Give the value of $\nu_{\sigma}$.

We assume that, for all non-zero natural integer $n$, $X$ admits a moment of order $n$ and that the power series $\sum _ { n \geqslant 0 } m _ { n } ( X ) \frac { t ^ { n } } { n ! }$ has a radius of convergence $R _ { X } > 0$. For all $t \in ] - R _ { X } , R _ { X } [$, we denote $M _ { X } ( t ) = \sum _ { n = 0 } ^ { + \infty } m _ { n } ( X ) \frac { t ^ { n } } { n ! }$.
Justify that knowledge of the function $M _ { X }$ allows us to determine uniquely the sequence $\left( m _ { n } ( X ) \right) _ { n \in \mathbb { N } ^ { * } }$.

Let $t$ be the function defined on $I = ]-\pi/2, \pi/2[$ by $t(x) = \tan(x)$. For every natural integer $n$, express $t^{(n)}(0)$ as a function of the reals $(\alpha_i)_{i \in \mathbb{N}}$.

We define a sequence $(a_n)_{n \geqslant 1}$ by setting $$\forall n \in \mathbb{N}^*, \quad a_n = \frac{(-n)^{n-1}}{n!}.$$ We define, when possible, $S(x) = \sum_{n=1}^{+\infty} a_n x^n$, with radius of convergence $R$. Justify that the function $S$ is of class $\mathcal{C}^\infty$ on $]-R, R[$ and, for every integer $n \in \mathbb{N}$, express $S^{(n)}(0)$ as a function of $n$.

For every integer $k \geqslant 2$, we set $\zeta(k) = \sum_{n=1}^{+\infty} n^{-k}$.
Let $h$ be the function from $\mathbb{R}$ to $\mathbb{R}$ defined by $$\forall x \in \mathbb{R}, \quad h(x) = \begin{cases} \frac{x}{e^x - 1} & \text{if } x \neq 0 \\ 1 & \text{if } x = 0 \end{cases}$$
Show that the function $h$ is of class $C^\infty$ on $\mathbb{R}$ and that, for all $n \in \mathbb{N}^*$, we have $$h^{(2n)}(0) = \frac{(-1)^{n-1}(2n)!}{\pi^{2n} 2^{2n-1}} \zeta(2n).$$

For every integer $k \geqslant 2$, we set $\zeta(k) = \sum_{n=1}^{+\infty} n^{-k}$. Let $h$ be the function from $\mathbb{R}$ to $\mathbb{R}$ defined by $$\forall x \in \mathbb{R}, \quad h(x) = \begin{cases} \frac{x}{e^x - 1} & \text{if } x \neq 0 \\ 1 & \text{if } x = 0 \end{cases}$$ Show that the function $h$ is of class $C^\infty$ on $\mathbb{R}$ and that, for all $n \in \mathbb{N}^*$, we have $$h^{(2n)}(0) = \frac{(-1)^{n-1}(2n)!}{\pi^{2n} 2^{2n-1}} \zeta(2n)$$

Starting from the relation $(1-x)s(x) = e^{-x}$ for $x \in ]-1,1[$, express $\frac{d_n}{n!}$ for natural number $n$, in the form of a sum.

We denote $\omega(t) = e^{2i\pi t}$ for $t \in [0,1]$, and for all $p \in \mathbb{Z}$, $$I_{p} = \int_{0}^{1} \frac{\omega(t)^{p+1}}{\mathrm{e}^{\omega(t)} - 1} \,\mathrm{d}t.$$ Show that $I_{0} = 1$ and that, for all $p \in \mathbb{N}^{*}$, $I_{p} = 0$.

Let $\log x = g(x) = x f(x)$. Find $f^{(n)}(1)$, the $n$-th derivative of $f$ evaluated at $x = 1$.