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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?

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Yes, you did it correctly. – Sasha Aug 14 '11 at 19:42
Looks right to me. If you want you can take out the sign and switch $1$ and $e^{-y_c}$ around. – anon Aug 14 '11 at 20:24

Yes, the substitution have been correctly done.

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HINT: first of all I think that your substitution isn't correct.

$ e^{e^{-y}}=t$ and if we apply ln on both sides we get

$e^{-y}=\ln(t)\Rightarrow -y=\ln(\ln(t)) \Rightarrow dy=\frac{-1}{t\ln(t)}dt$ and $y^2=(\ln(\ln(t)))^2$

If we substitute this into integral we get $I=-\int t^{a-1}(\ln(\ln(t)))^2 dt$ ,you can calculate bounds....

Now make substitution $\ln(t)=p \Rightarrow \frac{1}{t}dt=dp \Rightarrow I=-\int e^ {ap}(\ln(p))^2dp$

The last integral we can solve by partial integration method with $u=e^{ap}\ln(p)$ and $dv=\ln(p)dp$

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