Computing Fresnel integrals via modified trapezium rules

Abstract

In this paper we propose methods for computing Fresnel integrals based on truncated trapezium rule approximations to integrals on the real line, these trapezium rules modified to take into account poles of the integrand near the real axis. Our starting point is a method for computation of the error function of complex argument due to Matta and Reichel (J Math Phys 34:298–307, 1956) and Hunter and Regan (Math Comp 26:539–541, 1972). We construct approximations which we prove are exponentially convergent as a function of $N$, the number of quadrature points, obtaining explicit error bounds which show that accuracies of $10^{-15}$ uniformly on the real line are achieved with $N=12$, this confirmed by computations. The approximations we obtain are attractive, additionally, in that they maintain small relative errors for small and large argument, are analytic on the real axis (echoing the analyticity of the Fresnel integrals), and are straightforward to implement.

Appendix: bounds on \(\mathrm {erfc}\)

In this appendix we prove Theorem 4 as a corollary of bounds on $\mathrm {erfc}$ in the right hand complex plane contained in Theorem 6 below. In particular (51) follows immediately from (4) and the first bound in (73), while (52) follows from (4), (20), and the second of the bounds (73). The bounds in Theorem 6 are well-known in the case $z\ge 0$ [1, (7.8.2–7.8.3)], and the second bound (equivalent by (4) to the bound $|w(z)| \le 1$ for $\mathrm {Im}(z) \ge 0$) is recently proved by an alternative argument on p. 413 of [3].

Theorem 6

For $z=x+{\mathrm {i}}y$ with $x\ge 0$, $y\in \mathbb {R}$, we have that

$$\begin{aligned} |\mathrm {erfc}(z)|\ge \frac{{\mathrm {e}}^{y^2-x^2}}{\sqrt{(1+\sqrt{\pi }\,x)^2 + \pi y^2}}\ge \frac{{\mathrm {e}}^{y^2-x^2}}{1+\sqrt{\pi }\,|z|} \; \text{ and } \; |\mathrm {erfc}(z)| \le {\mathrm {e}}^{y^2-x^2}. \end{aligned}$$

(73)

Proof

The first of the bounds (73) is equivalent to the bound

$$\begin{aligned} |\mathcal {G}(z)| \ge 1, \quad \text{ for } \; \mathrm {Re}(z) \ge 0, \end{aligned}$$

(74)

where $\mathcal {G}(z) = (1+\sqrt{\pi }\, z) {\mathrm {e}}^{z^2}\mathrm {erfc}(z)$ is an entire function which has the properties that $\mathcal {G}(0)=1$ and $\mathcal {G}(z)\rightarrow 1$ as $|z|\rightarrow \infty$ in the right hand plane, uniformly in $\arg (z)$ [2, (7.1.23)]. (These properties imply that the first of the bounds (73) is sharp for $z=0$ and in the limit $|z|\rightarrow \infty$.) We will show (74) by showing that (74) holds for all $z$ in the right hand plane if it holds on the imaginary axis, and then showing that (74) holds on the imaginary axis.

To see that it is enough to prove that (74) holds for imaginary $z$, observe that, since $\mathrm {erfc}(z)$ has no zeros in the right hand complex plane [12, 28] [or on the imaginary axis where ${\mathrm {Re}}(\mathrm {erfc}(z)) = 1$, see (76)], the function $\mathcal H(z) := 1/\mathcal {G}(z)$ is also analytic in the right hand complex plane and is continuous up to the imaginary axis. Moreover, $\mathcal {H}(z)$ is bounded in the right hand plane since, as observed above, $\mathcal {G}(z)\rightarrow 1$ as $|z|\rightarrow \infty$ in the right hand plane (uniformly in $\arg (z)$). Since $\mathcal {H}(z)$ is bounded in the right hand plane, it follows from the maximum principle that

$$\begin{aligned} \sup _{\mathrm {Re}(z) \ge 0} |\mathcal {H}(z)| = \sup _{\mathrm {Re}(z) = 0} |\mathcal {H}(z)|. \end{aligned}$$

(75)

To see this, note that this equality holds for $\mathcal {H}_\alpha (z) := 1/\mathcal {G}_\alpha (z)$, with $\alpha >1$, where $\mathcal {G}_\alpha (z) := (1+\sqrt{\pi }\, z)^\alpha {\mathrm {e}}^{z^2}\mathrm {erfc}(z)$ with the branch cut taken as the negative real axis. This is clear since $\mathcal {H}_\alpha (z)$ is analytic in the right half-plane, continuous up to the imaginary axis, and vanishes at infinity, so that the standard maximum principle implies that $\mathcal {H}_\alpha (z)$ takes its maximum value on the imaginary axis. But then (75) follows by taking the limit $\alpha \rightarrow 1^+$.

In view of (75), to establish (74) we need only show that it holds for $z={\mathrm {i}}y$ with $y\in \mathbb {R}$; indeed, establishing this bound for $y\ge 0$ is sufficient since $\mathrm {erfc}(-{\mathrm {i}}y)=\overline{\mathrm {erfc}({\mathrm {i}}y)}$. Now, for $z={\mathrm {i}}y$ with $y\ge 0$, using [1, (7.5.1)], which implies

$$\begin{aligned} {\mathrm {e}}^{z^2}\mathrm {erfc}(z)= {\mathrm {e}}^{-y^2}\left( 1 -\frac{2{\mathrm {i}}}{\sqrt{\pi }}\int _0^y {\mathrm {e}}^{t^2} \mathrm {d}t\right) \end{aligned}$$

(76)

we see that

$$\begin{aligned} |\mathcal {G}({\mathrm {i}}y)|^2&= (1+\pi y^2) {\mathrm {e}}^{-2y^2}\left( 1 + \frac{4}{\pi }\left( \int _0^y {\mathrm {e}}^{t^2} \mathrm {d}t\right) ^2\right) \\&\ge (1+\pi y^2) {\mathrm {e}}^{-2y^2}\left( 1 +\frac{4}{\pi }y^2\right) \nonumber \\&= \left( 1+ \left( \pi + \frac{4}{\pi }\right) y^2 + 4 y^4\right) {\mathrm {e}}^{-2y^2}.\nonumber \end{aligned}$$

(77)

It is an easy calculus exercise to show the right hand side takes its minimum value on $[0,1]$ at either 0 or 1, and hence that $|\mathcal {G}({\mathrm {i}}y)| \ge 1$, for $0\le y \le 1$, since $|\mathcal {G}({\mathrm {i}})|^2 > (5+\pi )/{\mathrm {e}}^2>8/2.8^2>1$. Further, (77) implies that

$$\begin{aligned} |\mathcal {G}({\mathrm {i}}y)| \ge 2 y {\mathrm {e}}^{-y^2}\int _0^y {\mathrm {e}}^{t^2} \mathrm {d}t \end{aligned}$$

and, for $y\ge 1$, it follows on integrating by parts that

$$\begin{aligned} \int _0^y {\mathrm {e}}^{t^2} \mathrm {d}t = \int _0^1 {\mathrm {e}}^{t^2} \mathrm {d}t + \int _1^y e^{t^2} \mathrm {d}t&= \int _0^1{\mathrm {e}}^{t^2}\mathrm {d}t + \frac{{\mathrm {e}}^{y^2}}{2y} - \frac{{\mathrm {e}}}{2} + \int _1^y \frac{{\mathrm {e}}^{t^2}}{2t^2}\mathrm {d}t\\&> \int _0^1(1+t^2 + \tfrac{1}{2}t^4)\mathrm {d}t + \frac{ {\mathrm {e}}^{y^2}}{2y} - \frac{{\mathrm {e}}}{2} > \frac{{\mathrm {e}}^{y^2}}{2y}, \end{aligned}$$

since ${\mathrm {e}}< 2.8 < 2(1+1/3+1/10)$. Thus $|\mathcal {G}({\mathrm {i}}y)|\ge 1$ on $[1,\infty )$ and the bound (74) is proved.

Similarly,

$$\begin{aligned} \sup _{\mathrm {Re}(z) \ge 0} |{\mathrm {e}}^{-z^2}\mathrm {erfc}(z)| = \sup _{\mathrm {Re}(z) = 0} |{\mathrm {e}}^{-z^2}\mathrm {erfc}(z)| = \sup _{y\ge 0} |{\mathrm {e}}^{-y^2}\mathrm {erfc}({\mathrm {i}}y)|. \end{aligned}$$

(78)

Further, (76) implies that, for $y\ge 0$,

$$\begin{aligned} |\mathrm {erfc}({\mathrm {i}}y)|^2-1&= \frac{4}{\pi }\left( \int _0^y {\mathrm {e}}^{t^2} \mathrm {d}t\right) ^2 = \frac{4y^2}{\pi }\left( \sum _{n=0}^\infty \frac{y^{2n}}{n!(2n+1)}\right) ^2 \\&= \frac{2y^2}{\pi } \sum _{n=0}^\infty a_n y^{2n} \le \frac{2}{\pi }\left( {\mathrm {e}}^{2y^2}-1\right) \end{aligned}$$

where

$$\begin{aligned} a_n \!=\! \sum _{m=0}^n \frac{2}{m!(n-m)!(2m+1)(2(n-m)+1)} \!\le \! \frac{2}{n+1}\sum _{m=0}^n\frac{1}{m!(n-m)!} \!=\! \frac{2^{n+1}}{(n+1)!}. \end{aligned}$$

Thus, for $y\ge 0$,

$$\begin{aligned} |{\mathrm {e}}^{-y^2}\mathrm {erfc}({\mathrm {i}}y)|^2 \le \frac{2}{\pi } + \left( 1-\frac{2}{\pi }\right) {\mathrm {e}}^{-2y^2} \le 1. \end{aligned}$$

Combining this with (78) we see that the second of the bounds (73) holds. $\square$