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..
 A: 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 $\mu_4$.
