$$ \int_{-\infty}^{+\infty} e^{-t^2}\ln(1+e^t) \, dt $$
Hello everyone,i would like to know the result of the above integral and how to calculate or estimate it.
background and progress so far:(1)This is a transcendental integral,to calculate it,i used integration by Parts,but didn't work well.So i hope to find a way to give it an estimation or which is the better case,to calculate it in a closed form. (2)According to the work done by Jingeon An,this integral is convergent and can be estimated through a scaling method.
For the convergence, just find a function bigger than the given function and it's integral converges : Observe there exists $C>0$ such that $0<\ln(1+e^t)\leq Ct^2$ for all $t\in\mathbb{R}$ (check this claim by yourself). Then $$0\leq\bigg|\int_{-\infty}^\infty e^{-t^2}\ln(1+e^t)dt\bigg|=\int_{-\infty}^\infty e^{-t^2}\ln(1+e^t)dt\leq C\int_{-\infty}^\infty e^{-t^2}t^2<\infty$$ by Gaussian integral.
I believe calculating the exact result is more difficult, so if I can, I will add later.