Deduce that $$\int _ { 0 } ^ { 2 \pi } \ln \left( x ^ { 2 } - 2 x \cos \theta + 1 \right) d \theta = \begin{cases} 4 \pi \ln ( | x | ) & \text { if } | x | > 1 \\ 0 & \text { if } | x | < 1 \end{cases}$$

We denote $B$ the function defined on $\mathbb { R } ^ { + * }$ by $B ( x ) = \frac { \partial ^ { 2 } \beta } { \partial y ^ { 2 } } ( x , 1 )$, where $\beta ( x , y ) = \int _ { 0 } ^ { 1 } t ^ { x - 1 } ( 1 - t ) ^ { y - 1 } \mathrm {~d} t$.
Show that for every real $x > 0 , B ( x ) = \int _ { 0 } ^ { 1 } ( \ln ( 1 - t ) ) ^ { 2 } t ^ { x - 1 } \mathrm {~d} t$.

We consider a function $f : ] 0 , + \infty [ \rightarrow \mathbb { R }$ continuous piecewise satisfying the two following properties: (a) there exist an integer $K \geqslant 0$ and a real $C > 0$ such that $| f ( t ) | \leqslant C t ^ { K }$ on $[ 1 , + \infty [$, (b) there exist an integer $N \geqslant 0$, two reals $\lambda > 0$ and $\mu > 0$ and reals $a _ { 0 } , \ldots , a _ { N }$ such that $$f ( t ) = \sum _ { k = 0 } ^ { N } a _ { k } t ^ { ( k + \lambda - \mu ) / \mu } + o \left( t ^ { ( N + \lambda - \mu ) / \mu } \right) \quad \text { when } t \rightarrow 0 .$$
We denote $F$ the function defined by: $$F ( x ) = \int _ { 0 } ^ { + \infty } e ^ { - t / x } f ( t ) d t$$
Show that $F$ is well defined on $] 0 , + \infty [$ and that it satisfies the following asymptotic formula: $$F ( x ) = \sum _ { k = 0 } ^ { N } a _ { k } \Gamma \left( \frac { k + \lambda } { \mu } \right) x ^ { ( k + \lambda ) / \mu } + o \left( x ^ { ( N + \lambda ) / \mu } \right) \quad \text { when } x \rightarrow 0 ^ { + } .$$

For $x \in \mathbb{R}$, $\Gamma(x) = \int_{0}^{+\infty} t^{x-1} \mathrm{e}^{-t} \mathrm{~d}t$. Show the existence of the two integrals $\int_{0}^{+\infty} e^{-t^{2}} \mathrm{~d}t$ and $\int_{0}^{+\infty} e^{-t^{4}} \mathrm{~d}t$ and express them using $\Gamma$.

For $x \in \mathbb{R}^{+}$, we define $$f(x) = \int_{0}^{\infty} \frac{1 - \cos t}{t^{2}} \mathrm{e}^{-xt} \mathrm{~d}t$$ Show $$\forall s \in \mathbb{R}, \quad |s| = \frac{2}{\pi} \int_{0}^{\infty} \frac{1 - \cos(st)}{t^{2}} \mathrm{~d}t$$

Let $\Gamma$ be the Gamma function defined as the pointwise limit on $]0, +\infty[$ of $\Gamma_n(x) = \frac{1}{x} e^{-\gamma x} \prod_{k=1}^{n} \frac{e^{x k^{-1}}}{1 + x k^{-1}}$.
Show that for all $a \in ]0, +\infty[$ and $x \in ]0, +\infty[$: $$\int_0^{+\infty} \frac{t^{x-1}}{(1+t)^{x+a}} dt = \frac{\Gamma(x)\Gamma(a)}{\Gamma(x+a)}.$$ Hint: you may set, for $x \in ]0, +\infty[$, $f(x) = \frac{\Gamma(x+a)}{\Gamma(a)} \int_0^{+\infty} \frac{t^{x-1}}{(1+t)^{x+a}} dt$.

Let $\Gamma$ be the Gamma function. Show that for all $x \in ]0,1[$: $$\int_0^{+\infty} \frac{t^{x-1}}{1+t} dt = \frac{\pi}{\sin(\pi x)}.$$

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 $]0;1[$. Show that for all $\theta \in ] 0 ; \pi [$,
$$g ( \theta ) \sin ( x \theta ) = \frac { 1 } { 2 \mathrm { i } } \left( g ( - \theta ) e ^ { \mathrm { i } x \theta } - g ( \theta ) e ^ { - \mathrm { i } x \theta } \right) = \sin ( \theta ) \int _ { 0 } ^ { + \infty } \frac { t ^ { x } } { t ^ { 2 } + 2 t \cos ( \theta ) + 1 } \mathrm {~d} t$$

Show that for all $s \in \mathbf { R }$
$$\int _ { 0 } ^ { + \infty } \frac { 1 - \cos ( s t ) } { t ^ { 2 } } \mathrm {~d} t = \frac { \pi } { 2 } | s |$$

Prove that $$\int _ { 0 } ^ { + \infty } \frac { \sin ( x ) } { x } d x = \frac { \pi } { 2 }$$