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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?

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    $\begingroup$ i might be wrong but this could be a candidat for ramanujans master theorem... $\endgroup$
    – tired
    Oct 7 '15 at 9:12
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For a start, I would do

$\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 $x=1$.

Further than that I know not.

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    $\begingroup$ I don't think you can say much more than this - except perhaps an elaboration of the fact that it is coming from the lower incomplete gamma function. $\endgroup$ Nov 19 '15 at 12:39
  • $\begingroup$ than you very much for the contribution @martycohen $\endgroup$ Nov 26 '15 at 2:31

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