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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.

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  • $\begingroup$ Mathematica couldn't figure it out, FWIW. $\endgroup$ Commented Jul 10, 2023 at 9:19
  • $\begingroup$ With merely an extra $x$ in the integrand, it becomes trivial (even when the limits are not $\pm \infty$). $\endgroup$ Commented Jul 10, 2023 at 10:27
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    $\begingroup$ What is the motivation for solving this integral? Where does it come from? $\endgroup$ Commented Jul 10, 2023 at 11:44
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    $\begingroup$ The motivation is a bit tricky here: I am analyzing how the database of someone who observes draws from a normal distribution (thus, the $e^{-x^2}$) but overweights (or remembers or oversamples) "extreme" observations (thus, the Gumbel-distribution) looks like. $\endgroup$
    – max1993
    Commented Jul 10, 2023 at 14:52

4 Answers 4

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

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  • $\begingroup$ Stefan, many thanks for your reply! Could you briefly elaborate on the series expansion of $e^{-e^{-x^2}}$ that you used? I am not sure I can follow it, since the derivative of this expression seems more complex. $\endgroup$
    – max1993
    Commented Jul 10, 2023 at 14:49
  • $\begingroup$ I'm using the power series for the exponential function $$e^z = \sum \frac{z^n}{n!}$$ and substitute $z=e^{-z^2}$. $\endgroup$ Commented Jul 10, 2023 at 16:06
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    $\begingroup$ Ah, I see. My bad. Thanks for the explanation! $\endgroup$
    – max1993
    Commented Jul 10, 2023 at 16:20
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    $\begingroup$ This is a very nice result. (+1). The convergence is extremely fast. $\endgroup$ Commented Jul 11, 2023 at 5:55
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    $\begingroup$ Thank you. I thought this might help OP in case they wanted a numerical computation. $\endgroup$ Commented Jul 11, 2023 at 12:48
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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.

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  • $\begingroup$ Using your last integral and expanding $e^{-y}$ as a series, you have the same result as @Stefan Lafon since, for $n>-1$ $$\int_0^1\frac{y^n}{\sqrt{-\log (y)}}\,dy=\frac{\sqrt{\pi }}{\sqrt{n+1}}$$ $\endgroup$ Commented Jul 12, 2023 at 5:17
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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 $3.62\times 10^{-8}$.

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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:

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

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  • $\begingroup$ Thanks, Laura. I actually tried using polar coordinates, but eventually got stuck on $I^2 = \int_{\mathbb{R}\geq 0} e^{- (x^2 + y^2)} \; e^{-(e^{-x^2} + e^{-y^2})} dx dy$, since the transformation of $x = r \, cos \theta$ and $y = r \, sin \theta$ does not seem to simplify the second part. $\endgroup$
    – max1993
    Commented Jul 10, 2023 at 14:54

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