We denote for $x \in \mathbb{R}_{+}$ and $m \in \mathbb{R}$ $$T_{m}(x) = \int_{0}^{\infty} t^{m} e^{-\left(t^{2}+x/t\right)} dt$$
a) Let $x \in \mathbb{R}_{+}^{*}$ and $m \in \mathbb{R}$. Perform the change of variable $t = 1/u$ in the integral defining $T_{m}$. Justify the calculation carefully.
b) Let $n \in \mathbb{N}$. Justify the existence of the quantity $\int_{0}^{\infty} u^{n} e^{-u} du$ and calculate it.
c) Show that for $m \in \mathbb{N} - \{0,1\}$, $T_{-m}(1) \leq (m-2)!$.
d) Let $k \in \mathbb{N}$. Show that the radius of convergence $R$ of the power series $\sum_{n=0}^{\infty} \frac{(-1)^{n}}{n!} T_{k-n}(1) x^{n}$ satisfies $R \geq 1$.
e) Let $k \in \mathbb{N}$. Show that for $x \in ]-1,1[$, $$T_{k}(1+x) = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{n!} T_{k-n}(1) x^{n}$$

Show that for all $y > 0$, we have $\Gamma ( y ) = y ^ { - 1 } \int _ { 0 } ^ { + \infty } e ^ { - t } t ^ { y } d t$, then that
$$\Gamma ( y ) = e ^ { - y } y ^ { y } \int _ { - 1 } ^ { + \infty } e ^ { - y \phi ( s ) } d s$$
where $\phi$ is the function defined on $] - 1 , + \infty [$ by $\phi ( s ) = s - \ln ( 1 + s )$.
Recall that $\Gamma : ] 0 , + \infty [ \rightarrow \mathbb { R }$ is defined by $\Gamma ( y ) = \int _ { 0 } ^ { \infty } e ^ { - t } t ^ { y - 1 } d t$.

We study the sequence $\left(u_{n}\right)_{n \in \mathbb{N}^{*}}$ defined by $$\forall n \in \mathbb{N}^{*}, \quad u_{n} = \int_{0}^{\infty} \frac{1 - (\cos t)^{n}}{t^{2}} \mathrm{~d}t$$ Show that $$\forall n \in \mathbb{N}^{*}, \quad u_{n} = \frac{\sqrt{n}}{2\sqrt{2}} v_{n} \quad \text{with} \quad v_{n} = \int_{0}^{\infty} \frac{1 - (\cos(\sqrt{2u/n}))^{n}}{u\sqrt{u}} \mathrm{~d}u$$

For $n \in \mathbb{N}$, we set $I_{n} = \int_{0}^{+\infty} x^{n} e^{-x}\, dx$.
Show that for $n \geqslant 1$, we have $$I_{n} = \left(\frac{n}{e}\right)^{n} \sqrt{n} \int_{-\sqrt{n}}^{+\infty} \left(1 + \frac{x}{\sqrt{n}}\right)^{n} e^{-x\sqrt{n}}\, dx.$$

Let $g$ be the function defined by
$$\begin{aligned} g : ] - \pi ; \pi [ & \longrightarrow \mathbf { C } \\ \theta & \longmapsto e ^ { \mathrm { i } x \theta } \int _ { 0 } ^ { + \infty } \frac { t ^ { x - 1 } } { 1 + t e ^ { \mathrm { i } \theta } } \mathrm {~d} t \end{aligned}$$
where $x$ is a fixed element of $]0;1[$. Deduce that:
$$\forall \theta \in ] 0 ; \pi \left[ , \quad g ( \theta ) \sin ( \theta x ) = \int _ { \cot ( \theta ) } ^ { + \infty } \frac { ( u \sin ( \theta ) - \cos ( \theta ) ) ^ { x } } { 1 + u ^ { 2 } } \mathrm {~d} u , \right.$$
where $\cot ( \theta ) = \frac { \cos ( \theta ) } { \sin ( \theta ) }$.