TITLE: Integral $\int_{-\infty}^{\infty}\frac{\cos(s \arctan(ax))}{(1+x^2)(1+a^2x^2)^{s/2}}dx$
QUESTION [13 upvotes]: Prove that:
$$
\int_{-\infty}^{\infty}{\cos\left(\vphantom{\Large A}
s\ \arctan\left(\vphantom{\large A}ax\right)\right)\over
\left(1 + x^{2}\right)\left(1 + a^{2}x^{2}\right)^{s/2}}\,{\rm d}x
={\pi \over \left(1 + a\right)^{s}}$$
where $a,s \in \mathbb{R}^{+}$.
This looks difficult. What would be a good start? Can we use the Residue Theorem?

REPLY [11 votes]: I can't resist the temptation to answer this problem. The integral really makes me excited! ٩(˘◡˘)۶

Here is a non-contour/ residue approach. We have a classic result
\begin{equation}
 \int_{0}^{\infty} y^{s-1}\cos (a y)\,e^{-by}\,dy = \Gamma (s)\frac{\cos \left(\! s \arctan \left(\frac{a}{b}\right)\right)}{\left(a^2+b^2\right)^{s/2}},\quad\mbox{for}\quad \Re(s),\, a,\,b>0.\tag{1}
\end{equation}
where it can easily be proved by using
\begin{equation}
\Re\int_{0}^{\infty} y^{s-1}e^{-(b-ia)y}\,dyt = \Re\left[ \frac{\Gamma (s)}{(b-ia)^s}\right]\tag{2}
\end{equation}
Putting $a=ax$ and $b=1$ to $(1)$, then dividing by $\dfrac{1}{1+x^2}$ and taking the integral both of sides, we have
\begin{equation}
\int_{-\infty}^{\infty}\int_{0}^{\infty} \frac{y^{s-1}\cos (axy)\,e^{-y}}{1+x^2}\,dy\,dx = \Gamma (s)\int_{-\infty}^{\infty}\frac{\cos \left(\! s \arctan \left(ax\right)\right)}{(1+x^2)\left(1+a^2x^2\right)^{s/2}}\,dx\tag{3}
\end{equation}
Changing the order of integration on the LHS which can be justified by using Fubini's theorem we have
\begin{equation}
\int_{0}^{\infty}y^{s-1}\,e^{-y}\int_{-\infty}^{\infty} \frac{\cos (ay\,x)}{1+x^2}\,dx\,dy = \Gamma (s)\int_{-\infty}^{\infty}\frac{\cos \left(\! s \arctan \left(ax\right)\right)}{(1+x^2)\left(1+a^2x^2\right)^{s/2}}\,dx\tag{4}
\end{equation}
Using another classic result
\begin{equation}
\int_{-\infty}^{\infty} \frac{\cos nx}{1+x^2}\,dx=\pi e^{-n}\tag{5}
\end{equation}
then we obtain
\begin{align}
\int_{0}^{\infty}y^{s-1}\,e^{-y}\int_{-\infty}^{\infty} \frac{\cos (ay\,x)}{1+x^2}\,dx\,dy &=\pi\int_{0}^{\infty}y^{s-1}\,e^{-(1+a)y}\,dy\\
&=\frac{\pi\,\Gamma (s)}{(1+a)^s}\tag{6}
\end{align}
Thus, by putting $(6)$ to $(4)$, we obtain
\begin{equation}
\int_{-\infty}^{\infty}\frac{\cos \left(\! s \arctan \left(ax\right)\right)}{(1+x^2)\left(1+a^2x^2\right)^{s/2}}\,dx=\frac{\pi}{(1+a)^s}
\end{equation}
which is the announced result.$\qquad\square$