Does the following integral converge, and if yes, to what?

Question

Can someone help me solve the following definite integral?

$$\mathcal I(a) = \int_{-\infty}^\infty \frac{\mathrm{e}^{-i |t| \sqrt{a^2+x^2}}} {\sqrt{a^2+x^2}}\,dx\,. $$

This can be solved, I believe, by analytic continuation to complex numbers and noting that there are two poles of the integrand at $x = \pm i a$. Integrating over a semi-circular contour going therough the lower half plane, the solution should be:

$$ \mathcal I(a) = -2\pi i \times Residue\left(\frac{\mathrm{e}^{-i |t| \sqrt{a^2+x^2}}} {\sqrt{a^2+x^2}}\right)_{x=-ia} \,. $$

Can somebody help me calculate the residue? Thanks in advance.

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

I'll assume $\ds{a, t \in \mathbb{R}}$.

\begin{align} \mc{I}\pars{a} &\equiv \int_{-\infty}^{\infty}{\expo{-\ic\verts{t}\root{a^{2} + x^{2}}} \over \root{a^{2} + x^{2}}}\,\dd x \,\,\,\stackrel{x\ =\ \verts{a}\sinh\pars{\theta}}{=}\,\,\, \int_{-\infty}^{\infty}\expo{-\ic\verts{t}\verts{a}\cosh\pars{\theta}}\dd\theta \\[5mm] & = \bbx{-\pi\ic\,\mrm{H}_{0}^{\mrm{\pars{2}}}\pars{\verts{at}}\,,\qquad a t \not= 0} \end{align}

$\ds{\mrm{H}_{\nu}^{\mrm{\pars{2}}}}$ is a Hankel Function. Note that $\ds{\,\mc{I}\pars{a} \sim -2\ln\pars{\verts{at}}}$ as $\ds{at \to 0}$

and $\ds{\,\mc{I}\pars{a} \sim -\ic\root{2\pi}\expo{-\ic\pi/4}{\expo{\ic\verts{at}} \over \verts{at}^{1/2}}}$ as $\ds{\verts{at} \to \infty}$.

The following picture is a drawing of $\ds{\verts{\pi\,\mrm{H}_{0}^{\mrm{\pars{2}}}\pars{x}}}$: