Extensions of Clenshaw-Curtis-type rules to integrals over a semi-infinite interval

Abstract

The Clenshaw-Curtis (C-C) rule is a quadrature formula for integrals on an interval [− 1,1] and efficient for smooth integrands f(x). Analogous rules exist: Fejér’s first and second kind, Basu and corrected C-C rules. We attempt to extend these five rules to integrals over a semi-infinite interval $[0,\infty )$ to develop corresponding formulae. Developing contour integration representations of the errors of the formulae, we prove that for f(z) analytic in a region containing $[0,\infty )$ in the complex plane z, the errors are of O(h^je^−c/h) ($j=1,2,\dots ,5$), respectively, with a constant c > 0 as step size h → + 0. The extension of Fejér’s second rule (the case j = 1) agrees with a formula based on the Sinc interpolation. Numerical experiments show that new formulae inherit nice features of the C-C rule and its four analogs. For large h, the convergence rates are twice as fast as the asymptotic rates for small h. The kink phenomenon that is explained in the C-C rule and Fejér’s first rule appears in the convergence curves.

Appendix 1: Sinc integral

We show that the LFII formula (1.2) is derived from the Sinc interpolation and give the asymptotic convergence rate.

We start with a change of variables x = t² in the integral Qf,

$$ Qf={\int}_0^{\infty} f(x) dx={\int}_0^{\infty} g(t) dt, $$

(A.1)

where g(t) = 2tf(t²). We then interpolate g(t) in terms of the Sinc function $\text {Sinc}(z)=\sin \limits (\pi z)/(\pi z)$ (cf. [11]) as follows

$$ g(t)\approx\sum\limits_{k=-\infty}^{\infty} g(kh) \text{Sinc}((t-kh)/h). $$

(A.2)

Using the right-hand side of (A.2) in (A.1), we have the Sinc formula $I_{h}^{(\text {Sinc})}f$,

$$ Qf\approx I_{h}^{(\text{Sinc})}f=\sum\limits_{k=-\infty}^{\infty} g(kh) {\int}_0^{\infty}\text{Sinc}\left( \frac{t-kh}{h}\right) dt. $$

(A.3)

Since by a change of variables u = (t − kh)/h, we have

$$ {\int}_0^{\infty}\text{Sinc}\left( \frac{t-kh}{h}\right) dt =h{\int}_{-k}^{\infty}\frac{\sin \pi u}{\pi u} du =\frac{h}{\pi}{\int}_{-\pi k}^{\infty}\frac{\sin u}{u} du, $$

in view of (A.3), it follows that

$$ I^{(\text{Sinc})}_{h} f=\sum\limits_{k=-\infty}^{\infty} g(kh) \frac{h}{\pi}[\text{Si}(\infty) -\text{Si}(-\pi k)]. $$

(A.4)

In view of (A.4) and the relations g(−x) = −g(x) and Si(−x) = −Si(x), recalling that g(t) = 2tf(t²), we obtain

$$ I_{h}^{(\text{Sinc})}f=\frac{2h}{\pi}\sum\limits_{k=1}^{\infty} g(kh) \text{Si}(\pi k)= \frac{4h}{\pi}\sum\limits_{k=1}^{\infty} kh \text{Si}(\pi k) f(k^{2}h^{2}). $$

Then, for $I_{h}^{(\text {LFII})}f$ in (1.2), we see that

$$ I_{h}^{(\text{Sinc})}f=I_{h}^{(\text{LFII})}f. $$

Proof of the asymptotic convergence rate for the Sinc (LFII) formula

The mapping z = w² is a conformal map of the infinite strip domain of width 2d (d > 0), given by $D_{d}=\{w\in \mathbb {C} | |\Im w|<d\}$ onto the domain P_d in (1.7). Since f(z) is a function analytic in P_d, the function g(w) = 2wf(w²) in (A.1) is analytic in D_d. Then, Lemma 3.6.4 in [10, p.171] gives the asymptotic convergence rate O(he^−πd/h) (h → 0).