$$I=\large \int_{0}^{\infty}\frac{\operatorname{arccot}\left(\sqrt{1+x}+\sqrt{2+x}\right)}{1+x}\mathrm dx$$

$\large u=\sqrt{1+x}$

$\large \mathrm du=\frac{1}{2\sqrt{1+x}}\mathrm dx$

$$I=2\large \int_{1}^{\infty}\frac{\operatorname{arccot}\left(u+\sqrt{1+u^2}\right)}{u}\mathrm du$$

$$I=\large 2\sum_{k=0}^{\infty}\frac{(-1)^k}{2k+1}\int_{1}^{\infty}\frac{\mathrm du}{u(u+\sqrt{1+u})^{2k+1}}$$

Applying partial fraction decomposition would be very long.

How would we evaluate $I$?

  • $\begingroup$ The thing inside $\text{arccot}$ should be $u+\sqrt{u^2+1}$. $\endgroup$ – Szeto Jul 27 '18 at 9:28
  • $\begingroup$ Also, differentiation under the integral sign might be a good option. $\endgroup$ – Szeto Jul 27 '18 at 9:34

Please first do what Szeto has mentioned, it's $u+\sqrt{u^2+1}$ and not $1+u+u^2$.

Then use $\,u:=\cot t\,$ and $\,\displaystyle\cot\frac{t}{2}=\cot t + \sqrt{1+\cot^2 t}\,$ for $\,0<t<\pi\,$.

After some small manipulations (e.g. $\,x:=2t\,$) you will get

$\displaystyle I=\frac{1}{2}\int\limits_0^{\pi/2} \frac{x}{\sin x}dx = G\enspace$ where $\,G\,$ is the Catalan constant .

  • $\begingroup$ Sounds nice but any numerical evidence? $\endgroup$ – Szeto Jul 27 '18 at 12:29
  • 1
    $\begingroup$ @Szeto: That's the exact result, no need for something numerical. The OP's integral is reduced to a well known integral, which can be seen in the literature or e.g. here . $\endgroup$ – user90369 Jul 27 '18 at 13:46

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