How can I found 3rd and 4th central moments of gumbel distribution with characteristic function? I tried many ways(using digamma function, partial integration etc.) to find moments using by characteristic function but I couldn't. Is there any trick or suggestion? (I couldn't calculate integral of 2nd, 3nd and 4th derivatives of gamma function.)
Thanks in advance..

The characteristic function for the Gumbel distribution with location parameter $\alpha$ and scale parameter $\beta$ is
$$  \varphi_X(t) = \mathrm{e}^{\mathrm{i} t \alpha} \Gamma(1+\mathrm{i}t \beta)  \text{.}  $$
The $n^\text{th}$ (noncentral) moment can be found from the characteristic function using
$$  E[X^n] = \mathrm{i}^{-n} \left. \frac{\mathrm{d}^n}{\mathrm{d}t^n} \varphi_X(t) \right|_{t = 0}  \text{.}  $$
So, \begin{align*}
E[X^0] &= 1 \cdot \mathrm{e}^0 \Gamma(1) = 1  \\
E[X^1] &= -\mathrm{i} \cdot \left( \mathrm{e}^{\mathrm{i} \cdot 0 \cdot \alpha} \Gamma'(1 + 0)(\mathrm{i}\beta) + \mathrm{i}\alpha \mathrm{e}^{\mathrm{i}\cdot 0 \cdot \alpha} \Gamma(1+0)\right)  \\
    &= - \gamma \beta + \alpha  \\
E[X^2] &= \cdots  \\
    &= \alpha ^2-2 \gamma  \alpha  \beta +\frac{1}{6} \left(6 \gamma ^2+\pi ^2\right) \beta ^2  \\
E[X^3] &= \cdots  \\
    &= \alpha ^3-3 \gamma  \alpha ^2 \beta +\frac{1}{2} \left(6 \gamma ^2+\pi ^2\right) \alpha  \beta ^2-\frac{1}{2} \beta ^3 \left(2 \gamma ^3+\gamma  \pi ^2-2 \psi^{(2)}(1)\right)  \\
E[X^4] &= \cdots  \\
    &= \alpha ^4-4 \gamma  \alpha ^3 \beta +\left(6 \gamma ^2+\pi ^2\right) \alpha ^2 \beta ^2-2 \alpha  \beta ^3 \left(2 \gamma ^3+\gamma  \pi ^2-2 \psi^{(2)}(1)\right)+\frac{1}{20} \beta ^4 \left(20 \gamma ^4+20 \gamma^2 \pi^2 + 3 \pi^4 - 80 \gamma \psi^{(2)}(1)\right)  \text{,}
\end{align*}
where

*

*$\gamma$ is the Euler-Mascheroni constant and

*$\psi^{(2)}$ is the polygamma function of order $2$.

Now we compute the central moments from the noncentral moments.  The $n^\text{th}$ central moment, $\mu_n$ is
$$  \mu_n = \begin{cases}
1 ,& n = 0  \\
E[X^1] ,& n = 1  \\
\sum_{j=0}^n \binom{n}{j} (-1)^{n-j} E[X^j] E[X^1]^{n-j}  ,&  n \geq 2
\end{cases}  \text{.}  $$
So,
\begin{align*}
\mu_1 &= E[X^1]  \\
    &= -\gamma \beta + \alpha  \\
\mu_2 &= \alpha ^2-2 \gamma  \alpha  \beta +(i \alpha -i \gamma  \beta )^2+\frac{\pi^2 \beta ^2}{6}+\gamma ^2 \beta ^2  \\
    &= \frac{\pi^2 \beta^2}{6}  \\
\mu_3 &= 3 i \left(\alpha ^2-2 \gamma  \alpha  \beta +\frac{\pi ^2 \beta ^2}{6}+\gamma ^2 \beta ^2\right) (i \alpha -i \gamma  \beta )+i \left(-i \alpha ^3+3 i \gamma  \alpha ^2 \beta +3 i \alpha  \left(-\frac{1}{6} \pi ^2 \beta ^2-\gamma ^2 \beta ^2\right)+\frac{1}{2} i \gamma  \pi ^2 \beta ^3+i \gamma ^3 \beta ^3-i \beta ^3 \psi ^{(2)}(1)\right)+2 i (i \alpha -i \gamma  \beta )^3  \\
    &= \beta ^3 \psi ^{(2)}(1)
\end{align*}
and you should see how to compute $\mu_4$.
