I have some difficulty with the following exercise:
Calculate $\int_{0}^{\infty}\frac{\sin(xy)}{1+x^2+y^2}dx$.
I tried to differentiate $\frac{\sin(xy)}{1+x^2+y^2}$ to simplify the integral, but the result seems to be more difficult.
I don't have any idea on how to solve this. Can anyone help me or give me a hint?
Consider the more general integral for variable $a>0$ and fixed $b>0$: $$I(a)=\int_0^{\infty} \frac{\sin(ax)}{x^2+b^2}~dx. \tag{1}$$ We will show that $I$ satisfies a certain ODE, and then solve it. First, note by integration by parts that $$\begin{align*} I(a)&=\left[-\frac{\cos(ax)}{a(x^2+b^2)}\right]_0^{\infty}-\frac{2}{a}\int_0^{\infty} \frac{x\cos(ax)}{(x^2+b^2)^2}~dx\\&=\frac{1}{ab^2}-\frac{2}{a}\int_0^{\infty} \frac{x\cos(ax)}{(x^2+b^2)^2}~dx. \tag{2} \end{align*}$$ Thus it follows that $$\int_0^{\infty} \frac{x\cos(ax)}{(x^2+b^2)^2}~dx=\frac{1-ab^2I(a)}{2b^2}. \tag{3}$$ Note that $1/a$ is a common factor in equation $(2)$, so multiplying by $a$ and differentiating w.r.t. $a$ gives $$\begin{align*} I(a)+aI'(a)&=2\int_0^{\infty} \frac{x^2\sin(ax)}{(x^2+b^2)^2}~dx\\&=2\int_0^{\infty} \frac{\sin(ax)}{x^2+b^2}~dx-2b^2\int_0^{\infty} \frac{\sin(ax)}{(x^2+b^2)^2}~dx\\&=2I(a)-2b^2\int_0^{\infty} \frac{\sin(ax)}{(x^2+b^2)^2}~dx. \end{align*}$$ Differentiating with respect to $a$ one last time and using equation $(3)$ gives $$\begin{align*} I'(a)+I'(a)+aI''(a)&=2I'(a)-2b^2\int_0^{\infty} \frac{x\cos(ax)}{(x^2+b^2)^2}~dx\\&=2I'(a)-1+ab^2 I(a). \end{align*}$$ Simplifying gives the ODE $$I''(a)-b^2 I(a)=-1/a. \tag{4}$$ We now proceed to solve this differential equation. We can do this using variation of parameters. A fundamental set of solutions for the homogeneous equation is given by $\{e^{ab},e^{-ab}\}$. The method results in the need to consider the exponential integral $\operatorname{Ei}$ defined for nonzero real values of $x$ by $$\operatorname{Ei}(x):=-\int_{-x}^{\infty} \frac{e^{-t}}{t}~dt.$$ The general solution can then be written as $$I(a)=C_1 e^{ab}+C_2 e^{-ab}+\frac{e^{-ab}\operatorname{Ei}(ab)}{2b}-\frac{e^{ab} \operatorname{Ei}(-ab)}{2b}.$$ We now claim that $C_1=C_2=0$. Note that $$|I(a)|\leq \int_0^{\infty} \frac{1}{x^2+b^2}~dx=\frac{\pi}{2b},$$ hence for fixed $b$ we have that $I$ is bounded. Taking the limit as $a\to \infty$ and using Proof of $\lim_{x\to\infty}\frac{\operatorname{Ei}(x)}{e^x}=0$ and showing that $\lim_{x\to\infty}\frac{\operatorname{Ei}(-x)}{e^{-x}}=0$ proves that $C_1=0$. Finally, taking the limit as $a\to 0^+$ shows that $C_2=0$ since $\lim_{a\to 0^+} I(a)=0$. Therefore, the solution to the integral is (if you wish, you can generalize to $a,b\neq 0$ using the symmetry of the integrand) $$I(a)=\frac{e^{-ab}\operatorname{Ei}(ab)}{2b}-\frac{e^{ab} \operatorname{Ei}(-ab)}{2b}.$$ Hence, it follows that for $y\geq 0$ $$\bbox[5px,border:2px solid #C0A000]{\int_{0}^{\infty}\frac{\sin(xy)}{1+x^2+y^2}dx=\frac{e^{-y\sqrt{1+y^2}}\operatorname{Ei}(y\sqrt{1+y^2})}{2\sqrt{1+y^2}}-\frac{e^{y\sqrt{1+y^2}} \operatorname{Ei}(-y\sqrt{1+y^2})}{2\sqrt{1+y^2}}.}$$
Note: An alternative form for $I(a)$ is $$I(a)=\frac{\operatorname{Shi}(ab)\cosh(ab)-\operatorname{Chi}(ab)\sinh(ab)}{b},$$ where $\operatorname{Shi}$ is the hyperbolic sine integral and $\operatorname{Chi}$ is the hyperbolic cosine integral. Alternatively, one can write the result in terms of the Meijer G function as written by @EldarSultanow.
The searched integral function involves the Meijer G function and it is given by:
$\int_{0}^{\infty} \frac{\sin(xy)}{1+x^2+y^2}dx=\frac{\sqrt{\pi } G_{1,3}^{2,1}\left(\frac{1}{4} \left(y^4+y^2\right)| \begin{array}{c} 1 \\ 1,1,\frac{1}{2} \\ \end{array} \right)}{y^3+y}$
where the following condition must be met:
$\Im(y)\leq 0\land \left(\Im\left(\sqrt{-y^2-1}\right)\neq 0\lor \left(\Re(y)=0\land \Re\left(y^2\right)+1>0\land \Im(y)+1>0\land \Re\left(\sqrt{-y^2-1}\right)\leq 0\right)\right)$
The plot of original function for $x,y\in[0,10]$ is:
and the plot of the resulting integrated function for $y\in[9,9.5]$ is:
