I'm stuck with an integral since a while: $$ \int_0^\infty (e^x-1)^\alpha e^{ -\frac{1}{2} (\frac{x-\mu}{\sigma})^2} dx\\ \alpha\in [0,1] $$
$\sim$expected value of $(e^X-1)^\alpha$ where X follows a normal distribution $\mu \, \sigma$
I've managed to find an approximation (via a approximation of $(e^x-1)^\alpha\sim x^\alpha+\frac{\alpha}{2}x^{\alpha+1}+...$) and it's sufficient for my need but I feel like the exact solution exists. I've learned a lot of new techniques when trying to compute this integral :)
I'm looking for a closed formula made up of combination/composition of (most probably) special functions.
Do you have some tips?
maybe starting with
$$ \int_0^\infty (e^x-1)^\alpha e^{ -x^2} dx\\ $$
is a good start point but I'm not more successful.
Any idea?
Thanks!