How to evaluate the integral $\int_0^{\infty}\mathrm{d}x\frac{\sin(x)\sin(ax)}{\pi^2-x^2}e^{-ibx^2}$?

Question

Note that $a$ and $b$ are positive constants. Can this integral be evaluated in closed form ?

$$\int_0^{\infty}\mathrm{d}x\frac{\sin(x)\sin(ax)}{\pi^2-x^2}e^{-ibx^2}$$

Answer 1

After some toil I managed to evaluate this integral analytically. I am posting the detailed derivation below for the benefit of other users.

I. Background

We wish to evaluate $$I=\int_0^{\infty}\!\!\!\mathrm{d}x\frac{\sin(x)\sin(ax)}{\pi^2-x^2}e^{-ibx^2}=\frac{1}{2}\int_{-\infty}^{\infty}\!\!\!\mathrm{d}x\frac{\sin(x)\sin(ax)}{\pi^2-x^2}e^{-ibx^2}.\qquad(1)$$ Writing $$\sin(x)\sin(ax)=\frac{\left(e^{ix}-e^{-ix}\right)}{2i}\frac{\left(e^{iax}-e^{-iax}\right)}{2i}=\frac{1}{4}\left(e^{i(a-1)x}+e^{-i(a-1)x}-e^{i(a+1)x}-e^{-i(a+1)x}\right)$$ we have $$I=\frac{1}{8}\left\{\mathscr{D}\left(\frac{a-1}{2}\right)+\mathscr{D}\left(\frac{1-a}{2}\right)-\mathscr{D}\left(\frac{a+1}{2}\right)-\mathscr{D}\left(\frac{-a-1}{2}\right)\right\},\qquad(2)$$ where $$\mathscr{D}(\alpha):=\int_{-\infty}^{\infty}\!\!\!\mathrm{d}x~\frac{e^{-ibx^2+2i\alpha x}}{\pi^2-x^2}.\qquad(3)$$

II. prerequisites

In order to obtain $\mathscr{D}(\alpha)$ analytically, we allude to a special function $$w(z)=\frac{i}{\pi}\int_{-\infty}^{\infty}\!\!\!\mathrm{d}x\frac{e^{-x^2}}{z-x},\qquad\mathrm{Im}[z]>0\qquad(4)$$ called the Faddeeva function (after Soviet mathematician Vera Faddeeva), or the complex complementary error function, due to the identity $$w(z)=e^{-z^2}\mathrm{erfc}(-iz)\qquad z\in\mathbb{C}.\qquad(5)$$ Since the integral representation $(4)$ is only valid in the upper half of the complex plane, for $\mathrm{Im}[z]<0$, letitng $x=-y$ we have $$\frac{i}{\pi}\int_{-\infty}^{\infty}\!\!\!\mathrm{d}x\frac{e^{-x^2}}{z-x}=\frac{i}{\pi}\int_{-\infty}^{\infty}\!\!\!\mathrm{d}y\frac{e^{-y^2}}{z+y}=-\frac{i}{\pi}\int_{-\infty}^{\infty}\!\!\!\mathrm{d}y\frac{e^{-y^2}}{(-z)-y}=-w(-z),\qquad(6)$$ using equation $(4)$. Combining equations $(4)$ and $(6)$, we have $$\int_{-\infty}^{\infty}\!\!\!\mathrm{d}x\frac{e^{-x^2}}{z-x}=\frac{\pi}{i}\,\mathrm{sgn}\left(\mathrm{Im}[z]\right)w\left(\mathrm{sgn}\left(\mathrm{Im}[z]\right)z\right),\qquad(7)$$ which holds for any complex $z$ with nonzero imaginary part. We list the identity $$w(z)+w(-z)=2e^{-z^2}\qquad(8)$$ for later use.

III. Complex offset

We need one last ingredient for computing $\mathscr{D}(\alpha)$. Consider the integral $$I(\omega,z)=\int_{-\infty}^{\infty}\!\!\!\mathrm{d}x\frac{e^{-(x-\omega)^2}}{z-x},\qquad(9)$$ where $\omega$ is a complex offset. In order to evaluate this, we will apply the residue theorem. First, let

$\color{blue}{\mathrm{Im}[z]>0:}$

Next, choose the positively (negatively) oriented rectangular contour shown in the figure below, according as $\omega$ lies in the upper (lower) half of the complex plane.

Applying the residue theorem, we obtain $$\oint\mathrm{d}x\frac{e^{-(x-\omega)^2}}{z-x}=2\pi i\,\mathrm{Res}[z]\,\theta(\mathrm{Im}[\omega])\,\theta\left(\mathrm{Im}[\omega-z]\right),\qquad(10)$$ where $$\mathrm{Res}[z]=\lim_{x\to z}\,\,(x-z)\frac{e^{-(x-\omega)^2}}{z-x}=-e^{-(z-\omega)^2}\qquad(11)$$ is the residue of the simple pole at $z$. If $\omega$ lies in the lower half plane, the contour integral evaluates to zero, since $z$ lies outside the contour. This is ensured by the $\theta(\mathrm{Im}[\omega])$ term. Similarly, whem $\omega$ lies in the upper half plane, the $\theta\left(\mathrm{Im}[\omega-z]\right)$ evaluates to one (zero) according as $z$ lies inside (outside) the contour.

Now, the contour integral can be resolved into integrals over the edges of the rectangle. The edge $DA$ gives $$I_{DA}=\int_{-R}^R\!\!\mathrm{d}x\,\frac{e^{-(x-\omega)^2}}{z-x},\qquad(12)$$ which tends to $I(\omega,z)$ as $R\to\infty$. In this limit, the contributions from the side edges $I_{AB}$ and $I_{CD}$ go to zero, since the respective integrands go like $e^{-R^2}$.

Parameterizing the edge $BC$ by $\omega+y$, we have $$I_{BC}=\int_{R-\mathrm{Re}[\omega]}^{-R-\mathrm{Re}[\omega]}\mathrm{d}y\,\frac{e^{-y^2}}{z-(y+\omega)}\to-\int_{-\infty}^{\infty}\!\!\!\mathrm{d}y\,\frac{e^{-y^2}}{(z-\omega)-y}\qquad(13)$$ as $R\to\infty$. The resulting integral can be evaluated using equation $(7)$, $$I_{BC}=-\frac{\pi}{i}\mathrm{sgn}\left(\mathrm{Im}[z-\omega]\right)\,w\left(\mathrm{sgn}\left(\mathrm{Im}[z-\omega]\right)(z-\omega)\right).\qquad(14)$$ Finally, plugging equations $(11)$, $(12)$ and $(14)$ into $(10)$, we arrive at the desired result $$I(\omega,z)=\frac{\pi}{i}\left\{2\,\theta(\mathrm{Im}[\omega])\,\theta\left(\mathrm{Im}[\omega-z]\right)\,e^{-(z-\omega)^2}\right.\\+\mathrm{sgn}\left(\mathrm{Im}[z-\omega]\right)\,w\left(\mathrm{sgn}\left(\mathrm{Im}[z-\omega]\right)(z-\omega)\right)\Big\}\qquad(15)$$ Note that, for $\omega=0$ we recover the result from equation $(4)$, viz. $I(0,z)=w(z)$.

Equation $(15)$ holds for any complex $\omega$, with $z$ in the upper half plane. Next, consider the case

$\color{blue}{\mathrm{Im}[z]<0}:$

We can use the same trick as equation $(6)$ to handle this case. Substituting $x=-y$, we have $$\int_{-\infty}^{\infty}\!\!\!\mathrm{d}x\frac{e^{-(x-\omega)^2}}{z-x}=\int_{-\infty}^{\infty}\!\!\!\mathrm{d}y\frac{e^{-(y+\omega)^2}}{z+y}=-\int_{-\infty}^{\infty}\!\!\!\mathrm{d}x\frac{e^{-(y-(-\omega))^2}}{(-z)-y}=-I(-\omega,-z)\\=-\frac{\pi}{i}\left\{2\,\theta(-\mathrm{Im}[\omega])\,\theta\left(\mathrm{Im}[z-\omega]\right)\,e^{-(z-\omega)^2}\right.\\+\mathrm{sgn}\left(\mathrm{Im}[\omega-z]\right)\,w\left(\mathrm{sgn}\left(\mathrm{Im}[z-\omega]\right)(z-\omega)\right)\Big\}\qquad(16)$$ Finally, equations $(15)$ and $(16)$ can be combined into a single equation, viz. $$\int_{-\infty}^{\infty}\!\!\!\mathrm{d}x\frac{e^{-(x-\omega)^2}}{z-x}=\frac{2\pi}{i}\color{blue}{\mathrm{sgn}\left(\mathrm{Im}[z]\right)}\,\theta(\color{blue}{\mathrm{sgn}\left(\mathrm{Im}[z]\right)}\mathrm{Im}[\omega])\,\theta\left(\color{blue}{\mathrm{sgn}\left(\mathrm{Im}[z]\right)}\mathrm{Im}[\omega-z]\right)\,e^{-(z-\omega)^2}\\+\frac{\pi}{i}\mathrm{sgn}\left(\mathrm{Im}[z-\omega]\right)\,w\left(\mathrm{sgn}\left(\mathrm{Im}[z-\omega]\right)(z-\omega)\right),\qquad(17)$$ which holds for any complex offset $\omega$ and complex $z$ with nonzero imaginary part.

IV. Evaluation of $\mathscr{D}(\alpha)$

Since $b>0$, the 'complex Gaussian' $e^{-ibx^2}$ goes to zero, as $|x|\to\infty$ in the second and fourth quadrants of the complex $x$-plane. To use this property to our advantage, consider the oriented contour shown below. Applying the residue theorem, we have $$\oint\mathrm{d}x~\frac{e^{-ibx^2+2i\alpha x}}{\pi^2-x^2}=-2\pi i\,\mathrm{Res}[\pi]+2\pi i\,\mathrm{Res}[-\pi],\qquad(18)$$ where the real poles at $x=\pm\pi$ yield the following residues: $$\mathrm{Res}[\pm\pi]=\lim_{x\to\pm\pi}~(x\mp\pi)\frac{e^{-ibx^2+2i\alpha x}}{\pi^2-x^2}=\mp\frac{e^{-ib\pi^2\pm2i\alpha\pi}}{2\pi}.\qquad(19)$$ Thus, equation $(18)$ simplifies to $$\oint\mathrm{d}x~\frac{e^{-ibx^2+2i\alpha x}}{\pi^2-x^2}=2i\,e^{-ib\pi^2}\cos(2\pi\alpha)\qquad(20)$$ The 'arcs at infinity' (shown by dashed lines in the figure) make no contribution to the contour integral. The small semicircle around $x=\pi$, which may be parametrized as $x=\pi+\varepsilon\,e^{i\phi}$, $0<\phi<\pi$, gives $$i\varepsilon\int_0^{\pi}\!\!\!\mathrm{d}\phi\frac{e^{-ib(\pi+\varepsilon\,e^{i\phi})^2+2i\alpha(\pi+\varepsilon\,e^{i\phi})}}{(2\pi+\varepsilon\,e^{i\phi})(-\varepsilon\,e^{i\phi})}\to\frac{e^{-ib\pi^2+2i\alpha\pi}}{2i}$$ as $\varepsilon\to0$. Similarly, the other semicircular piece around $x=-\pi$ gives $\mathrm{exp}(-ib\pi^2\color{red}{-}2i\alpha\pi)/2i$. The line segments coincident on the real line yield $\mathscr{D}(\alpha)$. One is thus, left with the $\frac{\pi}{4}$-line, viz. $BC$.

At this stage, equation $(20)$ can be rewritten as $$\mathscr{D}(\alpha)+\int_{BC}\!\!\!\mathrm{d}x\,\frac{e^{-ibx^2+2i\alpha x}}{\pi^2-x^2}+\frac{e^{-ib\pi^2}}{2i}\left(e^{2i\pi\alpha}+e^{-2i\pi\alpha}\right)=2i\cos(2\pi\alpha),$$ which further yields $$\mathscr{D}(\alpha)=3i\cos(2\pi\alpha)e^{-i\pi^2b}\color{red}{+}\int_{\color{red}{CB}}\!\!\!\mathrm{d}x\,\frac{e^{-ibx^2+2i\alpha x}}{\pi^2-x^2}\qquad(21)$$

Now we compute the integral on the line $CB$, parameterizing it as $x=u\,e^{-i\frac{\pi}{4}}$, we have $$\int_{CB}\!\!\!\mathrm{d}x\,\frac{e^{-ibx^2+2i\alpha x}}{\pi^2-x^2}=\int_{-\infty}^{\infty}\!\!\!\mathrm{d}u\,\frac{e^{-bu^2+2\alpha\sqrt{i}u}}{(\pi\,e^{i\frac{\pi}{4}}-u)(\pi\,e^{i\frac{\pi}{4}}-u)}$$ $$=\int_{-\infty}^{\infty}\frac{\mathrm{d}u}{2\pi\sqrt{i}}\left\{\frac{1}{\pi\sqrt{i}-u}+\frac{1}{\pi\sqrt{i}+u}\right\}\,e^{-bu^2+2\alpha\sqrt{i}u},$$ which letting $u=\frac{s}{\sqrt{b}}$, and completing squares on the Gaussian factor, equals $$\frac{e^{i\frac{\alpha^2}{b}}}{2\pi\sqrt{i}}\int_{-\infty}^{\infty}\!\!\!\mathrm{d}s\,\left\{\frac{1}{\pi\sqrt{ib}-s}-\frac{1}{(-\pi\sqrt{ib})-s}\right\}\,e^{-\left(s-\alpha\sqrt{\frac{i}{b}}\right)^2}=\frac{e^{i\frac{\alpha^2}{b}}}{2\pi\sqrt{i}}\left\{I\left(\alpha\sqrt{i/b},\pi\sqrt{ib}\right)-I\left(\alpha\sqrt{i/b},-\pi\sqrt{ib}\right)\right\},\qquad(22)$$ where $I(\omega,z)$ was obtained in equation $(17)$. Applying equation $(17)$ and plugging into equation $(21)$, yields the desired result $$\mathscr{D}(\alpha)=3i\cos(2\pi\alpha)e^{-i\pi^2b}+\frac{e^{i\frac{\alpha^2}{b}}}{2i^{3/2}}\left\{w\left(\alpha\sqrt{i/b}+\pi\sqrt{ib}\right)+2\theta(\alpha)\theta(\alpha/b-\pi)\,e^{-i(\pi\sqrt{b}-\alpha/\sqrt{b})^2}+\mathrm{sgn}(\pi-\alpha/2b)w\left(\mathrm{sgn}(\pi-\alpha/2b)(\pi\sqrt{ib}-\alpha\sqrt{i/b})\right)\right\}$$ I am dead :p

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Since $w(z)$ is related to the error function via equation $(5)$, it might turn out, for $b=1$ this solution agrees with the Wolfram alpha solution proposed in one of the comments.