real analysis - Evaluation of an improper integral: $int_0^infty frac{tsin{t}}{u^2+t^2,}dt$

I want to calculate the integral below for positive $u$.
$$\int_0^\infty \dfrac{t\sin{t}}{u^2+t^2}dt$$

Wolframalpha gives me a simple answer :
$\dfrac{\pi}{2}e^{-u}$.

but I cannot approach to that.
Can anyone solve above without using complex integral (ex.residue integration.. because I'm beginner of analysis)?

Thanks.

Answer

Hint. Let's consider the Laplace transform of $\displaystyle I(a):=\int_{0}^{\infty}\frac{\cos(ax)}{x^2+1}\:dx$. We have
$$

\begin{aligned}
\mathcal{L}\left(I(a)\right)(s)&=\mathcal{L}\left(\int_{0}^{\infty}\frac{\cos(ax)}{x^2+1}\:dx\right)(s)
\\& = \int_{0}^{\infty}\int_{0}^{\infty}\frac{\cos(ax)}{x^2+1}e^{-as}\:da\:dx
\\&= \int_{0}^{\infty}\frac{s}{(x^2+1)(s^2+x^2)}\;{dx}
\\&= \frac{\pi}{2(s+1)}
\end{aligned}
$$giving
$$
I(a)=\int_{0}^{\infty}\frac{\cos(ax)}{x^2+1}\:dx=\mathcal{L}^{-1}\left(\frac{\pi}{2(s+1)}\right) =\frac{\pi}{2}e^{-a},\qquad a>0, \tag2
$$ then by differentiating $(2)$ with respect to $a$, one gets

$$
\int_{0}^{\infty}\frac{x\sin(ax)}{x^2+1}\:dx=\frac{\pi}{2}e^{-a},\qquad a>0. \tag3
$$ as given by Wolfram alpha.