Integrate Gumbel times Normal

Question

I try to solve an integral of the following form:

$$ \int_{-\infty}^\infty e^{-x^2} \, e^{-e^{-x^2}} dx $$

Intuitively, the first term, $e^{-x^2}$, is related to the pdf of a standard-normal distribution, while the second term, $e^{-e^{-x^2}}$, is related to the pdf of a Gumbel-distribution (except for the square).

From plotting the function, it seems that the integral should be well defined, but I cannot find a solution yet. Any hint on how to solve this is highly appreciated.

Answer 1

I'd be surprised if there was a closed form for that integral. However, you can approximate it via the following series expansion: $$\begin{split} I &= \int_{\mathbb R} e^{-x^2}e^{-e^{-x^2}}dx\\ &= \int_{\mathbb R} e^{-x^2} \left( \sum_{n\geq 0} \frac{(-1)^n}{n!}e^{-nx^2}\right)dx\\ &= \sum_{n\geq 0} \frac{(-1)^n}{n!} \int_{\mathbb R}e^{-(n+1)x^2}dx\\ &= \sum_{n\geq 0} \frac{(-1)^n}{n!} \sqrt{\frac{\pi}{n+1}} \end{split}$$

Answer 2

You can use the symmetry of $e^{-x^2-e^{-x^2}}$ and reduce the integral to

$\int_{\mathbb{R}}e^{-x^2-e^{-x^2}}d\lambda{(x)} = 2\int_{\mathbb{R}_{\geq{0}}}e^{-x^2-e^{-x^2}}d\lambda{(x)}$.

Have you tried using polar coordinates to solve the integral?

I would start like this:

$I=\int_{\mathbb{R_{\geq{0}}}}e^{-x^2-e^{-x^2}}d\lambda{(x)}\Leftrightarrow I^2=(\int_{\mathbb{R_{\geq{0}}}}e^{-x^2-e^{-x^2}}d\lambda{(x)})(\int_{\mathbb{R_{\geq{0}}}}e^{-y^2-e^{-y^2}}d\lambda{(y)})$

Answer 3

Some related expressions:

• Use a transformation $z = x^2$

  $$\begin{array}{} \int_{-\infty}^\infty e^{-x^2} \, e^{-e^{-x^2}} dx &=& \int_{-\infty}^\infty \, e^{-\left(x^2+e^{-x^2}\right)} dx \\ &=& 2 \int_{0}^\infty \, e^{-\left(z+e^{-z}\right)} (dx/dz) dz\\ &=& \int_{0}^\infty \, \frac{1}{ \sqrt{z}} e^{-\left(z+e^{-z}\right)} dz\\ \end{array}$$

  where $z$ is Gumbel distributed.

  The Gumbel distribution is related to the limiting extreme value distribution of an exponential distribution, so the -0.5 power of a Gumbel distributed variable is related to the limiting extreme value distribution of a the -0.5 power of an exponential distributed variable. Possibly one can work from there.

• We can further transform the Gumbel distribution to an exponential distribution by taking the logarithm. If $Z$ is standard Gumbel distributed then $Y = \exp(-Z)$ is exponentially distributed and the integral is equivalent to an integral with $1/\sqrt{-\log(Y)}$ where $Y$ is exponentially distributed.

  $$\int_0^1 \frac{e^{-y}}{\sqrt{-\log(y)}} dy$$

• Another interesting relationship is that this integral is related to $E[e^{-f(X)}]$ where $X$ is normal distributed.

Answer 4

Too long for comments.

I do not remember where I saw the approximation of @Stefan Lafon's nice result. $$ \sum_{n= 0}^\infty \frac{(-1)^n}{n!} \sqrt{\frac{\pi}{n+1}}\sim \left(\lambda \,\Gamma \left(\frac{7}{12}\right)\right)^2$$ where $\lambda$ is the Golomb–Dickman constant

Remember that in $1966$, Shepp and Lloyd proved that $$\lambda=\int_0^\infty \exp\left(-x -\int_x^\infty \frac {e^{-y}} y \,dy \right)\,dx$$ which is of the "same" family of your integral.

The error is $3.62\times 10^{-8}$.