We define for all real $x > 0$ the sequence $(I_{n}(x))_{n \geqslant 1}$ by: $$I_{n}(x) = \int_{0}^{n} \left(1 - \frac{t}{n}\right)^{n} t^{x-1} dt$$ Show that, for all $x > 0$, $$\lim_{n \rightarrow +\infty} I_{n}(x) = \Gamma(x)$$

Two special cases. Let $d > 0$. Let $g \in \mathcal { C } ^ { 0 } ( [ 0 , d ] )$ such that $g ( 0 ) \neq 0$.
(a) Show that $$\int _ { 0 } ^ { d } e ^ { - t x } g ( x ) d x \underset { t \rightarrow + \infty } { \sim } \frac { g ( 0 ) } { t }$$ Hint. For $t > 0$, one can construct a function $g _ { t }$ piecewise continuous on $[ 0 , + \infty [$, bounded, such that $$\int _ { 0 } ^ { d } e ^ { - t x } g ( x ) d x = \frac { 1 } { t } \int _ { 0 } ^ { + \infty } e ^ { - x } g _ { t } ( x ) d x$$ (b) Show similarly that $$\int _ { 0 } ^ { d } e ^ { - t x ^ { 2 } } g ( x ) d x \underset { t \rightarrow + \infty } { \sim } \frac { \sqrt { \pi } } { 2 } \frac { g ( 0 ) } { \sqrt { t } }$$ Hint. We recall the equality $\int _ { 0 } ^ { + \infty } e ^ { - x ^ { 2 } } d x = \frac { \sqrt { \pi } } { 2 }$.

For $( x , y )$ in $\left( \mathbb { R } ^ { + * } \right) ^ { 2 }$, we define $\beta ( x , y ) = \int _ { 0 } ^ { 1 } t ^ { x - 1 } ( 1 - t ) ^ { y - 1 } \mathrm {~d} t$. Throughout the rest of this question we assume $x > 1$ and $y > 1$. We denote $F _ { x , y }$ the antiderivative on $\mathbb { R } ^ { + }$ of $t \mapsto e ^ { - t } t ^ { x + y - 1 }$ which vanishes at 0, and $G ( a ) = \int _ { 0 } ^ { + \infty } \frac { u ^ { x - 1 } } { ( 1 + u ) ^ { x + y } } F _ { x , y } ( ( 1 + u ) a ) \mathrm { d } u$.
Show that $\lim _ { a \rightarrow + \infty } G ( a ) = \Gamma ( x + y ) \beta ( x , y )$.

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 $\rho _ { N } ( t ) = f ( t ) - \sum _ { k = 0 } ^ { N } a _ { k } t ^ { ( k + \lambda - \mu ) / \mu }$ the remainder of the asymptotic expansion of $f$.
Deduce that $$\int _ { 0 } ^ { + \infty } e ^ { - t / x } \rho _ { N } ( t ) d t = o \left( x ^ { ( N + \lambda ) / \mu } \right) \quad \text { when } x \rightarrow 0 ^ { + }$$

We assume $\lambda < 1$. Determine $\lim_{n \rightarrow +\infty} \left((n\lambda)^{-n} \int_{0}^{n\lambda} (n\lambda - t)^{n} \mathrm{e}^{t} \mathrm{~d}t\right)$.

We assume $\lambda < 1$, and $D_{n} = \sum_{k=n+1}^{+\infty} \frac{(n\lambda)^{k}}{k!}$. Using Taylor's formula with integral remainder, deduce an equivalent of $D_{n}$ when $n \rightarrow +\infty$.

For $x \in \mathbb{R}^{+}$, we define $$f(x) = \int_{0}^{\infty} \frac{1 - \cos t}{t^{2}} \mathrm{e}^{-xt} \mathrm{~d}t$$ Determine the limits of $f$ and $f^{\prime}$ at $+\infty$.

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

We study the sequence $\left(u_{n}\right)_{n \in \mathbb{N}^{*}}$ defined by $$\forall n \in \mathbb{N}^{*}, \quad u_{n} = \int_{0}^{\infty} \frac{1 - (\cos t)^{n}}{t^{2}} \mathrm{~d}t$$ We admit the relation $\int_{0}^{\infty} \frac{\mathrm{e}^{-u}}{\sqrt{u}} \mathrm{~d}u = \sqrt{\pi}$.
Conclude that $u_{n} \sim \sqrt{\frac{n\pi}{2}}$.

For every natural integer $n$ and every real $x$, set $I_n(x) = \int_0^{\pi/2} \cos(2xt)(\cos t)^n \, \mathrm{d}t$. Using the inequality $t\cos(t) \leqslant \sin(t)$ for $t \in [0, \pi/2]$, deduce $$\forall n \in \mathbb{N}^{\star}, \forall x \in [0,1], \quad 0 \leqslant -I_n^{\prime}(x) \leqslant \frac{4x}{n} I_n(x)$$ then, for $x \in [0,1]$, the limit $\lim_{n \rightarrow +\infty} \frac{I_n^{\prime}(x)}{I_n(x)}$.

Using the results of Q26 and Q28, deduce the equality $$\forall x \in J, \quad \pi \tan(\pi x) = \sum_{p=1}^{+\infty} 2\left(2^{2p} - 1\right) \zeta(2p) x^{2p-1}$$

For $k \in \mathbf { N } ^ { * }$ and $t \in \mathbf { R } _ { + }$, we set
$$u _ { k } ( t ) = \int _ { k/2 } ^ { ( k + 1 ) / 2 } \frac { t q ( u ) } { e ^ { t u } - 1 } \mathrm {~d} u \quad \text { if } t > 0 , \quad \text { and } \quad u _ { k } ( t ) = \int _ { k/2 } ^ { ( k + 1 ) / 2 } \frac { q ( u ) } { u } \mathrm {~d} u \quad \text { if } t = 0$$
where $q ( x ) = x - \lfloor x \rfloor - \frac { 1 } { 2 }$.
Deduce that
$$\int _ { 1 } ^ { + \infty } \frac { t q ( u ) } { e ^ { t u } - 1 } \mathrm {~d} u \underset { t \rightarrow 0 ^ { + } } { \longrightarrow } \frac { \ln ( 2 \pi ) } { 2 } - 1 .$$

For $k \in \mathbf{N}^*$ and $t \in \mathbf{R}_+$, we set $$u_k(t) = \int_{k/2}^{(k+1)/2} \frac{tq(u)}{e^{tu}-1} \mathrm{du} \text{ if } t > 0 \text{, and } u_k(t) = \int_{k/2}^{(k+1)/2} \frac{q(u)}{u} \mathrm{du} \text{ if } t = 0.$$ We admit that the bound $\left|\sum_{k=n}^{+\infty} u_k(t)\right| \leq \frac{1}{2n}$ holds for $t = 0$.
Deduce that $$\int_{1}^{+\infty} \frac{tq(u)}{e^{tu}-1} \mathrm{~d}u \underset{t \rightarrow 0^+}{\longrightarrow} \frac{\ln(2\pi)}{2} - 1$$

If $n \in \mathbf{N}$, we denote by $D_n$ the improper integral $\int_0^{\pi/2} (\ln(\sin(t)))^n \mathrm{~d}t$.
Verify that if $n \in \mathbf{N}^*$, then $$(-1)^n D_n = \int_0^{+\infty} \frac{u^n}{\sqrt{\mathrm{e}^{2u} - 1}} \mathrm{~d}u$$ then that $$D_n \underset{n \to +\infty}{\sim} (-1)^n n!$$

Deduce that, as $n$ tends to $+ \infty$, $$I _ { n } \sim K _ { n } .$$ where $I _ { n } = \int _ { 0 } ^ { 1 } \frac { 1 } { \left( 1 + t ^ { 2 } \right) ^ { n } } \mathrm {~d} t$ and $K _ { n } = \int _ { 0 } ^ { + \infty } \frac { 1 } { \left( 1 + t ^ { 2 } \right) ^ { n } } \mathrm {~d} t$.