On a problem I'm working on, I've come across this integral: $$I=\int_{-1}^{1}e^{-\ln^2\left(\frac{1+x}{1-x}\right)}dx$$ and I'm wondering how to evaluate it analytically. I don't think the context would be much help.
A simple substitution $u=\frac{1+x}{1-x}$ yields $$I=2\int_0^{\infty}\frac{e^{-\ln^2(u)}}{(u+1)^2}du$$
But I don't know if this form is any easier to tackle.
Any idea?
I continue from @user170231's comment (dated Apr 3, 2024).
We have $$ I {=\int_{-\infty}^\infty \frac{2e^{-v^2+v}}{(e^v+1)^2}dv \\= \int_{-\infty}^0 \frac{2e^{-v^2+v}}{(e^v+1)^2}dv + \int_{0}^\infty \frac{2e^{-v^2+v}}{(e^v+1)^2}dv \\= \int_{0}^\infty \frac{2e^{-v^2-v}}{(e^{-v}+1)^2}dv + \int_{0}^\infty \frac{2e^{-v^2+v}}{(e^v+1)^2}dv \\= 4\int_{0}^\infty \frac{e^{-v^2-v}}{(e^{-v}+1)^2}dv \\= 4\int_{0}^\infty e^{-v^2-v}\sum_{n=0}^\infty (n+1)(-1)^ne^{-nv}dv \\= 4\sum_{n=0}^\infty (n+1)(-1)^n\int_{0}^\infty e^{-v^2-(n+1)v}dv \\= 4\sum_{n=0}^\infty (n+1)(-1)^n\int_{0}^\infty e^{-v^2-(n+1)v-\frac{(n+1)^2}{4}}e^{\frac{(n+1)^2}{4}}dv \\= 4\sum_{n=0}^\infty (n+1)(-1)^ne^{\frac{(n+1)^2}{4}}\int_{\frac{n+1}{2}}^\infty e^{-v^2}dv \\= 2\sqrt\pi\sum_{n=0}^\infty (n+1)(-1)^ne^{\frac{(n+1)^2}{4}}\text{erfc}\left(\frac{n+1}{2}\right), } $$ where $\text{erfc}\left(\cdot\right)$ is the complementary error function.
