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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.

Nanashi No Gombe
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  • We do not know what is $t$. (And $a$.) Assuming $t$ is a real number, and eliminating $t=0$, we can then consider only the case $t>0$. Now the integral is not absolutely convergent, since $\text{arcsinh}(x/|a|)$ tends to infinity for $x\to\infty$. So which is the definition of the integral explicitly? Is it a limit...? – dan_fulea May 02 '18 at 11:09
  • @dan_fulea Since we are using the absolute value of $t$ (similarly, $a^2$), it does not matter whether $t$ is positive or negative. Anyway, for the sake of clarity, assume that we are trying to find the Cauchy principal value of that integral. My idea was to do this by means of a contour integral. Is the answer zero? – Nanashi No Gombe May 02 '18 at 20:21
  • Plugging $x=\sinh u$, it seems that the result is $\mathcal{I}(a)=2K_0(ia|t|)$ where $K_0$ is the modified Bessel function of the second kind. – Sangchul Lee May 02 '18 at 20:42
  • So $t=0$ is excluded, and $a,t\in \Bbb R_{>0}$. The substitution $y=ax$ reduces the computation, we need $J(t)= \int_{-\infty}^{\infty}\frac{\exp(-i t \sqrt{1+x^2})} {\sqrt{1+x^2}},dx =2\int_0^\infty\frac{\exp(-i t \sqrt{1+x^2})} {\sqrt{1+x^2}},dx =2\lim_{R\to\infty}\int_0^R\frac{\exp(-i t \sqrt{1+x^2})} {\sqrt{1+x^2}},dx $ (for an other $t$).

    The complex function under the integral does not behave well (is not meromorphic) in $\pm i$, so there is no residue there.

     – dan_fulea May 02 '18 at 21:53
  • Also, in order to connect the value of the integral with a residue, one needs a way to bound the fraction (e.g.) on the half-circle $[0,\pi]\to\Bbb C$, $s\to g(s)=Re^{is}$. But this is not the case. The expression $\exp(-it\sqrt{1+g(s)^2})$ is "approximatively" $\exp(-it,g(s))$ with modulus $\exp(-it,iR\sin s)$, which is exponential in $R$. – dan_fulea May 02 '18 at 21:57
  • Finally, the substitution $x=\sinh u$ gives $J(t)= 2\int_0^\infty\frac{\exp(-i t \sqrt{1+x^2})} {\sqrt{1+x^2}},dx = 2\int_0^\infty\frac{\exp(-i t \cosh u)} {\cosh u},\cosh u, du = 2\int_0^\infty\exp(-i t \cosh u), du $. – dan_fulea May 02 '18 at 22:02
  • $\pm, a\mathrm{i}$ are $\texttt{branch points}$. There $\color{red}{\texttt{are not}}$ poles. – Felix Marin May 02 '18 at 23:19

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

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 $\ds{\verts{\pi\,\mrm{H}_{0}^{\mrm{\pars{2}}}\pars{x}}}$:

enter image description here

Felix Marin
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