I have tried to evaluate this integral :$$ \int_0^\frac{\pi}{2} [\text{chi}(\cot^2x)+\text{shi}(\cot^2 x)]\csc^2(x)e^{-\csc^2(x)}dx $$ where, $\text{shi}(x)=\int_{0}^{x}\frac{\sinh t}{t}dt$ , $\text{chi}(x)=\gamma +\log(x)+\int_{0}^{x}\frac{\cosh(t)-1}{t} dt $

using integral by parts but I didn't succeed as I crossed this:
$$\int_0^\frac{\pi}{2}\csc^2(x)e^{-\csc^2(x)}dx$$ which gives : $\frac {\sqrt{\pi}}{2e}.$

I don't know how I can complete integration by parts since hasn't closed form .

Note : I guess this integral gives 0 (integration over closed path)

Thank you for any help !!!!!

  • $\begingroup$ Is $\gamma$ Euler-Mascheroni constant? $\endgroup$ – Henricus V. Jan 1 '16 at 3:42
  • $\begingroup$ according to the definition of CHi(x) i think it is Euler-Mascheroni constant $\endgroup$ – zeraoulia rafik Jan 1 '16 at 14:13
  • $\begingroup$ seems that's gives 0 but proving it look hard integral $\endgroup$ – Salmahamizi Hamizi Jan 1 '16 at 15:23
  • $\begingroup$ $$\int_0^{\pi/2} e^{-a^2 \csc^2 x} dx=\frac{\pi}{2} \text{erfc}(a)$$ $\endgroup$ – Yuriy S May 31 at 14:17
  • $\begingroup$ I plotted the function and it appears to have a huge singularity at $x=0$, I'm not sure it's even integrable. $\endgroup$ – Yuriy S May 31 at 14:22

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