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?


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 $\color{red}{\frac{2\pi\sqrt{2}}{3}(a^2+c^2)^{3/4}}$.

  • $\begingroup$ @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 $\mathbb{R}$. $\endgroup$ – Jack D'Aurizio Jan 18 '17 at 14:31
  • $\begingroup$ Yes, thanks. I noticed it after I asked this. Thank you for a beautiful approach. $\endgroup$ – Lisa Jan 18 '17 at 14:32
  • $\begingroup$ 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. $\endgroup$ – Lisa Jan 18 '17 at 14:34
  • $\begingroup$ @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 $f(z)=O\left(\frac{1}{|z|^2}\right)$. $\endgroup$ – Jack D'Aurizio Jan 18 '17 at 14:39
  • $\begingroup$ 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? $\endgroup$ – Lisa Jan 18 '17 at 14:48

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