How to show that $\int_0^{+\infty} e^{-u} \frac{\sin(xu)}{u}\operatorname{d}u=\arctan(x)$? In the process of answering to this question, I got stuck in the calculation of the following integral that prevent me to give a complete answer: 
$$\int_0^{+\infty} e^{-u} \frac{\sin(xu)}{u}\operatorname{d}u.$$
WolframAlpha gives for this integral the result $\arctan(x)$. However, I can't see a path to deduce this equality.
Any hint?
 A: Here we will consider the integral:
\begin{equation}
I(x) = \int_0^\infty e^{-u} \frac{\sin(xu)}{u}\:du \nonumber 
\end{equation} 
To evaluate this integral, we employ Leibniz's Integral Rule and take the derivative under the curve with respect to $x$:
\begin{equation}
I'(x) = \int_0^\infty e^{-u}\frac{\frac{\partial}{\partial x} \sin(xu)}{u} \:du = \int_0^\infty e^{-u} \frac{u \cos(ux)}{u}\:du = \int_0^\infty e^{-u}\cos(xu)\:du \nonumber
\end{equation}
We now call upon the integral:
\begin{equation}
\int e^{at}\cos(bt) = \frac{e^{at}}{a^2 + b^2}\left(a\cos(bx) + b\sin(bx) \right)
\end{equation}
Note - this can be found by applying Integration by Parts Twice. 
Returning to our integral, we find:
\begin{equation}
I'(x) = \left[\frac{e^{-u}}{(-1)^2 + x^2} \left( -1 \cdot \cos(xu) + x \cdot \sin(xu) \right)\right]_0^\infty = \frac{1}{1 + x^2}
\end{equation}
And thus, 
\begin{equation}
I(x) = \int  \frac{1}{1 + x^2} \:dx = \arctan(x) + C
\end{equation}
Where $C$ is the constant of integration.  To resolve $C$ we return to our definition of $I(x)$:
\begin{equation}
I(0) = \int_0^\infty e^{-u} \frac{\sin(0 \cdot u)}{u}\:du = 0 = \arctan(0) + C \nonumber  
\end{equation}
And so $C = 0$. Thus:
\begin{equation}
I(x) = \arctan(x)
\end{equation}
A: A standard argument using two integrations by parts shows that $\frac 1 {1+x^{2}}=\int_0^{\infty} e^{-u} \cos(xu)\, du$. Integrate w.r.t. $x$ and note that both sides of the given equation vanish at $0$. 
