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.
On
Contour integration We are going to evaluate the integral using contour integration along anti-clockwise direction of the upper semi-circle $$\gamma=\gamma_{1} \cup \gamma_{2} \textrm{ where } \gamma_{1}(t)=t+i 0(-R \leq t \leq R) \textrm{ and } \gamma_{2}(t)=R e^{i t} (0<t<\pi) $$ $$ \begin{aligned} \int_0^{\infty} \frac{t \sin t}{u^2+t^2} d t=& \frac{1}{2} \Im \int_{-\infty}^{\infty} \frac{t e^{t i}}{u^2+t^2} d t \\ = & \frac{1}{2} \Im\oint_\gamma \frac{z e^{z i}}{u^2+z^2} d t \\ = & \frac{1}{2} \Im\left[2 \pi i\operatorname{Res} \left(\frac{z e^{z i}}{u^2+z^2}, z=ui\right)\right] \\ = & \frac{1}{2} \Im\left(2 \pi i \lim _{z \rightarrow u i}(z-u i) \frac{z e^{zi}}{z^2+u^2}\right) \\ = & \frac{1}{2} \Im\left(2 \pi i \frac{(u i) e^{(u i) i}}{2 u i}\right) \\ = & \frac{\pi }{2} e^{-u} \end{aligned} $$
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.