Evaluating a specific definite integral

Question

In a quantum mechanics project I encountered the following definite integral:

$$I(x) = \int_0^\infty \sin(xt) \frac{t^2}{(1 + t^2)^2} \ dt$$

It appears deceptively simple, but gives me headaches....

The integration over variable $t$ runs from zero to +infinity. It would be nice if one could extend the integration interval to $(-\infty, +\infty)$, since then we could perform contour integration, leading us straight to the required solution. Unfortunately, this method is not available due to the properties of the integrand at $t=0$.

I welcome your suggestions in solving the above integral !

Answer 1

This is what Mathematica finds:

1/4 Sqrt[\[Pi]] x MeijerG[{{-1}, {}}, {{0, 0}, {-(1/2)}}, x^2/4]

Answer 2

According to Maple, $$\int_0^\infty \dfrac{\sin(x t) t^2}{(1+t^2)^2}\; dt = \dfrac{\sinh \left( x \right) x+\cosh \left( x \right)}{2} {\it Shi} \left( x \right) - \dfrac{ \cosh \left( x \right) x +\sinh \left( x \right) }{4} {\it Ei} \left( x \right)- \dfrac{ \cosh \left( x \right) x +\sinh \left( x \right) }{4} {\it Ei} \left( -x \right) $$

Answer 3

Since the inverse Laplace transform of $\frac{t^2}{(1+t^2)^2}$ is $\frac{1}{2}\left(x\cos x +\sin x\right)$, we have:

$$ I = \frac{1}{2}\int_{0}^{+\infty}\sin(xt)\int_{0}^{+\infty}\left(y\cos y +\sin y\right)e^{-yt}\,dy\,dt$$

and switching the order of integration:

$$ I = \frac{1}{2}\int_{0}^{+\infty}\frac{x\left(y\cos y +\sin y\right)}{x^2+y^2}\,dy $$

that is a nice integral to be approximated. Its "closed expression", however, is quite complicated, as already noticed by Robert Israel and Juliàn Aguirre.

Answer 4

Let $~F(x)=\displaystyle\int_0^\infty\frac{\sin(xt)}{t^2+a}dt.\quad$ Then $~I(x)=\bigg[\dfrac{d^2}{dx^2}\dfrac{d}{da}F(x)\bigg]_{a~{\large=}~1}\quad$ Unfortunately, unlike

its cosine equivalent, $F(x)$ does not possess a closed form in terms of known constants. If,

on the other hand, your integral would've been $~J(x)=\displaystyle\int_0^\infty\cos(xt)\frac{t^2}{(1+t^2)^2}dt,~$ then the

situation would've been completely different, since $~G(x)=\displaystyle\int_0^\infty\frac{\cos(xt)}{t^2+a^2}dt~=~\frac\pi{2|a|}e^{-|ax|}.$

However, Mathematica does return an expression in terms of the Meijer G function for $F(x)$.