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(), 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.
$\endgroup$ 02 Answers
$\begingroup$For $\int_0^\infty e^{-u-e^{-ku}}~du$ , where $k>0$ ,
$\int_0^\infty e^{-u-e^{-ku}}~du$
$=\int_0^\infty e^{-u}e^{-e^{-ku}}~du$
$=\int_1^0u^\frac{1}{k}e^{-u}~d\left(-\dfrac{\ln u}{k}\right)$
$=\dfrac{1}{k}\int_0^1u^{\frac{1}{k}-1}e^{-u}~du$
$=\dfrac{1}{k}\gamma\left(\dfrac{1}{k},1\right)$
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 $\int_0^{\infty}\!e^{-x^2}e^{-ae^{bx^2}}\,dx$?
$=2\int_0^\infty e^{-\theta^2}\sum\limits_{n=0}^\infty\dfrac{(-1)^n(e^{-k\theta^2})^n}{n!}d\theta$
$=2\int_0^\infty\sum\limits_{n=0}^\infty\dfrac{(-1)^ne^{-(kn+1)\theta^2}}{n!}d\theta$
$=\sum\limits_{n=0}^\infty\dfrac{(-1)^n\sqrt\pi}{n!\sqrt{kn+1}}$
$\endgroup$ $\begingroup$Hint for the first: substitute $t = \exp(-ku)$.
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