How to prove that $\int_{0}^{\infty}\frac{t}{x^2+t^2}\cos(ax)dx=\frac{\pi}{2}e^{-at}$. I want to prove that for $t,a>0$:
$$f_a(t)=\int_{0}^{\infty}\frac{t}{x^2+t^2}\cos(ax)dx=\frac{\pi}{2}e^{-at}$$
It is easy to prove that 
$$f_a''(t)=a^2f_a(t),$$
then
$$f_a(t)=c_1e^{at}+c_2e^{-at}$$
Cleary $\lim\limits_{t\to 0}f_a(t)=0$ and therefore $c_1=0$. But I could not justify that $c_2=\frac{\pi}{2}$. Any idea?
Thank you very much.
 A: $c_{2}=f_{0}(1)=\displaystyle\int_{0}^{\infty}\dfrac{1}{x^{2}+1}dx=\tan^{-1}x\bigg|_{x=0}^{x=\infty}=\dfrac{\pi}{2}$.
A: Another perspective is the change of variables Enforcing $x=yt$ then from this we get  
$$\int_{0}^{\infty}\frac{t}{x^2+t^2}\cos(ax)dx=\frac{\pi}{2}e^{-at}=\int_{0}^{\infty}\frac{t^2\cos(aty)}{t^2(y^2+1)}dy=\frac{\pi}{2}e^{-|at|} $$ 
A: There is another nice solution going through the complex via a residue:
$$I = \int_{0}^{\infty}\frac{t}{x^2+t^2}\cos(ax)dx= \frac{1}{2}\int_{-\infty}^{\infty}\frac{t}{x^2+t^2}\cos(ax)dx = \frac{1}{2}Re\left( \int_{-\infty}^{\infty}\frac{t}{x^2+t^2}e^{iax}dx \right)$$
Here I wrote $Re$ for real part, but it turns out that the integral on the right is real:
$$\int_{-\infty}^{\infty}\frac{t}{x^2+t^2}e^{iax}dx = 2\pi iRes_{it}\left( \frac{t}{z^2+t^2}e^{iaz}\right) = 2\pi i \lim_{z\rightarrow it}\left( \frac{t(z-it)}{(z+it)(z-it)}e^{iaz}\right)= 2\pi i\frac{t}{2it}e^{-at} = \pi e^{-at}$$
All together:
$$I = \int_{0}^{\infty}\frac{t}{x^2+t^2}\cos(ax)dx=\frac{\pi}{2}e^{-at}$$
