New conformal map for the trapezoidal formula for infinite integrals of unilateral rapidly decreasing functions☆

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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.

Section snippets

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(kh),where h is a mesh size. This approximation formula is fairly accurate if the integrand f(x) is analytic, which has been known since several decades ago [1], [2]. For example, the approximation ∫−∞∞e−x2dx≈h∑k=−∞∞e−(kh)2gives the correct answer in double-precision with h=1∕2, and the approximation ∫−∞∞14+x2dx≈h∑k=−∞∞14+(kh)2gives the correct answer in double-precision with h

Summary of existing and new results

Sections 2.1 Error analysis of Stenger’s formula, 2.2 Error bound for the formula by Okayama and Hanada describe the existing results, and Sections 2.3 General error bound for the proposed formula, 2.4 Special error bound for the proposed formula describe the new results. First, the relevant notations are introduced. Let Dd be a strip domain defined by Dd={ζ∈ℂ:|Imζ|<d} for d>0. Furthermore, let Dd−={ζ∈Dd:Reζ<0} and Dd+={ζ∈Dd:Reζ≥0}.

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)2exp−x2−1+x22dx=3−4eE1(1),∫−∞∞14+x2exp−x2−1+x22dx=Ci(1)sin1−si(1)cos1,∫−∞∞121+x4+x211+e(π∕2)xdx=1.136877446810281077257⋯, where E1(x) is the exponential integral defined by E1(x)=∫1∞(e−tx∕t)dt, Ci(x) is the cosine integral defined by Ci(x)=−∫x∞(cost∕t)dt, and si(x) is

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: Lemma 4.2, Lemma 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.

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: Lemma 5.1, Lemma 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.

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This work was partially supported by JSPS, Japan Grant-in-Aid for Young Scientists (B) JP17K14147.

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