Improper integral of $\frac{\sin x}{x}e^{-ax}$ For $a>0$ define $$I(a)=\int_0^\infty \frac{\sin x}{x}e^{-ax} \, dx,$$ I can show it is continuous at $0$, but by differentiating in $a$, I can't see why $$I(a)=\frac{\pi}{2}-\arctan(a).$$
Thanks for any hints!
 A: Note that
\begin{align}
 {dI \over da} & = {d \over da}\left( \int_0^\infty  {\sin x \over x} e^{-ax} \, dx \right) = \int_0^\infty  {\sin x \over x}{d \over {da}} (e^{-ax}) \, dx  = \int_0^\infty {\sin x \over x} (-x e^{-ax}) \, dx \\[10pt]
  & =  - \int_0^\infty (\sin x) e^{ - ax} \, dx \\[10pt]
I'(a) & =  - \int_0^\infty  (\sin x) e^{-ax} \, dx  = \left. {e^{-ax} (a\sin x + \cos x) \over 1 + a^2} \right|_0^\infty  = 0 - {1 \over 1 + a^2} \\[10pt]
  I'(a) & = - {1 \over 1 + a^2} \\[10pt]
  I(a) & =  - \arctan (a) + C
\end{align}
Also we have
$$\mathop {\lim }\limits_{a \to \infty } I(a) = 0$$
and hence
$$ - \lim_{a \to \infty } \arctan (a) + C = 0\,\,\, \to \,\,\,\, - {\pi  \over 2} + C = 0\,\,\,\, \to \,\,\,\,C = {\pi  \over 2}$$
A: Notice, we know the Laplace Transform of $\sin x$ 
$$L[\sin x]=\int_{0}^{\infty}e^{-st}\sin t\ dt=\frac{1}{s^2+1}$$
Using property of Laplace transform as follows
$$I(a)=\int_{0}^{\infty}e^{-ax}\ \frac{\sin x}{x}\ dx$$$$=\int_{a}^{\infty}L[\sin x]\ ds$$  $$=\int_{a}^{\infty}\frac{1}{s^2+1}\ ds$$
$$=[\tan^{-1}(s)]_{a}^{\infty}$$ $$=\tan^{-1}(\infty)-\tan^{-1}(a)$$
 $$=\frac{\pi}{2}-\tan^{-1}(a)$$
