Definite Integral with Branch Points

Question

I am trying to solve the integral

$$ \int_{k_x =-\infty}^\infty\int_{k_y=-\infty}^\infty \, \frac{\sin\left(\,k_x \, l\,\right)}{k_x\left[\,\left(\,k_x^2+k_y^2\,\right)^2-a^4\,\right]}\exp(i \, k_x \, x)\exp(i \, k_y \, y) \, \mathrm{d}k_x \, \mathrm{d}k_y$$

with $x>0$ and $y>0$ and $l>0$ and $a^4=\tilde{a}^4 \, \exp(-i\,\gamma)$ and $\tilde{a}>0$ and $0<\gamma<\pi/2$

for quite a while now. I always started with a integration with respect to $k_x$ which lead me to an integral (See here) I was not able to solve yet.

Now I tried a differnt approach starting with an integration with respect to $k_y$. Therefore I have to integrate

$\int_{k_y=-\infty}^\infty \, \frac{\exp(i \, k_y \, y)}{(k_x^2+k_y^2)^2-a^4} \mathrm{d}k_y$

I have four complexe first order poles

$k_{y1} = i \, \sqrt{k_x^2-a^2}$ with Im$(k_{y1})>0$

$k_{y2} = -i \, \sqrt{k_x^2-a^2}$ with Im$(k_{y2})<0$

$k_{y3} = i \, \sqrt{k_x^2+a^2}$ with Im$(k_{y3})>0$

$k_{y4} = -i \, \sqrt{k_x^2+a^2}$ with Im$(k_{y4})<0$

with $a^2=\tilde{a}^2 \, \exp(-i\,\gamma/2)$

using the residue theorem and Jordans lemma I end up with

$\int_{k_x=-\infty}^\infty \frac{\pi}{2\,a^2} \, \left(\frac{\exp\left(-\sqrt{k_x^2-a^2}\,y\right)}{\sqrt{k_x^2-a^2}}-\frac{\exp\left(-\sqrt{k_x^2+a^2}\,y\right)}{\sqrt{k_x^2+a^2}}\right)\,\frac{\sin(k_x \, l)}{k_x} \exp(i \, k_x \, x) \, \mathrm{d}k_x$

Using

$\sin(k_x \, l)=\frac{\exp(i \, k_x \, l)-\exp(-i \, k_x \, l)}{2\,i}$

I finally end up with

$\frac{\pi}{4\,i\,a^2}\,\int_{k_x=-\infty}^\infty \left(\frac{\exp\left(-\sqrt{k_x^2-a^2}\,y\right)}{k_x \, \sqrt{k_x^2-a^2}}-\frac{\exp\left(-\sqrt{k_x^2+a^2}\,y\right)}{k_x \, \sqrt{k_x^2+a^2}}\right)\,\left(\exp(i \, k_x \, (x+l))-\exp(i \, k_x \, (x-l))\right) \, \mathrm{d}k_x$

It seems to me that these are four integrals which can be evaluated using branch cuts and the residue theorem. Sadly I have hardly any experience with branch points and branch cuts.

Does anybody have an idea how to solve this integral or at least to convert it to an simpler integral using contour integration? Thank you.

Answer 1

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

Lets $\ds{\vec{r}_{\pm} \equiv \pars{x \pm \ell}\hat{x} + y\,\hat{y}}$ such that

\begin{align} &\int_{k_{x} = -\infty}^{\infty}\int_{k_{y} = -\infty}^{\infty} {\sin\pars{k_{x}x} \over k_{x}\bracks{\pars{k_{x}^{2} + k_{y}^{2}}^{2}- a^{4}}}\, \exp\pars{\ic k_{x}x}\exp\pars{\ic k_{y}y}\dd k_{x}\,\dd k_{y} \\[5mm] = &\ \iint_{\large\mathbb{R}^{2}}\pars{{1 \over 2}\,x\int_{-1}^{1}\exp\pars{\ic k_{x}xq}\,\dd q} {\exp\pars{\ic\vec{k}\cdot\vec{r}} \over k^{4} - a^{4}}\,\dd^{2}\vec{k} = {1 \over 2}\,x\int_{-1}^{1}\ \underbrace{\iint_{\large\mathbb{R}^{2}} {\exp\pars{\ic\vec{k}\cdot\vec{R}} \over k^{4} - a^{4}}\,\dd^{2}\vec{k}} _{\ds{\equiv\ \,\mc{I}\pars{\vec{R}}}}\,\ \dd q \\[5mm] &\ \mbox{where}\quad \vec{R} \equiv \pars{q + 1}x\,\hat{x} + y\,\hat{y}. \end{align}

---

\begin{align} \mc{I}\pars{\vec{R}} & \equiv \iint_{\large\mathbb{R}^{2}} {\exp\pars{\ic\vec{k}\cdot\vec{R}} \over k^{4} - a^{4}}\,\dd^{2}\vec{k} = \int_{0}^{\infty}{k \over k^{4} - a^{4}} \int_{0}^{2\pi}\exp\pars{\ic kR\cos\pars{\theta}}\,\dd\theta\,\dd k \end{align}

Note that $\ds{\int_{0}^{2\pi}\exp\pars{\ic kR\cos\pars{\theta}}\,\dd\theta = 2\int_{0}^{\pi}\cos\pars{kR\cos\pars{\theta}}\,\dd\theta = 2\pi\,\mrm{J}_{0}\pars{kR}}$ where $\ds{\,\mrm{J}_{\nu}}$ is a Bessel Function of the First Kind.

Then, \begin{align} \mc{I}\pars{\vec{R}} & = {\pi \over a^{2}}\int_{0}^{\infty}{\mrm{J}_{0}\pars{kR} \over k^{2} - a^{2}}\,k\,\dd k - {\pi \over a^{2}}\int_{0}^{\infty}{\mrm{J}_{0}\pars{kR} \over k^{2} + a^{2}}\,k\,\dd k \\[5mm] = &\ {\pi \over a^{2}}\,\mrm{K}_{0}\pars{-\ic \verts{a}R} - {\pi \over a^{2}}\,\mrm{K}_{0}\pars{\verts{a}R} \end{align} $\ds{\mrm{K}_{\nu}}$ is a Modified Bessel Function.

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\begin{align} &\int_{k_{x} = -\infty}^{\infty}\int_{k_{y} = -\infty}^{\infty} {\sin\pars{k_{x}x} \over k_{x}\bracks{\pars{k_{x}^{2} + k_{y}^{2}}^{2}- a^{4}}}\, \exp\pars{\ic k_{x}x}\exp\pars{\ic k_{y}y}\dd k_{x}\,\dd k_{y} \\[5mm] = &\ {\pi x \over 2a^{2}} \int_{0}^{2}\mrm{K}_{0}\pars{-\ic\verts{a}\root{q^{2}x^{2} + y^{2}}}\,\dd q - {\pi x \over 2a^{2}} \int_{0}^{2}\mrm{K}_{0}\pars{\verts{a}\root{q^{2}x^{2} + y^{2}}}\,\dd q \\[1cm] = &\ -\,{\pi\,\mrm{sgn}\pars{x} \over 2\verts{a}^{3}}\,\ic \int_{-\ic\,\verts{ay}}^{-\ic\,\verts{a}\root{4x^{2} + y^{2}}}{q\,\mrm{K}_{0}\pars{q} \over \root{q^{2} + \pars{ay}^{2}}}\,\dd q \\[5mm] &\ -{\pi\,\mrm{sgn}\pars{x} \over 2\verts{a}^{3}} \int_{\verts{ay}}^{\verts{a}\root{4x^{2} + y^{2}}}{q\,\mrm{K}_{0}\pars{q} \over \root{q^{2} - \pars{ay}^{2}}}\,\dd q \end{align}

So far, it's our result.