Evaluate $\int_{0}^{\infty} \left(1 + x^{2}\right)^{-(m+1)} dx$, where $m$ is a natural number.

The Euler Gamma function is defined, for all real $x > 0$, by: $$\Gamma(x) = \int_{0}^{+\infty} e^{-t} t^{x-1} dt$$ Calculate $\Gamma(n)$ for all natural integers $n$, $n \geqslant 1$.

We define for all real $x > 0$ the sequence $(J_{n}(x))_{n \geqslant 0}$ by: $$J_{n}(x) = \int_{0}^{1} (1-t)^{n} t^{x-1} dt$$ Deduce that, for all $x > 0$, $$J_{n}(x) = \frac{n!}{x(x+1) \cdots (x+n-1)(x+n)}$$

Application. For all $n \in \mathbb { N } ^ { * }$, we denote $\Gamma ( n ) = \int _ { 0 } ^ { + \infty } x ^ { n - 1 } e ^ { - x } d x$.
(a) Calculate $\Gamma ( n )$ for all $n \in \mathbb { N } ^ { * }$. One will use induction.
(b) Deduce the following asymptotic equivalent $$n ! \underset { n \rightarrow + \infty } { \sim } \sqrt { 2 \pi } n ^ { n + 1 / 2 } e ^ { - n }$$ Hint. First rewrite $\Gamma ( n + 1 )$ in the form $$\Gamma ( n + 1 ) = n ^ { n + 1 } \int _ { 0 } ^ { + \infty } e ^ { - n ( x - \ln x ) } d x$$

For $( x , y )$ in $\left( \mathbb { R } ^ { + * } \right) ^ { 2 }$, we define $\beta ( x , y ) = \int _ { 0 } ^ { 1 } t ^ { x - 1 } ( 1 - t ) ^ { y - 1 } \mathrm {~d} t$.
Deduce that for $x > 0 , y > 0 , \beta ( x + 1 , y + 1 ) = \frac { x y } { ( x + y ) ( x + y + 1 ) } \beta ( x , y )$.

For $( x , y )$ in $\left( \mathbb { R } ^ { + * } \right) ^ { 2 }$, we define $\beta ( x , y ) = \int _ { 0 } ^ { 1 } t ^ { x - 1 } ( 1 - t ) ^ { y - 1 } \mathrm {~d} t$. We want to show that for $x > 0$ and $y > 0$, $$\beta ( x , y ) = \frac { \Gamma ( x ) \Gamma ( y ) } { \Gamma ( x + y ) }$$ which will be denoted $(\mathcal{R})$. Throughout the rest of this question we assume $x > 1$ and $y > 1$. We denote $G ( a ) = \int _ { 0 } ^ { + \infty } \frac { u ^ { x - 1 } } { ( 1 + u ) ^ { x + y } } F _ { x , y } ( ( 1 + u ) a ) \mathrm { d } u$.
Deduce from the above the relation $(\mathcal{R})$.

Throughout the problem, we denote for every integer $n \geqslant 1$, $H _ { n } = \sum _ { k = 1 } ^ { n } \frac { 1 } { k }$. We define $\psi ( x ) = \frac { \Gamma ^ { \prime } ( x ) } { \Gamma ( x ) }$ on $\mathbb{R}^{+*}$, satisfying $\psi ( x + 1 ) - \psi ( x ) = \frac{1}{x}$ for all $x > 0$.
Show that for every real $x > - 1$ and for every integer $n \geqslant 1$ $$\psi ( 1 + x ) - \psi ( 1 ) = \psi ( n + x + 1 ) - \psi ( n + 1 ) + \sum _ { k = 1 } ^ { n } \left( \frac { 1 } { k } - \frac { 1 } { k + x } \right)$$

We define $\psi ( x ) = \frac { \Gamma ^ { \prime } ( x ) } { \Gamma ( x ) }$ on $\mathbb{R}^{+*}$, satisfying $\psi ( x + 1 ) - \psi ( x ) = \frac{1}{x}$ for all $x > 0$.
Deduce that, for every real $x > - 1$, $$\psi ( 1 + x ) = \psi ( 1 ) + \sum _ { n = 1 } ^ { + \infty } \left( \frac { 1 } { n } - \frac { 1 } { n + x } \right)$$

We denote $S _ { r } = \sum _ { n = 1 } ^ { + \infty } \frac { H _ { n } } { ( n + 1 ) ^ { r } }$ for $r \geqslant 2$, $\zeta ( x ) = \sum _ { n = 1 } ^ { + \infty } \frac { 1 } { n ^ { x } }$ for $x > 1$, and $B ( x ) = \int _ { 0 } ^ { 1 } ( \ln ( 1 - t ) ) ^ { 2 } t ^ { x - 1 } \mathrm {~d} t$. We have $S _ { r } = \frac { ( - 1 ) ^ { r } } { 2 ( r - 2 ) ! } \lim _ { x \rightarrow 0 ^ { + } } B ^ { ( r - 2 ) } ( x )$.
Find again the value of $S _ { 2 }$ already calculated in I.F.3.

We consider the function $F : ] 0 , + \infty [ \rightarrow \mathbb { R }$ defined by $F ( x ) = \int _ { 1 } ^ { + \infty } e ^ { - t / x } t ^ { - 1 } d t$.
For $N \in \mathbb { N } ^ { * }$ and $x > 0$, we set $$\begin{aligned} r _ { N } ( x ) & = ( - 1 ) ^ { N } N ! x ^ { N + 1 } e ^ { - 1 / x } \\ S _ { N } ( x ) & = \sum _ { k = 1 } ^ { N } ( - 1 ) ^ { k - 1 } ( k - 1 ) ! x ^ { k } e ^ { - 1 / x } \\ R _ { N } ( x ) & = ( - 1 ) ^ { N } N ! x ^ { N } \int _ { 1 } ^ { + \infty } e ^ { - t / x } t ^ { - ( N + 1 ) } d t \end{aligned}$$
Show that, for all $N \geqslant 1$ and all $x > 0 , F ( x ) = S _ { N } ( x ) + R _ { N } ( x )$.

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 $$\left\{ \begin{array}{l} \forall x > 0, \quad f(x) = x \ln(x) - \frac{1}{2} x \ln\left(x^{2} + 1\right) - \arctan(x) + \frac{\pi}{2} \\ f(0) = \frac{\pi}{2} \end{array} \right.$$

We denote $M_{p} = \int_{-\infty}^{+\infty} x^{2p} \exp\left(-x^{2}\right) \mathrm{d}x$. For $p$ a natural integer, give a relation between $M_{p+1}$ and $M_{p}$ and deduce that, for all $p \in \mathbb{N}$, $$M_{p} = \frac{\sqrt{\pi}(2p)!}{2^{2p} p!}$$

For $n \in \mathbb{N}$, we set $$I_{n} = \int_{0}^{+\infty} x^{n} e^{-x}\, dx.$$ Determine by induction $I_{n}$ for all $n \in \mathbb{N}$.

We define, for all $x \in \mathbb{R}^{+*}$, $$\Gamma(x) = \int_0^{+\infty} t^{x-1} \mathrm{e}^{-t} \, \mathrm{d}t$$ Determine the value of $\Gamma(n)$, for $n \in \mathbb{N}^*$.

Throughout this problem, $I = ]-1, +\infty[$, and $f(x) = \int_0^{\pi/2} (\sin(t))^x \mathrm{~d}t$.
Show that for every natural number $n$, $$f(n)f(n+1) = \frac{\pi}{2(n+1)}$$ then that: $$f(x) \underset{x \to +\infty}{\sim} \sqrt{\frac{\pi}{2x}}$$

Using the recurrence relation $K _ { n } = K _ { n + 1 } + \frac { 1 } { 2 n } K _ { n }$ and the fact that $I_n \sim K_n$, deduce a simple equivalent of $I _ { n }$ as $n$ tends to $+ \infty$, where $I _ { n } = \int _ { 0 } ^ { 1 } \frac { 1 } { \left( 1 + t ^ { 2 } \right) ^ { n } } \mathrm {~d} t$.

Deduce that:
$$\int _ { 0 } ^ { + \infty } \frac { 1 - ( \cos ( t ) ) ^ { 2 p + 1 } } { t ^ { 2 } } \mathrm {~d} t = \frac { \pi } { 2 } \frac { ( 2 p + 1 ) ! } { 2 ^ { 2 p } \cdot ( p ! ) ^ { 2 } }$$

Given the function in $x$
$$f _ { n } ( x ) = \sin ^ { n } x \quad ( n = 1,2,3 , \cdots ) ,$$
answer the following questions.
(1) Consider the cases in which the equality
$$\lim _ { x \rightarrow 0 } \frac { a - x ^ { 2 } - \left( b - x ^ { 2 } \right) ^ { 2 } } { f _ { n } ( x ) } = c$$
holds for three real numbers $a , b$ and $c$.
(i) We have $a = b$.
(ii) When $n = 2$, if $c = 6$, then $b = \frac { \mathbf { P } } { \mathbf { Q } }$.
(iii) When $n = 4$, then $b = \frac { \mathbf { R } } { \mathbf { S } }$ and $c = - \mathbf { T }$.
(2) For this $f _ { n } ( x )$, consider the definite integral
$$I _ { n } = \int _ { 0 } ^ { \frac { \pi } { 2 } } f _ { n } ( x ) \sin 2 x \, d x \quad ( n = 1,2,3 , \cdots )$$
When the integral is calculated, we have
$$I _ { n } = \frac { \mathbf { U } } { n + \mathbf { V } } .$$
Hence we obtain
$$\lim _ { n \rightarrow \infty } \left( I _ { n - 1 } + I _ { n } + I _ { n + 1 } + \cdots + I _ { 2 n - 2 } \right) = \int _ { 0 } ^ { \mathbf { W } } \frac { \mathbf { X } } { \mathbf { Y } + x } \, dx = \log \mathbf { Z }$$

For every integer $m$ greater than 1
$$\int \tan ^ { m } x d x = \frac { 1 } { m - 1 } \tan ^ { m - 1 } x - \int \tan ^ { m - 2 } x d x$$
the equality is satisfied. Accordingly, what is the value of the integral $\int _ { 0 } ^ { \frac { \pi } { 4 } } \tan ^ { 4 } \mathrm { xdx }$?
A) $\frac { 2 \pi + 3 } { 4 }$
B) $\frac { 4 \pi - 3 } { 8 }$
C) $\frac { 3 \pi - 8 } { 12 }$
D) $\pi + 2$
E) $2 \pi + 1$