# Integral $\int_0^\frac{\pi}{2} \left(\operatorname{chi}(\cot^2x)+\text{shi}(\cot^2x)\right)\csc^2(x)e^{-\csc^2(x)}dx$

The following problem was posted here a while ago by Cornel Ioan Valean.

Evaluate: $$\int_0^\frac{\pi}{2} [\text{chi}(\cot^2x)+\text{shi}(\cot^2 x)]\csc^2(x)e^{-\csc^2(x)}dx$$ where $$\operatorname{shi}(x)=\int_{0}^{x}\frac{\sinh t}{t}dt$$ and $$\operatorname{chi}(x)=\gamma +\log(x)+\int_{0}^{x}\frac{\cosh(t)-1}{t} dt.$$

I have tried to use 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 is: $$\frac {\sqrt{\pi}}{2e}.$$

I don't know how I can complete integration by parts since it doesn't have a closed form.

Note: I guess this integral is $$0$$ (integration over closed path).

• Is $\gamma$ Euler-Mascheroni constant? Commented Jan 1, 2016 at 3:42
• according to the definition of CHi(x) i think it is Euler-Mascheroni constant Commented Jan 1, 2016 at 14:13
• seems that's gives 0 but proving it look hard integral Commented Jan 1, 2016 at 15:23
• $$\int_0^{\pi/2} e^{-a^2 \csc^2 x} dx=\frac{\pi}{2} \text{erfc}(a)$$ Commented May 31, 2019 at 14:17
• I plotted the function and it appears to have a huge singularity at $x=0$, I'm not sure it's even integrable. Commented May 31, 2019 at 14:22

Two solutions can be found in this pdf. Here is another way: $$\operatorname{chi}(x)+\operatorname{shi}(x):=\gamma+\ln x+\int_0^x \frac{\cosh t+\sinh t-1}{t}=\gamma +\ln x+\int_0^1 \frac{e^{tx}-1}{t}dt$$ $$\int_0^\frac{\pi}{2}\left(\operatorname{chi}(\cot^2 x)+\operatorname{shi}(\cot^2 x)\right)\csc^2 x\,e^{-(1+\cot^2 x)}dx\overset{\cot x=y}=\frac{1}{e}\int_0^\infty\left(\operatorname{chi}(y^2)+\operatorname{shi}(y^2)\right)e^{-y^2}dy$$ $$=\frac{1}{e}\int_0^\infty\left(\gamma +\ln y^2\right)e^{-y^2}dy+\frac1e\int_0^\infty e^{-y^2}\int_0^{1}\frac{e^{ty^2}-1}{t}dtdy=\frac1e\left(-\sqrt \pi \ln 2+\sqrt \pi \ln 2\right)=0$$
$$\int_0^\infty e^{-y^2}\ln y^2 dy = -\frac{\sqrt \pi}{2}\left(\gamma +2\ln 2 \right)\Rightarrow \int_0^\infty\left(\gamma +\ln y^2\right)e^{-y^2}dy=-\sqrt{\pi}\ln 2$$
$$\int_0^\infty e^{-y^2}\int_0^1 \frac{e^{ty^2}-1}{t}dtdy=\int_0^1 \frac{1}{t}\int_0^\infty \left(e^{-y^2(1-t)}-e^{-y^2}\right)dydt$$ $$=\frac{\sqrt\pi}{2}\int_0^1\frac{1}{t}\left(\frac{1}{\sqrt{1-t}}-1\right)dt\overset{1-t=x^2}=\sqrt \pi \int_0^1 \frac{1}{1+x}dx=\sqrt \pi \ln 2$$