𝒟d\mathscr{D}_{d}

We prepare four lemmas here.

Lemma 4.10 (Okayama et al. [5, Lemma 4.6])

It holds for all ζ∈𝒟π¯\zeta\in\overline{\mathscr{D}_{\uppi}} that

|log⁡(1+eζ)1+log⁡(1+eζ)⋅e−l+eζeζ|\displaystyle\left|\frac{\log(1+\operatorname{\mathrm{e}}^{\zeta})}{1+\log(1+\operatorname{\mathrm{e}}^{\zeta})}\cdot\frac{\operatorname{\mathrm{e}}^{-l}+\operatorname{\mathrm{e}}^{\zeta}}{\operatorname{\mathrm{e}}^{\zeta}}\right| ≤1,\displaystyle\leq 1, (26)

where l=log⁡(e/(e−1))l=\log(\operatorname{\mathrm{e}}/(\operatorname{\mathrm{e}}-1)).

Lemma 4.11 (Okayama et al. [4, Lemma 4.21])

For all x∈ℝx\in\mathbb{R} and y∈(−π,π)y\in(-\uppi,\uppi), putting ζ=x+i⁡y\zeta=x+\operatorname{\mathrm{i}}y, we have

1|1+eζ|\displaystyle\frac{1}{|1+\operatorname{\mathrm{e}}^{\zeta}|} ≤1(1+ex)​cos⁡(y/2),\displaystyle\leq\frac{1}{(1+\operatorname{\mathrm{e}}^{x})\cos(y/2)}, (27)
1|1+e−ζ|\displaystyle\frac{1}{|1+\operatorname{\mathrm{e}}^{-\zeta}|} ≤1(1+e−x)​cos⁡(y/2).\displaystyle\leq\frac{1}{(1+\operatorname{\mathrm{e}}^{-x})\cos(y/2)}. (28)
Lemma 4.12 (Three lines lemma, cf. [Stein-Shakarchi, p. 133])

Let gg be analytic and bounded in 𝒟d\mathscr{D}_{d} and continuous on 𝒟d¯\overline{\mathscr{D}_{d}}. Let Mg​(y)=supx∈ℝ|g​(x+i⁡y)|M_{g}(y)=\sup_{x\in\mathbb{R}}|g(x+\operatorname{\mathrm{i}}y)|. Then, we have

{Mg​(y)}2​d≤{Mg​(−d)}d−y​{Mg​(d)}y+d(−d≤y≤d).\{M_{g}(y)\}^{2d}\leq\{M_{g}(-d)\}^{d-y}\{M_{g}(d)\}^{y+d}\quad(-d\leq y\leq d).
Lemma 4.13

Let dd be a constant satisfying 0<d<π0<d<\uppi. For all ζ∈𝒟d\zeta\in\mathscr{D}_{d} and x∈ℝx\in\mathbb{R}, we have

|1+{log⁡(1+eζ)}2(1+e−ζ)2​{log⁡(1+eζ)}2|\displaystyle\left|\frac{1+\{\log(1+\operatorname{\mathrm{e}}^{\zeta})\}^{2}}{(1+\operatorname{\mathrm{e}}^{-\zeta})^{2}\{\log(1+\operatorname{\mathrm{e}}^{\zeta})\}^{2}}\right| ≤1+{log⁡(2+cd)}2{log⁡(2+cd)}2​(1+cd)2,\displaystyle\leq\frac{1+\{\log(2+c_{d})\}^{2}}{\{\log(2+c_{d})\}^{2}}(1+c_{d})^{2}, (29)
1+{log⁡(1+ex)}2(1+e−x)2​{log⁡(1+ex)}2\displaystyle\frac{1+\{\log(1+\operatorname{\mathrm{e}}^{x})\}^{2}}{(1+\operatorname{\mathrm{e}}^{-x})^{2}\{\log(1+\operatorname{\mathrm{e}}^{x})\}^{2}} ≤e1/π3,\displaystyle\leq\operatorname{\mathrm{e}}^{1/\uppi^{3}}, (30)

where cd=1/cos⁡(d/2)c_{d}=1/\cos(d/2).

Proof 6

First, consider (30), which is proved by showing

p​(t)=1+t2t2​(1−e−t)2≤e1/π3p(t)=\frac{1+t^{2}}{t^{2}}(1-\operatorname{\mathrm{e}}^{-t})^{2}\leq\operatorname{\mathrm{e}}^{1/\uppi^{3}}

for all t>0t>0 (put t=log⁡(1+ex)t=\log(1+\operatorname{\mathrm{e}}^{x})). The derivative of p​(t)p(t) is expressed as

p′​(t)=−2​(et−1)​(et−t3−t−1)t3​e2​t.p^{\prime}(t)=-\frac{2(\operatorname{\mathrm{e}}^{t}-1)(\operatorname{\mathrm{e}}^{t}-t^{3}-t-1)}{t^{3}\operatorname{\mathrm{e}}^{2t}}.

Let κ\kappa be a value that satisfies p′​(κ)=0p^{\prime}(\kappa)=0 and log⁡(108)<κ<log⁡(109)\log(108)<\kappa<\log(109), i.e., p​(t)p(t) has its maximum at t=κt=\kappa. Using eκ=κ3+κ+1\operatorname{\mathrm{e}}^{\kappa}=\kappa^{3}+\kappa+1, we have

p​(κ)=1+κ2κ2​((κ3+κ+1)−1eκ)2=(1+κ2)3e2​κ.p(\kappa)=\frac{1+\kappa^{2}}{\kappa^{2}}\left(\frac{(\kappa^{3}+\kappa+1)-1}{\operatorname{\mathrm{e}}^{\kappa}}\right)^{2}=\frac{(1+\kappa^{2})^{3}}{\operatorname{\mathrm{e}}^{2\kappa}}.

Since the function q​(x)=(1+x2)3/e2​xq(x)=(1+x^{2})^{3}/\operatorname{\mathrm{e}}^{2x} monotonically decreases for x≥(3+5)/2x\geq(3+\sqrt{5})/2, q​(log⁡(109))<q​(κ)<q​(log⁡(108))q(\log(109))<q(\kappa)<q(\log(108)) holds (note that log⁡(108)>(3+5)/2\log(108)>(3+\sqrt{5})/2). Thus, it holds that

1<q(log(109))<p(κ)=q(κ)<q(log(108))<e1/π3.1<q(\log(109))<p(\kappa)=q(\kappa)<q(\log(108))<\operatorname{\mathrm{e}}^{1/\uppi^{3}}.

Next, we show (29). Let

g(ζ)=1+{log(1+eζ)}2(1+e−ζ)2{log(1+eζ)}2.g(\zeta)=\frac{1+\{\log(1+\operatorname{\mathrm{e}}^{\zeta})\}^{2}}{(1+\operatorname{\mathrm{e}}^{-\zeta})^{2}\{\log(1+\operatorname{\mathrm{e}}^{\zeta})\}^{2}}.

Since the function g(ζ)g(\zeta) is analytic and bounded in 𝒟d\mathscr{D}_{d} and continuous on 𝒟d¯\overline{\mathscr{D}_{d}}, by Lemma 4.12, we obtain (29) if we show the following two inequalities:

Mg(d)≤1+{log(2+cd)}2{log(2+cd)}2(1+cd)2,Mg(−d)≤1+{log(2+cd)}2{log(2+cd)}2(1+cd)2,\displaystyle M_{g}(d)\leq\frac{1+\{\log(2+c_{d})\}^{2}}{\{\log(2+c_{d})\}^{2}}(1+c_{d})^{2},\quad M_{g}(-d)\leq\frac{1+\{\log(2+c_{d})\}^{2}}{\{\log(2+c_{d})\}^{2}}(1+c_{d})^{2},

where Mg(y)=supx∈ℝ|g(x+iy)|M_{g}(y)=\sup_{x\in\mathbb{R}}|g(x+\operatorname{\mathrm{i}}y)|. We show only the first one, because the second one is also shown in the same way. Putting ξ=log(1+ex+id)\xi=\log(1+\operatorname{\mathrm{e}}^{x+\operatorname{\mathrm{i}}d}), g(x+id)=p(ξ)g(x+\operatorname{\mathrm{i}}d)=p(\xi) holds, and thus, in what follows we prove

|p(ξ)|≤1+{log(2+cd)}2{log(2+cd)}2(1+cd)2.|p(\xi)|\leq\frac{1+\{\log(2+c_{d})\}^{2}}{\{\log(2+c_{d})\}^{2}}(1+c_{d})^{2}. (31)

We consider the following two cases: (a) |ξ|≤log(2+cd)|\xi|\leq\log(2+c_{d}) and (b) |ξ|>log(2+cd)|\xi|>\log(2+c_{d}). In case (a), we have

|p(ξ)|=|1+ξ2ξ2(−∑k=1∞(−ξ)kk!)2|≤(1+|ξ|2)(∑k=1∞|ξ|k−1k!)2=1+|ξ|2|ξ|2(∑n=1∞|ξ|kk!)2=1+|ξ|2|ξ|2(e|ξ|−1)2.\displaystyle|p(\xi)|=\left|\frac{1+\xi^{2}}{\xi^{2}}\left(-\sum_{k=1}^{\infty}\frac{(-\xi)^{k}}{k!}\right)^{2}\right|\leq(1+|\xi|^{2})\left(\sum_{k=1}^{\infty}\frac{|\xi|^{k-1}}{k!}\right)^{2}=\frac{1+|\xi|^{2}}{|\xi|^{2}}\left(\sum_{n=1}^{\infty}\frac{|\xi|^{k}}{k!}\right)^{2}=\frac{1+|\xi|^{2}}{|\xi|^{2}}\left(\operatorname{\mathrm{e}}^{|\xi|}-1\right)^{2}.

Here, if we put q(x)=(1+x2)(ex−1)2/x2q(x)=(1+x^{2})(\operatorname{\mathrm{e}}^{x}-1)^{2}/x^{2}, then we have q′(x)=2(ex−1)r(x)/x3q^{\prime}(x)=2(\operatorname{\mathrm{e}}^{x}-1)r(x)/x^{3}, where r(x)=1+ex(x3+x−1)r(x)=1+\operatorname{\mathrm{e}}^{x}(x^{3}+x-1). Since r′(x)=xex{(x+1)2+x}≥0r^{\prime}(x)=x\operatorname{\mathrm{e}}^{x}\{(x+1)^{2}+x\}\geq 0 for x≥0x\geq 0, r(x)r(x) monotonically increases for x≥0x\geq 0. Therefore, r(x)≥r(0)=0r(x)\geq r(0)=0 holds, from which we have q′(x)≥0q^{\prime}(x)\geq 0 for x≥0x\geq 0, i.e., q(x)q(x) monotonically increases for x≥0x\geq 0. Thus, from |ξ|≤log(2+cd)|\xi|\leq\log(2+c_{d}), we have (31) as

|p(ξ)|≤q(|ξ|)≤1+{log(2+cd)}2{log(2+cd)}2(elog(2+cd)−1)2=1+{log(2+cd)}2{log(2+cd)}2(1+cd)2.|p(\xi)|\leq q(|\xi|)\leq\frac{1+\{\log(2+c_{d})\}^{2}}{\{\log(2+c_{d})\}^{2}}(\operatorname{\mathrm{e}}^{\log(2+c_{d})}-1)^{2}=\frac{1+\{\log(2+c_{d})\}^{2}}{\{\log(2+c_{d})\}^{2}}(1+c_{d})^{2}.

In case (b), from (27), it holds that

Re(ξ)=Re(log(1+ex+id))=log|1+ex+id|≥log[(1+ex)cos(d/2)]≥log(cos(d/2)).\operatorname{Re}(\xi)=\operatorname{Re}(\log(1+\operatorname{\mathrm{e}}^{x+\operatorname{\mathrm{i}}d}))=\log|1+\operatorname{\mathrm{e}}^{x+\operatorname{\mathrm{i}}d}|\geq\log[(1+\operatorname{\mathrm{e}}^{x})\cos(d/2)]\geq\log(\cos(d/2)). (32)

Using this, we have

|p(ξ)|≤1+|ξ|2|ξ|2(1+|e−ξ|)2=1+|ξ|2|ξ|2(1+e−Re(ξ))2≤1+|ξ|2|ξ|2(1+e−log(cos(d/2)))2=1+|ξ|2|ξ|2(1+cd)2.|p(\xi)|\leq\frac{1+|\xi|^{2}}{|\xi|^{2}}(1+|\operatorname{\mathrm{e}}^{-\xi}|)^{2}=\frac{1+|\xi|^{2}}{|\xi|^{2}}(1+\operatorname{\mathrm{e}}^{-\operatorname{Re}(\xi)})^{2}\leq\frac{1+|\xi|^{2}}{|\xi|^{2}}(1+\operatorname{\mathrm{e}}^{-\log(\cos(d/2))})^{2}=\frac{1+|\xi|^{2}}{|\xi|^{2}}(1+c_{d})^{2}.

Furthermore, since (1+x2)/x2(1+x^{2})/x^{2} decreases monotonically for x>0x>0, we have (31). This completes the proof.

4.6 Estimation of the discretization error (proof of Lemma 4.2)

Lemma 4.2 is shown as follows.

Proof 7

Let F(ζ)=f(ϕ(ζ))ϕ′(ζ)F(\zeta)=f(\phi(\zeta))\phi^{\prime}(\zeta). Since ff is analytic in ϕ(𝒟d)\phi(\mathscr{D}_{d}), f(ϕ(⋅))f(\phi(\cdot)) is analytic in 𝒟d\mathscr{D}_{d}. In addition, since ϕ′\phi^{\prime} is analytic in 𝒟π\mathscr{D}_{\uppi}, FF is analytic in 𝒟d\mathscr{D}_{d} (note that d<πd<\uppi). Therefore, the remaining task is to show 𝒩1(F,d)≤2KC3\mathcal{N}_{1}(F,d)\leq 2KC_{3}. From (9), by using (23) and (24), it holds for all ζ∈𝒟d+\zeta\in\mathscr{D}_{d}^{+} that

|F(ζ)|\displaystyle|F(\zeta)| ≤K|e−ϕ(ζ)|β|ϕ′(ζ)|\displaystyle\leq K|\operatorname{\mathrm{e}}^{-\phi(\zeta)}|^{\beta}|\phi^{\prime}(\zeta)|
=K|e1/log(1+eζ)|β|11+eζ|β|1+{log(1+eζ)}2||1+e−ζ||log(1+eζ)|2\displaystyle=K\left|\operatorname{\mathrm{e}}^{1/\log(1+\operatorname{\mathrm{e}}^{\zeta})}\right|^{\beta}\left|\frac{1}{1+\operatorname{\mathrm{e}}^{\zeta}}\right|^{\beta}\frac{|1+\{\log(1+\operatorname{\mathrm{e}}^{\zeta})\}^{2}|}{|1+\operatorname{\mathrm{e}}^{-\zeta}||\log(1+\operatorname{\mathrm{e}}^{\zeta})|^{2}}
≤K(e1/log2)β1|1+eζ|β1|1+e−ζ|{1+1(log2)2}.\displaystyle\leq K\left(\operatorname{\mathrm{e}}^{1/\log 2}\right)^{\beta}\frac{1}{|1+\operatorname{\mathrm{e}}^{\zeta}|^{\beta}}\frac{1}{|1+\operatorname{\mathrm{e}}^{-\zeta}|}\left\{1+\frac{1}{(\log 2)^{2}}\right\}. (33)

Furthermore, from (10), by using (25), (26), and (29), it holds for all ζ∈𝒟d−\zeta\in\mathscr{D}_{d}^{-} that

|F(ζ)|\displaystyle|F(\zeta)| ≤K1|ϕ(ζ)|α+1|ϕ′(ζ)|\displaystyle\leq K\frac{1}{|\phi(\zeta)|^{\alpha+1}}\left|\phi^{\prime}(\zeta)\right|
=K|log(1+eζ)1+log(1+eζ)|α+1|1+e−ζ||−1+log(1+eζ)|α+1|1+{log(1+eζ)}2||1+e−ζ|2|log(1+eζ)|2\displaystyle=K\left|\frac{\log(1+\operatorname{\mathrm{e}}^{\zeta})}{1+\log(1+\operatorname{\mathrm{e}}^{\zeta})}\right|^{\alpha+1}\frac{|1+\operatorname{\mathrm{e}}^{-\zeta}|}{|-1+\log(1+\operatorname{\mathrm{e}}^{\zeta})|^{\alpha+1}}\frac{|1+\{\log(1+\operatorname{\mathrm{e}}^{\zeta})\}^{2}|}{|1+\operatorname{\mathrm{e}}^{-\zeta}|^{2}|\log(1+\operatorname{\mathrm{e}}^{\zeta})|^{2}}
≤K1|1+e−ζ−l|α+11+|e−ζ|(1−log2)α+11+{log(2+cd)}2{log(2+cd)}2(1+cd)2,\displaystyle\leq K\frac{1}{|1+\operatorname{\mathrm{e}}^{-\zeta-l}|^{\alpha+1}}\frac{1+|\operatorname{\mathrm{e}}^{-\zeta}|}{(1-\log 2)^{\alpha+1}}\frac{1+\{\log(2+c_{d})\}^{2}}{\{\log(2+c_{d})\}^{2}}(1+c_{d})^{2}, (34)

where cd=1/cos(d/2)c_{d}=1/\cos(d/2) and l=log(e/(e−1))l=\log(\operatorname{\mathrm{e}}/(\operatorname{\mathrm{e}}-1)). By definition, 𝒩1(F,d)\mathcal{N}_{1}(F,d) is expressed as

𝒩1(F,d)\displaystyle\mathcal{N}_{1}(F,d)
=limϵ→0{∫−1/ϵ1/ϵ|F(x−id)|dx+∫−d(1−ϵ)d(1−ϵ)|F(1/ϵ+iy)|dy+∫−1/ϵ1/ϵ|F(x+id)|dx+∫−d(1−ϵ)d(1−ϵ)|F(−1/ϵ+iy)|dy}.\displaystyle=\lim_{\epsilon\to 0}\left\{\int_{-1/\epsilon}^{1/\epsilon}|F(x-\operatorname{\mathrm{i}}d)|\mathrm{d}x+\int_{-d(1-\epsilon)}^{d(1-\epsilon)}|F(1/\epsilon+\operatorname{\mathrm{i}}y)|\mathrm{d}y+\int_{-1/\epsilon}^{1/\epsilon}|F(x+\operatorname{\mathrm{i}}d)|\mathrm{d}x+\int_{-d(1-\epsilon)}^{d(1-\epsilon)}|F(-1/\epsilon+\operatorname{\mathrm{i}}y)|\mathrm{d}y\right\}. (35)

Using (27), (28), and (33), we can bound the second term as

∫−d(1−ϵ)d(1−ϵ)|F(1/ϵ+iy)|dy\displaystyle\int_{-d(1-\epsilon)}^{d(1-\epsilon)}|F(1/\epsilon+\operatorname{\mathrm{i}}y)|\mathrm{d}y ≤K(e1/log2)β(1+1(log2)2)∫−d(1−ϵ)d(1−ϵ)1|1+e1/ϵ+iy|β|1+e−(1/ϵ+iy)|dy\displaystyle\leq K\left(\operatorname{\mathrm{e}}^{1/\log 2}\right)^{\beta}\left(1+\frac{1}{(\log 2)^{2}}\right)\int_{-d(1-\epsilon)}^{d(1-\epsilon)}\frac{1}{|1+\operatorname{\mathrm{e}}^{1/\epsilon+\operatorname{\mathrm{i}}y}|^{\beta}|1+\operatorname{\mathrm{e}}^{-(1/\epsilon+\operatorname{\mathrm{i}}y)}|}\mathrm{d}y
≤K(e1/log2)β(1+1(log2)2)(1+e1/ϵ)β(1+e−1/ϵ)∫−d(1−ϵ)d(1−ϵ)1cosβ(y/2)cos(y/2)dy,\displaystyle\leq\frac{K\left(\operatorname{\mathrm{e}}^{1/\log 2}\right)^{\beta}\left(1+\frac{1}{(\log 2)^{2}}\right)}{(1+\operatorname{\mathrm{e}}^{1/\epsilon})^{\beta}(1+\operatorname{\mathrm{e}}^{-1/\epsilon})}\int_{-d(1-\epsilon)}^{d(1-\epsilon)}\frac{1}{\cos^{\beta}(y/2)\cos(y/2)}\mathrm{d}y,

from which we have

limϵ→0∫−d(1−ϵ)d(1−ϵ)|F(1/ϵ+iy)|dy=0.\lim_{\epsilon\to 0}\int_{-d(1-\epsilon)}^{d(1-\epsilon)}|F(1/\epsilon+\operatorname{\mathrm{i}}y)|\mathrm{d}y=0.

In the same manner, with regard to the fourth term of (35), using (27), (28), and (34), we have

limϵ→0∫−d(1−ϵ)d(1−ϵ)|F(−1/ϵ+iy)|dy=0.\lim_{\epsilon\to 0}\int_{-d(1-\epsilon)}^{d(1-\epsilon)}|F(-1/\epsilon+\operatorname{\mathrm{i}}y)|\mathrm{d}y=0.

Therefore, 𝒩1(F,d)\mathcal{N}_{1}(F,d) is expressed as

𝒩1(F,d)\displaystyle\mathcal{N}_{1}(F,d) =∫−∞∞|F(x−id)|dx+∫−∞∞|F(x+id)|dx\displaystyle=\int_{-\infty}^{\infty}|F(x-\operatorname{\mathrm{i}}d)|\mathrm{d}x+\int_{-\infty}^{\infty}|F(x+\operatorname{\mathrm{i}}d)|\mathrm{d}x
=∫−∞0|F(x−id)|dx+∫0∞|F(x−id)|dx+∫−∞0|F(x+id)|dx+∫0∞|F(x+id)|dx.\displaystyle=\int_{-\infty}^{0}|F(x-\operatorname{\mathrm{i}}d)|\mathrm{d}x+\int_{0}^{\infty}|F(x-\operatorname{\mathrm{i}}d)|\mathrm{d}x+\int_{-\infty}^{0}|F(x+\operatorname{\mathrm{i}}d)|\mathrm{d}x+\int_{0}^{\infty}|F(x+\operatorname{\mathrm{i}}d)|\mathrm{d}x.

With regard to the first term, using (27), (28), and (34), we have

∫−∞0|F(x−id)|dx\displaystyle\int_{-\infty}^{0}|F(x-\operatorname{\mathrm{i}}d)|\mathrm{d}x ≤K(1−log2)α+11+{log(2+cd)}2{log(2+cd)}2(1+cd)2∫−∞01+|e−x+id||1+e−x−l+id|α+1dx\displaystyle\leq\frac{K}{(1-\log 2)^{\alpha+1}}\frac{1+\{\log(2+c_{d})\}^{2}}{\{\log(2+c_{d})\}^{2}}(1+c_{d})^{2}\int_{-\infty}^{0}\frac{1+|\operatorname{\mathrm{e}}^{-x+\operatorname{\mathrm{i}}d}|}{|1+\operatorname{\mathrm{e}}^{-x-l+\operatorname{\mathrm{i}}d}|^{\alpha+1}}\mathrm{d}x
≤K(1−log2)α+11+{log(2+cd)}2{log(2+cd)}2(1+cd)2∫−∞01+|e−x+id|(1+e−x−l)α+1cosα+1(d/2)dx\displaystyle\leq\frac{K}{(1-\log 2)^{\alpha+1}}\frac{1+\{\log(2+c_{d})\}^{2}}{\{\log(2+c_{d})\}^{2}}(1+c_{d})^{2}\int_{-\infty}^{0}\frac{1+|\operatorname{\mathrm{e}}^{-x+\operatorname{\mathrm{i}}d}|}{(1+\operatorname{\mathrm{e}}^{-x-l})^{\alpha+1}\cos^{\alpha+1}(d/2)}\mathrm{d}x
=Kcdα+1(1−log2)α+11+{log(2+cd)}2{log(2+cd)}2(1+cd)2∫−∞01+e−x(1+e−x−l)α+1dx.\displaystyle=\frac{Kc_{d}^{\alpha+1}}{(1-\log 2)^{\alpha+1}}\frac{1+\{\log(2+c_{d})\}^{2}}{\{\log(2+c_{d})\}^{2}}(1+c_{d})^{2}\int_{-\infty}^{0}\frac{1+\operatorname{\mathrm{e}}^{-x}}{(1+\operatorname{\mathrm{e}}^{-x-l})^{\alpha+1}}\mathrm{d}x.

The integral is further bounded as

∫−∞01+e−x(1+e−x−l)α+1dx\displaystyle\int_{-\infty}^{0}\frac{1+\operatorname{\mathrm{e}}^{-x}}{(1+\operatorname{\mathrm{e}}^{-x-l})^{\alpha+1}}\mathrm{d}x =∫−∞0(e(α+1)x(ex+e−l)α+1+eαx(ex+e−l)α+1)dx\displaystyle=\int_{-\infty}^{0}\left(\frac{\operatorname{\mathrm{e}}^{(\alpha+1)x}}{(\operatorname{\mathrm{e}}^{x}+\operatorname{\mathrm{e}}^{-l})^{\alpha+1}}+\frac{\operatorname{\mathrm{e}}^{\alpha x}}{(\operatorname{\mathrm{e}}^{x}+\operatorname{\mathrm{e}}^{-l})^{\alpha+1}}\right)\mathrm{d}x
≤∫−∞0(e(α+1)x(0+e−l)α+1+eαx(0+e−l)α+1)dx\displaystyle\leq\int_{-\infty}^{0}\left(\frac{\operatorname{\mathrm{e}}^{(\alpha+1)x}}{(0+\operatorname{\mathrm{e}}^{-l})^{\alpha+1}}+\frac{\operatorname{\mathrm{e}}^{\alpha x}}{(0+\operatorname{\mathrm{e}}^{-l})^{\alpha+1}}\right)\mathrm{d}x
=(ee−1)α+1(1α+1+1α).\displaystyle=\left(\frac{\operatorname{\mathrm{e}}}{\operatorname{\mathrm{e}}-1}\right)^{\alpha+1}\left(\frac{1}{\alpha+1}+\frac{1}{\alpha}\right).

In the same manner, the third term is bounded as

∫−∞0|F(x+id)|dx≤Kcdα+1(1−log2)α+11+{log(2+cd)}2{log(2+cd)}2(1+cd)2(ee−1)α+1(1α+1+1α).\int_{-\infty}^{0}|F(x+\operatorname{\mathrm{i}}d)|\mathrm{d}x\leq\frac{Kc_{d}^{\alpha+1}}{(1-\log 2)^{\alpha+1}}\frac{1+\{\log(2+c_{d})\}^{2}}{\{\log(2+c_{d})\}^{2}}(1+c_{d})^{2}\left(\frac{\operatorname{\mathrm{e}}}{\operatorname{\mathrm{e}}-1}\right)^{\alpha+1}\left(\frac{1}{\alpha+1}+\frac{1}{\alpha}\right).

With regard to the second term, using (27), (28), and (33), we have

∫0∞|F(x−id)|dx\displaystyle\int_{0}^{\infty}|F(x-\operatorname{\mathrm{i}}d)|\mathrm{d}x ≤Keβ/log2{1+1(log2)2}∫0∞1|1+ex−id|β|1+e−x+id|dx\displaystyle\leq K\operatorname{\mathrm{e}}^{\beta/\log 2}\left\{1+\frac{1}{(\log 2)^{2}}\right\}\int_{0}^{\infty}\frac{1}{|1+\operatorname{\mathrm{e}}^{x-\operatorname{\mathrm{i}}d}|^{\beta}|1+\operatorname{\mathrm{e}}^{-x+\operatorname{\mathrm{i}}d}|}\mathrm{d}x
≤Keβ/log2{1+1(log2)2}∫0∞1(1+ex)β(1+e−x)cosβ+1(d/2)dx\displaystyle\leq K\operatorname{\mathrm{e}}^{\beta/\log 2}\left\{1+\frac{1}{(\log 2)^{2}}\right\}\int_{0}^{\infty}\frac{1}{(1+\operatorname{\mathrm{e}}^{x})^{\beta}(1+\operatorname{\mathrm{e}}^{-x})\cos^{\beta+1}(d/2)}\mathrm{d}x
=Keβ/log2{1+1(log2)2}cdβ+1∫0∞e−βx(1+e−x)β+1dx\displaystyle=K\operatorname{\mathrm{e}}^{\beta/\log 2}\left\{1+\frac{1}{(\log 2)^{2}}\right\}c_{d}^{\beta+1}\int_{0}^{\infty}\frac{\operatorname{\mathrm{e}}^{-\beta x}}{(1+\operatorname{\mathrm{e}}^{-x})^{\beta+1}}\mathrm{d}x
≤Keβ/log2{1+1(log2)2}cdβ+1∫0∞e−βx(1+0)β+1dx\displaystyle\leq K\operatorname{\mathrm{e}}^{\beta/\log 2}\left\{1+\frac{1}{(\log 2)^{2}}\right\}c_{d}^{\beta+1}\int_{0}^{\infty}\frac{\operatorname{\mathrm{e}}^{-\beta x}}{(1+0)^{\beta+1}}\mathrm{d}x
=Keβ/log2{1+1(log2)2}cdβ+1β.\displaystyle=K\operatorname{\mathrm{e}}^{\beta/\log 2}\left\{1+\frac{1}{(\log 2)^{2}}\right\}\frac{c_{d}^{\beta+1}}{\beta}.

In the same manner, the fourth term is bounded as

∫0∞|F(x+id)|dx≤Keβ/log2{1+1(log2)2}cdβ+1β.\int_{0}^{\infty}|F(x+\operatorname{\mathrm{i}}d)|\mathrm{d}x\leq K\operatorname{\mathrm{e}}^{\beta/\log 2}\left\{1+\frac{1}{(\log 2)^{2}}\right\}\frac{c_{d}^{\beta+1}}{\beta}.

Thus, we have 𝒩1(F,d)≤2KC3\mathcal{N}_{1}(F,d)\leq 2KC_{3}.

4.7 Estimation of the truncation error (proof of Lemma 4.3)

Lemma 4.3 is shown as follows.

Proof 8

Let F(t)=f(ϕ(t))ϕ′(t)F(t)=f(\phi(t))\phi^{\prime}(t). From (9), by using (23) and (24), it holds for all t≥0t\geq 0 that

|F(t)|\displaystyle|F(t)| ≤K(e−ϕ(t))βϕ′(t)\displaystyle\leq K\left(\operatorname{\mathrm{e}}^{-\phi(t)}\right)^{\beta}\phi^{\prime}(t)
=K(e1/log(1+et))β(e−t1+e−t)β1+{log(1+et)}2(1+e−t){log(1+et)}2\displaystyle=K\left(\operatorname{\mathrm{e}}^{1/\log(1+\operatorname{\mathrm{e}}^{t})}\right)^{\beta}\left(\frac{\operatorname{\mathrm{e}}^{-t}}{1+\operatorname{\mathrm{e}}^{-t}}\right)^{\beta}\frac{1+\{\log(1+\operatorname{\mathrm{e}}^{t})\}^{2}}{(1+\operatorname{\mathrm{e}}^{-t})\{\log(1+\operatorname{\mathrm{e}}^{t})\}^{2}}
≤Keβ/log2e−βt(1+0)β+1{1+1(log2)2}.\displaystyle\leq K\operatorname{\mathrm{e}}^{\beta/\log 2}\frac{\operatorname{\mathrm{e}}^{-\beta t}}{(1+0)^{\beta+1}}\left\{1+\frac{1}{(\log 2)^{2}}\right\}.

Using this estimate, we have

|h∑k=N+1∞F(kh)|\displaystyle\left|h\sum_{k=N+1}^{\infty}F(kh)\right| ≤h∑k=N+1∞|F(kh)|\displaystyle\leq h\sum_{k=N+1}^{\infty}|F(kh)|
≤Keβ/log2{1+1(log2)2}h∑k=N+1∞e−βkh\displaystyle\leq K\operatorname{\mathrm{e}}^{\beta/\log 2}\left\{1+\frac{1}{(\log 2)^{2}}\right\}h\sum_{k=N+1}^{\infty}\operatorname{\mathrm{e}}^{-\beta kh}
≤Keβ/log2{1+1(log2)2}∫Nh∞e−βxdx\displaystyle\leq K\operatorname{\mathrm{e}}^{\beta/\log 2}\left\{1+\frac{1}{(\log 2)^{2}}\right\}\int_{Nh}^{\infty}\operatorname{\mathrm{e}}^{-\beta x}\mathrm{d}x
=Keβ/log2{1+1(log2)2}e−βNhβ.\displaystyle=K\operatorname{\mathrm{e}}^{\beta/\log 2}\left\{1+\frac{1}{(\log 2)^{2}}\right\}\frac{\operatorname{\mathrm{e}}^{-\beta Nh}}{\beta}.

Next, from (6), using (20), (25), and (30), it holds for all t≤0t\leq 0 that

|F(t)|\displaystyle|F(t)| ≤K1|ϕ(t)|α+1ϕ′(t)\displaystyle\leq K\frac{1}{|\phi(t)|^{\alpha+1}}\phi^{\prime}(t)
=K|log(1+et)1+log(1+et)|α+11+etet|−1+log(1+et)|α+11+{log(1+et)}2(1+e−t)2{log(1+et)}2\displaystyle=K\left|\frac{\log(1+\operatorname{\mathrm{e}}^{t})}{1+\log(1+\operatorname{\mathrm{e}}^{t})}\right|^{\alpha+1}\frac{1+\operatorname{\mathrm{e}}^{t}}{\operatorname{\mathrm{e}}^{t}|-1+\log(1+\operatorname{\mathrm{e}}^{t})|^{\alpha+1}}\frac{1+\{\log(1+\operatorname{\mathrm{e}}^{t})\}^{2}}{(1+\operatorname{\mathrm{e}}^{-t})^{2}\{\log(1+\operatorname{\mathrm{e}}^{t})\}^{2}}
≤K(et1+et)α+11+etet(1−log2)α+1e1/π3\displaystyle\leq K\left(\frac{\operatorname{\mathrm{e}}^{t}}{1+\operatorname{\mathrm{e}}^{t}}\right)^{\alpha+1}\frac{1+\operatorname{\mathrm{e}}^{t}}{\operatorname{\mathrm{e}}^{t}(1-\log 2)^{\alpha+1}}\operatorname{\mathrm{e}}^{1/\uppi^{3}}
≤Keαt(1+0)α1(1−log2)α+1e1/π3.\displaystyle\leq K\frac{\operatorname{\mathrm{e}}^{\alpha t}}{(1+0)^{\alpha}}\frac{1}{(1-\log 2)^{\alpha+1}}\operatorname{\mathrm{e}}^{1/\uppi^{3}}.

Using this estimate, we have

|h∑k=−∞−M−1F(kh)|\displaystyle\left|h\sum_{k=-\infty}^{-M-1}F(kh)\right| ≤h∑k=−∞−M−1|F(kh)|\displaystyle\leq h\sum_{k=-\infty}^{-M-1}|F(kh)|
≤Ke1/π3(1−log2)α+1h∑k=−∞−M−1eαkh\displaystyle\leq K\frac{\operatorname{\mathrm{e}}^{1/\uppi^{3}}}{(1-\log 2)^{\alpha+1}}h\sum_{k=-\infty}^{-M-1}\operatorname{\mathrm{e}}^{\alpha kh}
≤Ke1/π3(1−log2)α+1∫−∞−Mheαxdx\displaystyle\leq K\frac{\operatorname{\mathrm{e}}^{1/\uppi^{3}}}{(1-\log 2)^{\alpha+1}}\int_{-\infty}^{-Mh}\operatorname{\mathrm{e}}^{\alpha x}\mathrm{d}x
=Ke1/π3(1−log2)α+1e−αMhα.\displaystyle=K\frac{\operatorname{\mathrm{e}}^{1/\uppi^{3}}}{(1-\log 2)^{\alpha+1}}\frac{\operatorname{\mathrm{e}}^{-\alpha Mh}}{\alpha}.

Thus, using (7), we have

|h∑k=−∞−M−1F(kh)+h∑k=N+1∞F(kh)|\displaystyle\left|h\sum_{k=-\infty}^{-M-1}F(kh)+h\sum_{k=N+1}^{\infty}F(kh)\right| ≤Ke1/π3α(1−log2)α+1e−αMh+Keβ/log2β{1+1(log2)2}e−βNh\displaystyle\leq\frac{K\operatorname{\mathrm{e}}^{1/\uppi^{3}}}{\alpha(1-\log 2)^{\alpha+1}}\operatorname{\mathrm{e}}^{-\alpha Mh}+\frac{K\operatorname{\mathrm{e}}^{\beta/\log 2}}{\beta}\left\{1+\frac{1}{(\log 2)^{2}}\right\}\operatorname{\mathrm{e}}^{-\beta Nh}
≤Ke1/π3α(1−log2)α+1e−μnh+Keβ/log2β{1+1(log2)2}e−μnh,\displaystyle\leq\frac{K\operatorname{\mathrm{e}}^{1/\uppi^{3}}}{\alpha(1-\log 2)^{\alpha+1}}\operatorname{\mathrm{e}}^{-\mu nh}+\frac{K\operatorname{\mathrm{e}}^{\beta/\log 2}}{\beta}\left\{1+\frac{1}{(\log 2)^{2}}\right\}\operatorname{\mathrm{e}}^{-\mu nh},

which is the desired estimate.

5 Proofs for Theorem 2.4

This section presents the proof of Theorem 2.4. It is organized as follows. In Section 5.1, the task is decomposed into two lemmas: Lemmas 5.1 and 5.2. To prove these lemmas, a useful inequality is presented in Section 5.2. Then, Lemma 5.1 is proved in Section 5.3, and Lemma 5.2 is proved in Section 5.4.

5.1 Sketch of the proof

The main strategy in the proof of Theorem 2.4 is identical to that of Theorem 2.3, that is, splitting the error into the discretization error and the truncation error as (19). For the discretization error, we show the following lemma. The proof is given in Section 5.3.

Lemma 5.1

Let the assumptions made in Theorem 2.4 be fulfilled. Then, the function F(ζ)=f(ϕ(ζ))ϕ′(ζ)F(\zeta)=f(\phi(\zeta))\phi^{\prime}(\zeta) belongs to 𝐇1(𝒟d)\mathbf{H}^{1}(\mathscr{D}_{d}), and 𝒩1(F,d)\mathcal{N}_{1}(F,d) is bounded as

𝒩1(F,d)≤2KC5,\mathcal{N}_{1}(F,d)\leq 2KC_{5},

where C5C_{5} is a constant defined as (14).

In addition, we bound the truncation error as follows. The proof is given in Section 5.4.

Lemma 5.2

Let the assumptions made in Theorem 2.4 be fulfilled. Then, setting F(ζ)=f(ϕ(ζ))ϕ′(ζ)F(\zeta)=f(\phi(\zeta))\phi^{\prime}(\zeta), we have

|h∑k=−∞−M−1F(kh)+h∑k=N+1∞F(kh)|≤KC6e−μnh,\displaystyle\left|h\sum_{k=-\infty}^{-M-1}F(kh)+h\sum_{k=N+1}^{\infty}F(kh)\right|\leq KC_{6}\operatorname{\mathrm{e}}^{-\mu nh},

where C6C_{6} is a constant defined as (15).

Setting hh as (8), the above estimates (Theorem 4.1, Lemmas 5.1, and 5.2) yield the desired result as

|∫−∞∞f(x)dx−h∑k=−MNf(ϕ(kh))ϕ′(kh)|\displaystyle\left|\int_{-\infty}^{\infty}f(x)\mathrm{d}x-h\sum_{k=-M}^{N}f(\phi(kh))\phi^{\prime}(kh)\right| ≤2KC51−e−2πd/he−2πd/h+KC6e−μnh\displaystyle\leq\frac{2KC_{5}}{1-\operatorname{\mathrm{e}}^{-2\uppi d/h}}\operatorname{\mathrm{e}}^{-2\uppi d/h}+KC_{6}\operatorname{\mathrm{e}}^{-\mu nh}
=K(2C51−e−2πdμn+C6)e−2πdμn\displaystyle=K\left(\frac{2C_{5}}{1-\operatorname{\mathrm{e}}^{-\sqrt{2\uppi d\mu n}}}+C_{6}\right)\operatorname{\mathrm{e}}^{-\sqrt{2\uppi d\mu n}}
≤K(2C51−e−2πdμ+C6)e−2πdμn.\displaystyle\leq K\left(\frac{2C_{5}}{1-\operatorname{\mathrm{e}}^{-\sqrt{2\uppi d\mu}}}+C_{6}\right)\operatorname{\mathrm{e}}^{-\sqrt{2\uppi d\mu n}}.

This completes the proof of Theorem 2.4.

5.2 Useful inequality on 𝒟d\mathscr{D}_{d}

We prepare the following lemma here.

Lemma 5.3

Let dd be a constant satisfying 0<d<π0<d<\uppi. For all ζ∈𝒟d\zeta\in\mathscr{D}_{d} and x∈ℝx\in\mathbb{R}, we have

|1(1+e−ζ)log(1+eζ)|\displaystyle\left|\frac{1}{(1+\operatorname{\mathrm{e}}^{-\zeta})\log(1+\operatorname{\mathrm{e}}^{\zeta})}\right| ≤1+cdlog(2+cd),\displaystyle\leq\frac{1+c_{d}}{\log(2+c_{d})}, (36)
1(1+e−x)log(1+ex)\displaystyle\frac{1}{(1+\operatorname{\mathrm{e}}^{-x})\log(1+\operatorname{\mathrm{e}}^{x})} ≤1,\displaystyle\leq 1, (37)

where cd=1/cos(d/2)c_{d}=1/\cos(d/2).

Proof 9

First, consider (37), which is proved by showing

p(t)=1−e−tt≤1p(t)=\frac{1-\operatorname{\mathrm{e}}^{-t}}{t}\leq 1

for all t>0t>0 (put t=log(1+ex)t=\log(1+\operatorname{\mathrm{e}}^{x})). Differentiating p(x)p(x), we have

p′(t)=−et−(1+t)ett2≤0,p^{\prime}(t)=-\frac{\operatorname{\mathrm{e}}^{t}-(1+t)}{\operatorname{\mathrm{e}}^{t}t^{2}}\leq 0,

since et≥1+t\operatorname{\mathrm{e}}^{t}\geq 1+t holds. Therefore, p(t)p(t) decreases monotonically, and thus, it holds that p(t)≤limt→0p(t)=1p(t)\leq\lim_{t\to 0}p(t)=1.

Next, we show (36). Let g(ζ)=1/{(1+e−ζ)log(1+eζ)}g(\zeta)=1/\{(1+\operatorname{\mathrm{e}}^{-\zeta})\log(1+\operatorname{\mathrm{e}}^{\zeta})\}. Since the function g(ζ)g(\zeta) is analytic and bounded in 𝒟d\mathscr{D}_{d} and continuous on 𝒟d¯\overline{\mathscr{D}_{d}}, by Lemma 4.12, we obtain (36) if we show the following two inequalities:

Mg(d)≤1+cdlog(2+cd),Mg(−d)≤1+cdlog(2+cd),\displaystyle M_{g}(d)\leq\frac{1+c_{d}}{\log(2+c_{d})},\quad M_{g}(-d)\leq\frac{1+c_{d}}{\log(2+c_{d})},

where Mg(y)=supx∈ℝ|g(x+iy)|M_{g}(y)=\sup_{x\in\mathbb{R}}|g(x+\operatorname{\mathrm{i}}y)|. We show only the first one, because the second one is also shown in the same way. Putting ξ=log(1+ex+id)\xi=\log(1+\operatorname{\mathrm{e}}^{x+\operatorname{\mathrm{i}}d}), g(x+id)=p(ξ)g(x+\operatorname{\mathrm{i}}d)=p(\xi) holds, and thus, in what follows we prove

|p(ξ)|≤1+cdlog(2+cd).|p(\xi)|\leq\frac{1+c_{d}}{\log(2+c_{d})}. (38)

We consider the following two cases: (a) |ξ|≤log(2+cd)|\xi|\leq\log(2+c_{d}) and (b) |ξ|>log(2+cd)|\xi|>\log(2+c_{d}). In case (a), we have

|p(ξ)|=|∑k=1∞(−ξ)k−1k!|≤∑k=1∞|ξ|k−1k!=e|ξ|−1|ξ|.\displaystyle|p(\xi)|=\left|\sum_{k=1}^{\infty}\frac{(-\xi)^{k-1}}{k!}\right|\leq\sum_{k=1}^{\infty}\frac{|\xi|^{k-1}}{k!}=\frac{\operatorname{\mathrm{e}}^{|\xi|}-1}{|\xi|}.

Here, if we put q(x)=(ex−1)/xq(x)=(\operatorname{\mathrm{e}}^{x}-1)/x, then we have q′(x)=r(x)/x2q^{\prime}(x)=r(x)/x^{2}, where r(x)=1+(x−1)exr(x)=1+(x-1)\operatorname{\mathrm{e}}^{x}. Since r′(x)=xex≥0r^{\prime}(x)=x\operatorname{\mathrm{e}}^{x}\geq 0 for x≥0x\geq 0, r(x)r(x) monotonically increases for x≥0x\geq 0. Therefore, r(x)≥r(0)=0r(x)\geq r(0)=0 holds, from which we have q′(x)≥0q^{\prime}(x)\geq 0 for x≥0x\geq 0, i.e., q(x)q(x) monotonically increases for x≥0x\geq 0. Thus, from |ξ|≤log(2+cd)|\xi|\leq\log(2+c_{d}), we have (38) as

|p(ξ)|≤q(|ξ|)≤elog(2+cd)−1log(2+cd)=1+cdlog(2+cd).|p(\xi)|\leq q(|\xi|)\leq\frac{\operatorname{\mathrm{e}}^{\log(2+c_{d})}-1}{\log(2+c_{d})}=\frac{1+c_{d}}{\log(2+c_{d})}.

In case (b), using (32), we have

|p(ξ)|≤1+|e−ξ||ξ|=1+e−Reξ|ξ|≤1+e−log(cos(d/2))|ξ|=1+cd|ξ|.|p(\xi)|\leq\frac{1+|\operatorname{\mathrm{e}}^{-\xi}|}{|\xi|}=\frac{1+\operatorname{\mathrm{e}}^{-\operatorname{Re}\xi}}{|\xi|}\leq\frac{1+\operatorname{\mathrm{e}}^{-\log(\cos(d/2))}}{|\xi|}=\frac{1+c_{d}}{|\xi|}.

Furthermore, since 1/x1/x decreases monotonically for x>0x>0, we have (38). This completes the proof.

5.3 Estimation of the discretization error (proof of Lemma 5.1)

Lemma 5.1 is essentially shown by the following lemma, which holds for 0<δ<π0<\delta<\uppi (not only 0<d<(1+π)/20<d<(1+\uppi)/2).

Lemma 5.4

Assume that FF is analytic in 𝒟δ\mathscr{D}_{\delta} with 0<δ<π0<\delta<\uppi, and that there exist positive constants K+K_{+}, K−K_{-}, α\alpha, and β\beta such that

|F(ζ)|\displaystyle|F(\zeta)| ≤K+|e1/log(1+eζ)1+eζ|β\displaystyle\leq K_{+}\left|\frac{\operatorname{\mathrm{e}}^{1/\log(1+\operatorname{\mathrm{e}}^{\zeta})}}{1+\operatorname{\mathrm{e}}^{\zeta}}\right|^{\beta} (39)
holds for all ζ∈𝒟δ+\zeta\in\mathscr{D}_{\delta}^{+}, and
|F(ζ)|\displaystyle|F(\zeta)| ≤K−|log(1+eζ){1+log(1+eζ)}{−1+log(1+eζ)}|α\displaystyle\leq K_{-}\left|\frac{\log(1+\operatorname{\mathrm{e}}^{\zeta})}{\{1+\log(1+\operatorname{\mathrm{e}}^{\zeta})\}\{-1+\log(1+\operatorname{\mathrm{e}}^{\zeta})\}}\right|^{\alpha} (40)

holds for all ζ∈𝒟δ−\zeta\in\mathscr{D}_{\delta}^{-}. Then, FF belongs to 𝐇1(𝒟δ)\mathbf{H}^{1}(\mathscr{D}_{\delta}), and 𝒩1(F,δ)\mathcal{N}_{1}(F,\delta) is bounded as

𝒩1(F,δ)≤2K−α{ecδ(1−log2)(e−1)}α+2K+β(e1/log2cδ)β,\mathcal{N}_{1}(F,\delta)\leq\frac{2K_{-}}{\alpha}\left\{\frac{\operatorname{\mathrm{e}}c_{\delta}}{(1-\log 2)(\operatorname{\mathrm{e}}-1)}\right\}^{\alpha}+\frac{2K_{+}}{\beta}\left(\operatorname{\mathrm{e}}^{1/\log 2}c_{\delta}\right)^{\beta}, (41)

where cδ=1/cos(δ/2)c_{\delta}=1/\cos(\delta/2).

Proof 10

Since FF is analytic on 𝒟δ\mathscr{D}_{\delta}, the remaining task is to show (41). From (39), by using (23), it holds for all ζ∈𝒟δ+\zeta\in\mathscr{D}_{\delta}^{+} that

|F(ζ)|\displaystyle|F(\zeta)| ≤K+|e1/log(1+eζ)|β|11+eζ|β\displaystyle\leq K_{+}\left|\operatorname{\mathrm{e}}^{1/\log(1+\operatorname{\mathrm{e}}^{\zeta})}\right|^{\beta}\left|\frac{1}{1+\operatorname{\mathrm{e}}^{\zeta}}\right|^{\beta}
≤K+(e1/log2)β1|1+eζ|β.\displaystyle\leq K_{+}\left(\operatorname{\mathrm{e}}^{1/\log 2}\right)^{\beta}\frac{1}{|1+\operatorname{\mathrm{e}}^{\zeta}|^{\beta}}. (42)

Furthermore, from (40), by using (25) and (26), it holds for all ζ∈𝒟δ−\zeta\in\mathscr{D}_{\delta}^{-} that

|F(ζ)|\displaystyle|F(\zeta)| ≤K−|log(1+eζ)1+log(1+eζ)|α1|−1+log(1+eζ)|α\displaystyle\leq K_{-}\left|\frac{\log(1+\operatorname{\mathrm{e}}^{\zeta})}{1+\log(1+\operatorname{\mathrm{e}}^{\zeta})}\right|^{\alpha}\frac{1}{|-1+\log(1+\operatorname{\mathrm{e}}^{\zeta})|^{\alpha}}
≤K−1|1+e−ζ−l|α1(1−log2)α,\displaystyle\leq K_{-}\frac{1}{|1+\operatorname{\mathrm{e}}^{-\zeta-l}|^{\alpha}}\frac{1}{(1-\log 2)^{\alpha}}, (43)

where l=log(e/(e−1))l=\log(\operatorname{\mathrm{e}}/(\operatorname{\mathrm{e}}-1)). As described earlier, 𝒩1(F,d)\mathcal{N}_{1}(F,d) is expressed as (35). Using (27) and (42), we have

∫−δ(1−ϵ)δ(1−ϵ)|F(1/ϵ+iy)|dy\displaystyle\int_{-\delta(1-\epsilon)}^{\delta(1-\epsilon)}|F(1/\epsilon+\operatorname{\mathrm{i}}y)|\mathrm{d}y ≤K+(e1/log2)β∫−δ(1−ϵ)δ(1−ϵ)1|1+e1/ϵ+iy|βdy\displaystyle\leq K_{+}\left(\operatorname{\mathrm{e}}^{1/\log 2}\right)^{\beta}\int_{-\delta(1-\epsilon)}^{\delta(1-\epsilon)}\frac{1}{|1+\operatorname{\mathrm{e}}^{1/\epsilon+\operatorname{\mathrm{i}}y}|^{\beta}}\mathrm{d}y
≤K+(e1/log2)β(1+e1/ϵ)β∫−δ(1−ϵ)δ(1−ϵ)1cosβ(y/2)dy,\displaystyle\leq\frac{K_{+}\left(\operatorname{\mathrm{e}}^{1/\log 2}\right)^{\beta}}{(1+\operatorname{\mathrm{e}}^{1/\epsilon})^{\beta}}\int_{-\delta(1-\epsilon)}^{\delta(1-\epsilon)}\frac{1}{\cos^{\beta}(y/2)}\mathrm{d}y,

from which we have

limϵ→0∫−δ(1−ϵ)δ(1−ϵ)|F(1/ϵ+iy)|dy=0.\lim_{\epsilon\to 0}\int_{-\delta(1-\epsilon)}^{\delta(1-\epsilon)}|F(1/\epsilon+\operatorname{\mathrm{i}}y)|\mathrm{d}y=0.

In the same manner, using (28) and (43), we have

limϵ→0∫−δ(1−ϵ)δ(1−ϵ)|F(−1/ϵ+iy)|dy=0.\lim_{\epsilon\to 0}\int_{-\delta(1-\epsilon)}^{\delta(1-\epsilon)}|F(-1/\epsilon+\operatorname{\mathrm{i}}y)|\mathrm{d}y=0.

Therefore, 𝒩1(F,δ)\mathcal{N}_{1}(F,\delta) is expressed as

𝒩1(F,δ)\displaystyle\mathcal{N}_{1}(F,\delta) =∫−∞∞|F(x−iδ)|dx+∫−∞∞|F(x+iδ)|dx\displaystyle=\int_{-\infty}^{\infty}|F(x-\operatorname{\mathrm{i}}\delta)|\mathrm{d}x+\int_{-\infty}^{\infty}|F(x+\operatorname{\mathrm{i}}\delta)|\mathrm{d}x
=∫−∞0|F(x−iδ)|dx+∫0∞|F(x−iδ)|dx+∫−∞0|F(x+iδ)|dx+∫0∞|F(x+iδ)|dx.\displaystyle=\int_{-\infty}^{0}|F(x-\operatorname{\mathrm{i}}\delta)|\mathrm{d}x+\int_{0}^{\infty}|F(x-\operatorname{\mathrm{i}}\delta)|\mathrm{d}x+\int_{-\infty}^{0}|F(x+\operatorname{\mathrm{i}}\delta)|\mathrm{d}x+\int_{0}^{\infty}|F(x+\operatorname{\mathrm{i}}\delta)|\mathrm{d}x.

With regard to the first term, using (28) and (43), we have

∫−∞0|F(x−iδ)|dx\displaystyle\int_{-\infty}^{0}|F(x-\operatorname{\mathrm{i}}\delta)|\mathrm{d}x ≤K−(1−log2)α∫−∞01|1+e−x−l+iδ|αdx\displaystyle\leq\frac{K_{-}}{(1-\log 2)^{\alpha}}\int_{-\infty}^{0}\frac{1}{|1+\operatorname{\mathrm{e}}^{-x-l+\operatorname{\mathrm{i}}\delta}|^{\alpha}}\mathrm{d}x
≤K−(1−log2)α∫−∞01(1+e−x−l)αcosα(δ/2)dx\displaystyle\leq\frac{K_{-}}{(1-\log 2)^{\alpha}}\int_{-\infty}^{0}\frac{1}{(1+\operatorname{\mathrm{e}}^{-x-l})^{\alpha}\cos^{\alpha}(\delta/2)}\mathrm{d}x
=K−cδα(1−log2)α∫−∞01(1+e−x−l)αdx.\displaystyle=\frac{K_{-}c_{\delta}^{\alpha}}{(1-\log 2)^{\alpha}}\int_{-\infty}^{0}\frac{1}{(1+\operatorname{\mathrm{e}}^{-x-l})^{\alpha}}\mathrm{d}x.

The integral is further bounded as

∫−∞01(1+e−x−l)αdx\displaystyle\int_{-\infty}^{0}\frac{1}{(1+\operatorname{\mathrm{e}}^{-x-l})^{\alpha}}\mathrm{d}x =∫−∞0eαx(ex+e−l)αdx\displaystyle=\int_{-\infty}^{0}\frac{\operatorname{\mathrm{e}}^{\alpha x}}{(\operatorname{\mathrm{e}}^{x}+\operatorname{\mathrm{e}}^{-l})^{\alpha}}\mathrm{d}x
≤∫−∞0eαx(0+e−l)αdx\displaystyle\leq\int_{-\infty}^{0}\frac{\operatorname{\mathrm{e}}^{\alpha x}}{(0+\operatorname{\mathrm{e}}^{-l})^{\alpha}}\mathrm{d}x
=(ee−1)α1α.\displaystyle=\left(\frac{\operatorname{\mathrm{e}}}{\operatorname{\mathrm{e}}-1}\right)^{\alpha}\frac{1}{\alpha}.

In the same manner, the third term is bounded as

∫−∞0|F(x+iδ)|dx≤K−cδαα(1−log2)α(ee−1)α.\int_{-\infty}^{0}|F(x+\operatorname{\mathrm{i}}\delta)|\mathrm{d}x\leq\frac{K_{-}c_{\delta}^{\alpha}}{\alpha(1-\log 2)^{\alpha}}\left(\frac{\operatorname{\mathrm{e}}}{\operatorname{\mathrm{e}}-1}\right)^{\alpha}.

With regard to the second term, using (27) and (42), we have

∫0∞|F(x−iδ)|dx\displaystyle\int_{0}^{\infty}|F(x-\operatorname{\mathrm{i}}\delta)|\mathrm{d}x ≤K+eβ/log2∫0∞1|1+ex−iδ|βdx\displaystyle\leq K_{+}\operatorname{\mathrm{e}}^{\beta/\log 2}\int_{0}^{\infty}\frac{1}{|1+\operatorname{\mathrm{e}}^{x-\operatorname{\mathrm{i}}\delta}|^{\beta}}\mathrm{d}x
≤K+eβ/log2∫0∞1(1+ex)βcosβ(δ/2)dx\displaystyle\leq K_{+}\operatorname{\mathrm{e}}^{\beta/\log 2}\int_{0}^{\infty}\frac{1}{(1+\operatorname{\mathrm{e}}^{x})^{\beta}\cos^{\beta}(\delta/2)}\mathrm{d}x
=K+eβ/log2cδβ∫0∞e−βx(1+e−x)βdx\displaystyle=K_{+}\operatorname{\mathrm{e}}^{\beta/\log 2}c_{\delta}^{\beta}\int_{0}^{\infty}\frac{\operatorname{\mathrm{e}}^{-\beta x}}{(1+\operatorname{\mathrm{e}}^{-x})^{\beta}}\mathrm{d}x
≤K+eβ/log2cδβ∫0∞e−βx(1+0)βdx\displaystyle\leq K_{+}\operatorname{\mathrm{e}}^{\beta/\log 2}c_{\delta}^{\beta}\int_{0}^{\infty}\frac{\operatorname{\mathrm{e}}^{-\beta x}}{(1+0)^{\beta}}\mathrm{d}x
=K+eβ/log2cδββ.\displaystyle=K_{+}\operatorname{\mathrm{e}}^{\beta/\log 2}\frac{c_{\delta}^{\beta}}{\beta}.

In the same manner, the fourth term is bounded as

∫0∞|F(x+iδ)|dx≤K+eβ/log2cδββ.\int_{0}^{\infty}|F(x+\operatorname{\mathrm{i}}\delta)|\mathrm{d}x\leq K_{+}\operatorname{\mathrm{e}}^{\beta/\log 2}\frac{c_{\delta}^{\beta}}{\beta}.

Thus, we obtain (41).

Using this lemma, Lemma 5.1 is shown as follows.

Proof 11

Let F(ζ)=f(ϕ(ζ))ϕ′(ζ)F(\zeta)=f(\phi(\zeta))\phi^{\prime}(\zeta). Since ff is analytic in ϕ(𝒟d)\phi(\mathscr{D}_{d}), f(ϕ(⋅))f(\phi(\cdot)) is analytic in 𝒟d\mathscr{D}_{d}. In addition, since ϕ′\phi^{\prime} is analytic in 𝒟π\mathscr{D}_{\uppi}, FF is analytic in 𝒟d\mathscr{D}_{d} (note that d<πd<\uppi). Therefore, the remaining task is to show 𝒩1(F,d)≤2KC5\mathcal{N}_{1}(F,d)\leq 2KC_{5}. Using (28), we have

1|1+e−ζ|≤1(1+e−Reζ)cos((Imζ)/2)≤1(1+0)cos(d/2)\frac{1}{|1+\operatorname{\mathrm{e}}^{-\zeta}|}\leq\frac{1}{(1+\operatorname{\mathrm{e}}^{-\operatorname{Re}\zeta})\cos((\operatorname{Im}\zeta)/2)}\leq\frac{1}{(1+0)\cos(d/2)}

for all ζ∈𝒟d\zeta\in\mathscr{D}_{d}. Therefore, from (9), by using (24), it holds for all ζ∈𝒟d+\zeta\in\mathscr{D}_{d}^{+} that

|F(ζ)|\displaystyle|F(\zeta)| ≤K|e−ϕ(ζ)|β|ϕ′(ζ)|\displaystyle\leq K|\operatorname{\mathrm{e}}^{-\phi(\zeta)}|^{\beta}|\phi^{\prime}(\zeta)|
=K|e1/log(1+eζ)|β|11+eζ|β|1+{log(1+eζ)}2||1+e−ζ||log(1+eζ)|2\displaystyle=K\left|\operatorname{\mathrm{e}}^{1/\log(1+\operatorname{\mathrm{e}}^{\zeta})}\right|^{\beta}\left|\frac{1}{1+\operatorname{\mathrm{e}}^{\zeta}}\right|^{\beta}\frac{|1+\{\log(1+\operatorname{\mathrm{e}}^{\zeta})\}^{2}|}{|1+\operatorname{\mathrm{e}}^{-\zeta}||\log(1+\operatorname{\mathrm{e}}^{\zeta})|^{2}}
≤K|e1/log(1+eζ)1+eζ|βcd{1+1(log2)2},\displaystyle\leq K\left|\frac{\operatorname{\mathrm{e}}^{1/\log(1+\operatorname{\mathrm{e}}^{\zeta})}}{1+\operatorname{\mathrm{e}}^{\zeta}}\right|^{\beta}c_{d}\left\{1+\frac{1}{(\log 2)^{2}}\right\},

where cd=1/cos(d/2)c_{d}=1/\cos(d/2). Furthermore, from (13), by using (36), it holds for all ζ∈𝒟d−\zeta\in\mathscr{D}_{d}^{-} that

|F(ζ)|\displaystyle|F(\zeta)| ≤K1|4+{ϕ(ζ)}2|1/2|ϕ(ζ)|α|ϕ′(ζ)|\displaystyle\leq K\frac{1}{|4+\{\phi(\zeta)\}^{2}|^{1/2}|\phi(\zeta)|^{\alpha}}\left|\phi^{\prime}(\zeta)\right|
=K|log(1+eζ)1+log(1+eζ)|α1|−1+log(1+eζ)|α1|(1+e−ζ)log(1+eζ)|\displaystyle=K\left|\frac{\log(1+\operatorname{\mathrm{e}}^{\zeta})}{1+\log(1+\operatorname{\mathrm{e}}^{\zeta})}\right|^{\alpha}\frac{1}{|-1+\log(1+\operatorname{\mathrm{e}}^{\zeta})|^{\alpha}}\frac{1}{|(1+\operatorname{\mathrm{e}}^{-\zeta})\log(1+\operatorname{\mathrm{e}}^{\zeta})|}
≤K|log(1+eζ){1+log(1+eζ)}{−1+log(1+eζ)}|α1+cdlog(2+cd).\displaystyle\leq K\left|\frac{\log(1+\operatorname{\mathrm{e}}^{\zeta})}{\{1+\log(1+\operatorname{\mathrm{e}}^{\zeta})\}\{-1+\log(1+\operatorname{\mathrm{e}}^{\zeta})\}}\right|^{\alpha}\frac{1+c_{d}}{\log(2+c_{d})}.

Thus, the assumptions of Lemma 5.4 are fulfilled with δ=d\delta=d and

K+\displaystyle K_{+} =Kcd{1+1(log2)2},\displaystyle=Kc_{d}\left\{1+\frac{1}{(\log 2)^{2}}\right\},
K−\displaystyle K_{-} =K1+cdlog(2+cd),\displaystyle=K\frac{1+c_{d}}{\log(2+c_{d})},

from which we have 𝒩1(F,d)≤2KC5\mathcal{N}_{1}(F,d)\leq 2KC_{5}.

5.4 Estimation of the truncation error (proof of Lemma 5.2)

Lemma 5.2 is essentially shown by the following lemma.

Lemma 5.5

Assume that there exist positive constants K+K_{+}, K−K_{-}, α\alpha, and β\beta such that

|F(x)|\displaystyle|F(x)| ≤K+|e1/log(1+ex)1+ex|β\displaystyle\leq K_{+}\left|\frac{\operatorname{\mathrm{e}}^{1/\log(1+\operatorname{\mathrm{e}}^{x})}}{1+\operatorname{\mathrm{e}}^{x}}\right|^{\beta} (44)
holds for all x≥0x\geq 0, and
|F(x)|\displaystyle|F(x)| ≤K−|log(1+ex){1+log(1+ex)}{−1+log(1+ex)}|α\displaystyle\leq K_{-}\left|\frac{\log(1+\operatorname{\mathrm{e}}^{x})}{\{1+\log(1+\operatorname{\mathrm{e}}^{x})\}\{-1+\log(1+\operatorname{\mathrm{e}}^{x})\}}\right|^{\alpha} (45)

holds for all x<0x<0. Let μ=min{α,β}\mu=\min\{\alpha,\beta\}, and let MM and NN be defined as (7). Then, we have

h∑k=−∞−M−1|F(kh)|+h∑k=N+1∞|F(kh)|≤{K−α(1−log2)α+K+β(e1/log2)β}e−μnh.h\sum_{k=-\infty}^{-M-1}|F(kh)|+h\sum_{k=N+1}^{\infty}|F(kh)|\leq\left\{\frac{K_{-}}{\alpha(1-\log 2)^{\alpha}}+\frac{K_{+}}{\beta}\left(\operatorname{\mathrm{e}}^{1/\log 2}\right)^{\beta}\right\}\operatorname{\mathrm{e}}^{-\mu nh}. (46)
Proof 12

From (44), by using (23), it holds for all x≥0x\geq 0 that

|F(x)|≤K+(e1/log(1+ex))β(e−x1+e−x)β≤K+eβ/log2e−βx(1+0)β.\displaystyle|F(x)|\leq K_{+}\left(\operatorname{\mathrm{e}}^{1/\log(1+\operatorname{\mathrm{e}}^{x})}\right)^{\beta}\left(\frac{\operatorname{\mathrm{e}}^{-x}}{1+\operatorname{\mathrm{e}}^{-x}}\right)^{\beta}\leq K_{+}\operatorname{\mathrm{e}}^{\beta/\log 2}\frac{\operatorname{\mathrm{e}}^{-\beta x}}{(1+0)^{\beta}}.

Using this estimate, we have

h∑k=N+1∞|F(kh)|≤K+eβ/log2h∑k=N+1∞e−βkh≤K+eβ/log2∫Nh∞e−βxdx=K+eβ/log2e−βNhβ.\displaystyle h\sum_{k=N+1}^{\infty}|F(kh)|\leq K_{+}\operatorname{\mathrm{e}}^{\beta/\log 2}h\sum_{k=N+1}^{\infty}\operatorname{\mathrm{e}}^{-\beta kh}\leq K_{+}\operatorname{\mathrm{e}}^{\beta/\log 2}\int_{Nh}^{\infty}\operatorname{\mathrm{e}}^{-\beta x}\mathrm{d}x=K_{+}\operatorname{\mathrm{e}}^{\beta/\log 2}\frac{\operatorname{\mathrm{e}}^{-\beta Nh}}{\beta}.

Next, from (45), using (20) and (25), it holds for all x<0x<0 that

|F(x)|≤K−|log(1+ex)1+log(1+ex)|α1|−1+log(1+ex)|α≤K−(ex1+ex)α1(1−log2)α≤K−eαx(1+0)α1(1−log2)α.\displaystyle|F(x)|\leq K_{-}\left|\frac{\log(1+\operatorname{\mathrm{e}}^{x})}{1+\log(1+\operatorname{\mathrm{e}}^{x})}\right|^{\alpha}\frac{1}{|-1+\log(1+\operatorname{\mathrm{e}}^{x})|^{\alpha}}\leq K_{-}\left(\frac{\operatorname{\mathrm{e}}^{x}}{1+\operatorname{\mathrm{e}}^{x}}\right)^{\alpha}\frac{1}{(1-\log 2)^{\alpha}}\leq K_{-}\frac{\operatorname{\mathrm{e}}^{\alpha x}}{(1+0)^{\alpha}}\frac{1}{(1-\log 2)^{\alpha}}.

Using this estimate, we have

h∑k=−∞−M−1|F(kh)|≤K−(1−log2)αh∑k=−∞−M−1eαkh≤K−(1−log2)α∫−∞−Mheαxdx=K−(1−log2)αe−αMhα.\displaystyle h\sum_{k=-\infty}^{-M-1}|F(kh)|\leq\frac{K_{-}}{(1-\log 2)^{\alpha}}h\sum_{k=-\infty}^{-M-1}\operatorname{\mathrm{e}}^{\alpha kh}\leq\frac{K_{-}}{(1-\log 2)^{\alpha}}\int_{-\infty}^{-Mh}\operatorname{\mathrm{e}}^{\alpha x}\mathrm{d}x=\frac{K_{-}}{(1-\log 2)^{\alpha}}\frac{\operatorname{\mathrm{e}}^{-\alpha Mh}}{\alpha}.

Thus, using (7), we have (46).

Using this lemma, Lemma 5.2 is shown as follows.

Proof 13

Let F(x)=f(ϕ(x))ϕ′(x)F(x)=f(\phi(x))\phi^{\prime}(x). From (9), by using (24), it holds for all x≥0x\geq 0 that

|F(x)|\displaystyle|F(x)| ≤K|e−ϕ(x)|β|ϕ′(x)|\displaystyle\leq K|\operatorname{\mathrm{e}}^{-\phi(x)}|^{\beta}|\phi^{\prime}(x)|
=K|e1/log(1+ex)|β|11+ex|β11+e−x1+{log(1+ex)}2{log(1+ex)}2\displaystyle=K\left|\operatorname{\mathrm{e}}^{1/\log(1+\operatorname{\mathrm{e}}^{x})}\right|^{\beta}\left|\frac{1}{1+\operatorname{\mathrm{e}}^{x}}\right|^{\beta}\frac{1}{1+\operatorname{\mathrm{e}}^{-x}}\frac{1+\{\log(1+\operatorname{\mathrm{e}}^{x})\}^{2}}{\{\log(1+\operatorname{\mathrm{e}}^{x})\}^{2}}
≤K|e1/log(1+ex)1+ex|β11+0{1+1(log2)2}.\displaystyle\leq K\left|\frac{\operatorname{\mathrm{e}}^{1/\log(1+\operatorname{\mathrm{e}}^{x})}}{1+\operatorname{\mathrm{e}}^{x}}\right|^{\beta}\frac{1}{1+0}\left\{1+\frac{1}{(\log 2)^{2}}\right\}.

Next, from (13), using (37), it holds for all x<0x<0 that

|F(x)|\displaystyle|F(x)| ≤K1|1+{ϕ(x)}2|1/2|ϕ(x)|αϕ′(x)\displaystyle\leq K\frac{1}{|1+\{\phi(x)\}^{2}|^{1/2}|\phi(x)|^{\alpha}}\phi^{\prime}(x)
=K|log(1+ex)1+log(1+ex)|α1|−1+log(1+ex)|α⋅1(1+e−x)log(1+ex)\displaystyle=K\left|\frac{\log(1+\operatorname{\mathrm{e}}^{x})}{1+\log(1+\operatorname{\mathrm{e}}^{x})}\right|^{\alpha}\frac{1}{|-1+\log(1+\operatorname{\mathrm{e}}^{x})|^{\alpha}}\cdot\frac{1}{(1+\operatorname{\mathrm{e}}^{-x})\log(1+\operatorname{\mathrm{e}}^{x})}
≤K|log(1+ex){1+log(1+ex)}{−1+log(1+ex)}|α⋅1.\displaystyle\leq K\left|\frac{\log(1+\operatorname{\mathrm{e}}^{x})}{\{1+\log(1+\operatorname{\mathrm{e}}^{x})\}\{-1+\log(1+\operatorname{\mathrm{e}}^{x})\}}\right|^{\alpha}\cdot 1.

Thus, the assumptions of Lemma 5.5 are fulfilled with

K+\displaystyle K_{+} =K(1+1(log2)2),\displaystyle=K\left(1+\frac{1}{(\log 2)^{2}}\right),
K−\displaystyle K_{-} =K,\displaystyle=K,

which completes the proof.

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