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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?

  • 2
    Yes, you did it correctly.2011-08-14
  • 1
    Looks right to me. If you want you can take out the sign and switch $1$ and $e^{-y_c}$ around.2011-08-14

2 Answers 2