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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?

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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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    @Lisa: no. If $x=z^2$, then $dx = 2z\,dz$. The factor $2$ is absorbed by the change of the integration range from $\mathbb{R}^+$ to $\mathbb{R}$. – 2017-01-18
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    Yes, thanks. I noticed it after I asked this. Thank you for a beautiful approach. – 2017-01-18
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    Why are we only computing polls in the upper half-plane? I know this has to do with how you construct the path of integration in complex plane. – 2017-01-18
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    @Lisa: because such integral is the limit as $R\to +\infty$ of the integral of the same function in the complex plane, with the integration path being a half-circle centered at the origin in the upper half-plane. The contribute given by the integral along the arc vanishes as $R\to +\infty$ due to $f(z)=O\left(\frac{1}{|z|^2}\right)$. – 2017-01-18
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    Ok. Thanks. This part is clear. Another, question when you combine $\arctan$'s you get $\tan^{-1} \frac{2ac}{c^2+x^2-a^2}$, right? But then the derivative of this is $ \frac{d}{dx} \tan^{-1} \frac{2ac}{c^2+x^2-a^2} = \frac{4a c x}{(c^2+x^2-a^2)^2+(2ac)^2}=\frac{4acx}{(a^2-2ac+c^2+x^2)(a^2+2ac+c^2+x^2)}$. For some reason my denominator is different then yours. Where am I making a mistake? – 2017-01-18
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    @Lisa: it is easier to first differentiate both arctangents, then factoring the difference of such derivatives. – 2017-01-18
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    Can you also give a little detail on how you compute the residues? – 2017-01-18
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    @Lisa: the poles are simple, hence the residues can be computed through: $$\text{Res}\left(\frac{f(z)}{g(z)},z=z_0\right) = \lim_{z\to z_0}\frac{(z-z_0)f(z)}{g(z)} = \frac{f(z_0)}{g'(z_0)}.$$ – 2017-01-18
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    Sure. But how do you get such a nice expression at the end. When I do this my expression is very messy. – 2017-01-18
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    For example, for $z_0=\sqrt(a+ic)$ I get that the residue is $Res= \frac{(a+ic)^{3-\frac{3}{2}}}{16 i ac}$. Is this correct? – 2017-01-18
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    @Lisa: it is. The remaining part is just simple algebra to simplify the sum of those residues. – 2017-01-18
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    Dea Jack after simplifications I arrive at $\sum Res(\cdot,z_k)= \frac{ (1+(-1)^{5/2} )}{16} \left( \frac{( -a+ ic)^{3/2} }{ (a^2 - i a c - c^2)} + \frac{( a+ ic)^{3/2} }{ (a^2 + i a c - c^2)} \right)$. Is this correct? I can't seem to get the same expression as you? – 2017-01-19
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    @Lisa: I do not have my original computations anymore, but the point is that you have to get the sum of two conjugates at the end, since the integral is real. – 2017-01-19