How to show that $\int_0^{+\infty} e^{-u} \frac{\sin(xu)}{u}\operatorname{d}u=\arctan(x)$?

Question

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?

Answer 1

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 $0$.

Answer 2

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}