<s>[INST] Find $\int_0^\infty \sqrt{x} \left( \tan^{-1} \left(\frac{x+a}{c} \right)- \tan^{-1} \left(\frac{x-a}{c} \right) \right)dx$ How to compute the integral
\begin{align}
\int_{0}^{\infty}\sqrt{x}\left[\tan^{-1}\left(\frac{x+a}{c} \right)- \tan^{-1} \left(\frac{x-a}{c} \right) \right]\mathrm{d}x
\end{align}
for $c,a>0$.
If the it the integration is to difficult can we give a good upper bound? 
 [/INST] By using integration by parts, the given integral equals
$$ \frac{2}{3}\int_{0}^{+\infty}\frac{4ac x^{5/2}}{\left(a^2+c^2-2 a x+x^2\right) \left(a^2+c^2+2 a x+x^2\right)}\,dx $$
and by substituting $x=z^2$, this integral becomes
$$ \frac{2}{3}\int_{-\infty}^{+\infty}\frac{4ac z^{6}}{\left(a^2+c^2-2 a z^2+z^4\right) \left(a^2+c^2+2 a z^2+z^4\right)}\,dz $$
where the integrand function is $O\left(\frac{1}{|z|^2}\right)$ as $|z|\to +\infty$. By the residue theorem, we just need to compute the residues of the integrand function at the poles in the upper half-plane, given by $\sqrt{\pm a+ ic}$ and $-\sqrt{\pm a-ic}$. It follows that the given integral equals:
$$\boxed{ \frac{2\pi}{3}\sqrt{c(3a^2-c^2)+(a^2+c^2)\sqrt{a^2+c^2}} }$$
and a simple upper bound (that is tight iff $a\approx 0$ or $a\approx c\sqrt{3}$) is given by $\color{red}{\frac{2\pi\sqrt{2}}{3}(a^2+c^2)^{3/4}}$.
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