I have the following integral:
$\int_0^{y_c} e^{-y} e^{-a e^{-y}}y^2 \operatorname dy$
My first attempt at solving in Mathematica was fruitless (I did not try using any assumptions though).
However, I think I can do it by hand when I make the substitution: $u = e^{-y}$ which means $\operatorname du = -e^{-y}dy = -u\operatorname dy$, and $y = -\ln u$. The integral then becomes
$-\int_1^{e^{-y_c}} u e^{-a u}(\ln u)^2 \frac{\operatorname du}{u} = -\int_1^{e^{-y_c}} e^{-a u}(\ln u)^2 \operatorname du$ Mathematica can then perform the integral in terms of exponential integrals and hypergeometric function.
Does it look like I've performed the $u$-substitution correctly?