# Mellin transform of Gumbel distribution

The probability density function (PDF) of Gumbel distribution is given as:

$$f\left(x\right)=\frac{\exp \left(-\left(\exp \left(-\frac{x-\mu}{\beta }\right)+\frac{x-\mu}{\beta }\right)\right)}{\beta },$$

where $\beta>0$. I would like to assume the translation $\mu$ to be $0$, which gives:

$$f\left(x\right)=\frac{\exp \left(-\left(\exp \left(-\frac{x}{\beta }\right)+\frac{x}{\beta }\right)\right)}{\beta }.$$

The Mellin transform of the PDF is therefore:

$$\int_0^{\infty } \frac{x^{s-1} \exp \left(-\left(\exp \left(-\frac{x}{\beta }\right)+\frac{x}{\beta }\right)\right)}{\beta } \ dx,$$ where $s>0$.

To solve this, I input the expression into Mathematica but no luck. Is there any other way to solve this?

• i might be wrong but this could be a candidat for ramanujans master theorem... Oct 7 '15 at 9:12

$\begin{array}\\ \int_0^{\infty } \frac{x^{s-1} \exp \left(-\left(\exp \left(-\frac{x}{\beta }\right)+\frac{x}{\beta }\right)\right)}{\beta } \ dx &=\int_0^{\infty } \frac{x^{s-1}\exp(-\frac{x}{\beta }) \exp \left(-\left(\exp \left(-\frac{x}{\beta }\right)\right)\right)}{\beta } \ dx\\ \text{Letting } y =\exp(-\frac{x}{\beta }), dy=-\frac1{\beta}\exp(-\frac{x}{\beta })dx, &x=\beta\ln(1/y)\\ &=\int_1^{0 } -(\beta\ln(1/y))^{s-1} \exp (-y)\ dy\\ &=\beta^{s-1}\int_0^1 (\ln(1/y))^{s-1} \exp (-y)\ dy\\ \end{array}$
This resembles the formula for the $s$-th derivative of $\Gamma(x)$ at $x=1$.