I would like to find the integral of $\int_0^\infty\exp(-u-\exp(-ku))\,du$ for $k>0$.

This is related to the gumbel distribution(http://en.wikipedia.org/wiki/Gumbel_distribution), which shows that this integral is one if k=1.

However, I would like to know how to integrate this without using the fact that this is a distribution, just so that I can see the method of integration.

Side question: Any suggestions on integrating $\int_{-\infty}^\infty\exp(-\theta^2-\exp(-k\theta^2))\,d\theta$

update: See my comment below on Roberts answer for a solution.


For $\int_0^\infty e^{-u-e^{-ku}}~du$ , where $k>0$ ,

$\int_0^\infty e^{-u-e^{-ku}}~du$

$=\int_0^\infty e^{-u}e^{-e^{-ku}}~du$

$=\int_1^0u^\frac{1}{k}e^{-u}~d\left(-\dfrac{\ln u}{k}\right)$



For $\int_{-\infty}^\infty e^{-\theta^2-e^{-k\theta^2}}~d\theta$ , where $k>0$ ,

$\int_{-\infty}^\infty e^{-\theta^2-e^{-k\theta^2}}~d\theta$

$=\int_{-\infty}^\infty e^{-\theta^2}e^{-e^{-k\theta^2}}~d\theta$

$=\int_{-\infty}^0e^{-\theta^2}e^{-e^{-k\theta^2}}~d\theta+\int_0^\infty e^{-\theta^2}e^{-e^{-k\theta^2}}~d\theta$

$=\int_\infty^0e^{-(-\theta)^2}e^{-e^{-k(-\theta)^2}}~d(-\theta)+\int_0^\infty e^{-\theta^2}e^{-e^{-k\theta^2}}~d\theta$

$=\int_0^\infty e^{-\theta^2}e^{-e^{-k\theta^2}}~d\theta+\int_0^\infty e^{-\theta^2}e^{-e^{-k\theta^2}}~d\theta$

$=2\int_0^\infty e^{-\theta^2}e^{-e^{-k\theta^2}}~d\theta$ , which is similar to What is $\int_0^{\infty}\!e^{-x^2}e^{-ae^{bx^2}}\,dx$?

$=2\int_0^\infty e^{-\theta^2}\sum\limits_{n=0}^\infty\dfrac{(-1)^n(e^{-k\theta^2})^n}{n!}d\theta$




Hint for the first: substitute $t = \exp(-ku)$.

  • $\begingroup$ Going by your suggestion, I get $$\frac{1}{k}\int_0^1 t^{1/k-1}\exp(-t) dt=\frac{1}{k}\gamma(\frac{1}{k},1)$$ for k=1 this approximately 0.63, am I doing something wrong? $\endgroup$ – sachinruk Oct 23 '13 at 1:57
  • $\begingroup$ No, that's correct. Note that for the Gumbel distribution, $u$ goes from $-\infty$ to $\infty$ so $t$ goes from $0$ to $\infty$. $\endgroup$ – Robert Israel Oct 23 '13 at 3:34

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