We denote for $x \in \mathbb{R}_{+}$ and $m \in \mathbb{R}$ $$T_{m}(x) = \int_{0}^{\infty} t^{m} e^{-\left(t^{2}+x/t\right)} dt$$
Let $L > 0$, $\rho \in C([0,L])$, and $g_{0}$ continuous and integrable on $\mathbb{R}_{+}$. We set for $x \in [0,L]$ and $v \in \mathbb{R}^{*}$: $$\begin{gathered} g(x,v) = \frac{2}{\sqrt{\pi}} \frac{e^{-v^{2}}}{|v|} \int_{x}^{L} \rho(y) e^{-\frac{x-y}{v}} dy, \quad \text{if} \quad v < 0, \\ g(x,v) = \frac{2}{\sqrt{\pi}} \frac{e^{-v^{2}}}{|v|} \int_{0}^{x} \rho(y) e^{-\frac{x-y}{v}} dy + g_{0}(v) e^{-\frac{x}{v}}, \quad \text{if} \quad v > 0. \end{gathered}$$
a) Show that $\alpha : x \in [0,L] \mapsto \int_{0}^{\infty} g_{0}(v) e^{-\frac{x}{v}} dv$ defines a function in $C([0,L])$.
b) Show that for $v \in \mathbb{R}^{*}$, the function $x \in [0,L] \mapsto g(x,v)$ is of class $C^{1}$ on $[0,L]$ and $$\begin{aligned} & \forall x \in [0,L], v \in \mathbb{R}^{*}, \quad v \frac{\partial g}{\partial x}(x,v) = \rho(x) \frac{2}{\sqrt{\pi}} e^{-v^{2}} - g(x,v) \\ & \forall v \in \mathbb{R}_{+}^{*}, \quad g(0,v) = g_{0}(v), \quad \forall v \in \mathbb{R}_{-}^{*}, \quad g(L,v) = 0 \end{aligned}$$

We denote for $x \in \mathbb{R}_{+}$ and $m \in \mathbb{R}$ $$T_{m}(x) = \int_{0}^{\infty} t^{m} e^{-\left(t^{2}+x/t\right)} dt$$
We admit in this question the following theorem: Let $L > 0$. If $H$ is a mapping from $C([0,L])$ to $C([0,L])$ satisfying $$\exists k \in [0,1[, \quad \forall \rho_{1}, \rho_{2} \in C([0,L]), \quad \|H(\rho_{1}) - H(\rho_{2})\|_{\infty} \leq k \|\rho_{1} - \rho_{2}\|_{\infty}$$ then there exists a unique $\rho \in C([0,L])$ such that $H(\rho) = \rho$.
a) Let $L > 0$ and $\tilde{\alpha} \in C([0,L])$. Show that if $L \in ]0, 1/20[$, there exists a unique function $\tilde{\rho} \in C([0,L])$ such that $$\forall x \in [0,L], \quad \tilde{\rho}(x) = \tilde{\alpha}(x) + \frac{2}{\sqrt{\pi}} \int_{0}^{L} \tilde{\rho}(y) T_{-1}(|x-y|) dy$$
b) Let $L \in ]0, 1/20[$, and $g_{0}$ continuous and integrable on $\mathbb{R}_{+}$. Show that there exists a function $\tilde{g}$ from $[0,L] \times \mathbb{R}$ to $\mathbb{R}$ such that

• $\forall v \in \mathbb{R}^{*}$, $\tilde{g}(\cdot, v)$ is of class $C^{1}$ on $[0,L]$
• $\forall x \in [0,L]$, $\tilde{g}(x, \cdot)$ is integrable on $\mathbb{R}_{+}^{*}$ and $\mathbb{R}_{-}^{*}$
• $\forall x \in [0,L], v \in \mathbb{R}^{*}$, $v \frac{\partial \tilde{g}}{\partial x}(x,v) = \left(\int_{\mathbb{R}_{+}^{*}} \tilde{g}(x,w) dw + \int_{\mathbb{R}_{-}^{*}} \tilde{g}(x,w) dw\right) \frac{2}{\sqrt{\pi}} e^{-v^{2}} - \tilde{g}(x,v)$
• $\forall v \in \mathbb{R}_{+}^{*}$, $\tilde{g}(0,v) = g_{0}(v)$, $\quad \forall v \in \mathbb{R}_{-}^{*}$, $\tilde{g}(L,v) = 0$

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 $\int _ { 0 } ^ { + \infty } e ^ { - x ^ { 2 } } d x = \frac { \sqrt { \pi } } { 2 }$.

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 $$\Gamma ( n + 1 ) = n ^ { n + 1 } \int _ { 0 } ^ { + \infty } e ^ { - n ( x - \ln x ) } d x$$

Throughout the problem, we denote for every integer $n \geqslant 1$, $H _ { n } = \sum _ { k = 1 } ^ { n } \frac { 1 } { k }$ and $S _ { r } = \sum _ { n = 1 } ^ { + \infty } \frac { H _ { n } } { ( n + 1 ) ^ { r } }$ when the series converges.
Deduce from the previous questions that for every integer $r \geqslant 2$, $$S _ { r } = \sum _ { n = 1 } ^ { + \infty } \frac { H _ { n } } { ( n + 1 ) ^ { r } } = \frac { ( - 1 ) ^ { r } } { ( r - 1 ) ! } \int _ { 0 } ^ { 1 } ( \ln t ) ^ { r - 1 } \frac { \ln ( 1 - t ) } { 1 - t } \mathrm {~d} t$$

Throughout the problem, we denote for every integer $n \geqslant 1$, $H _ { n } = \sum _ { k = 1 } ^ { n } \frac { 1 } { k }$ and $S _ { r } = \sum _ { n = 1 } ^ { + \infty } \frac { H _ { n } } { ( n + 1 ) ^ { r } }$ when the series converges.
Establish that we then have $S _ { r } = \frac { ( - 1 ) ^ { r } } { 2 ( r - 2 ) ! } \int _ { 0 } ^ { 1 } \frac { ( \ln t ) ^ { r - 2 } ( \ln ( 1 - t ) ) ^ { 2 } } { t } \mathrm {~d} t$.

Throughout the problem, we denote for every integer $n \geqslant 1$, $H _ { n } = \sum _ { k = 1 } ^ { n } \frac { 1 } { k }$, $S _ { r } = \sum _ { n = 1 } ^ { + \infty } \frac { H _ { n } } { ( n + 1 ) ^ { r } }$ when the series converges, and $\zeta ( x ) = \sum _ { n = 1 } ^ { + \infty } \frac { 1 } { n ^ { x } }$ for $x > 1$.
Deduce that $S _ { 2 } = \frac { 1 } { 2 } \int _ { 0 } ^ { 1 } \frac { ( \ln t ) ^ { 2 } } { 1 - t } \mathrm {~d} t$ then find the value of $S _ { 2 }$ as a function of $\zeta ( 3 )$.

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

For $( x , y )$ in $\left( \mathbb { R } ^ { + * } \right) ^ { 2 }$, we define $\beta ( x , y ) = \int _ { 0 } ^ { 1 } t ^ { x - 1 } ( 1 - t ) ^ { y - 1 } \mathrm {~d} t$.
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 $(\mathcal{R})$.
Explain why it suffices to show the relation $(\mathcal{R})$ for $x > 1$ and $y > 1$.

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 $\Gamma ( x ) = \int _ { 0 } ^ { + \infty } t ^ { x - 1 } e ^ { - t } \mathrm {~d} t$.
We denote $F _ { x , y }$ the antiderivative on $\mathbb { R } ^ { + }$ of $t \mapsto e ^ { - t } t ^ { x + y - 1 }$ which vanishes at 0. Show that $$\forall t \in \mathbb { R } ^ { + } , F _ { x , y } ( t ) \leqslant \Gamma ( x + y )$$

For $( x , y )$ in $\left( \mathbb { R } ^ { + * } \right) ^ { 2 }$, we define $\beta ( x , y ) = \int _ { 0 } ^ { 1 } t ^ { x - 1 } ( 1 - t ) ^ { y - 1 } \mathrm {~d} t$. 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.
Let $G ( a ) = \int _ { 0 } ^ { + \infty } \frac { u ^ { x - 1 } } { ( 1 + u ) ^ { x + y } } F _ { x , y } ( ( 1 + u ) a ) \mathrm { d } u$.
Show that $G$ is defined and continuous on $\mathbb { R } ^ { + }$.

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 $\lim _ { a \rightarrow + \infty } G ( a ) = \Gamma ( x + y ) \beta ( x , y )$.

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 $G$ is of class $\mathcal { C } ^ { 1 }$ on every segment $[ c , d ]$ included in $\mathbb { R } ^ { + * }$, then that $G$ is of class $\mathcal { C } ^ { 1 }$ on $\mathbb { R } ^ { + * }$.

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 $\Gamma ( x ) = \int _ { 0 } ^ { + \infty } t ^ { x - 1 } e ^ { - t } \mathrm {~d} t$, $F _ { x , y }$ the antiderivative on $\mathbb { R } ^ { + }$ of $t \mapsto e ^ { - t } t ^ { x + y - 1 }$ which vanishes at 0, and $G ( a ) = \int _ { 0 } ^ { + \infty } \frac { u ^ { x - 1 } } { ( 1 + u ) ^ { x + y } } F _ { x , y } ( ( 1 + u ) a ) \mathrm { d } u$.
Express for $a > 0$, $G ^ { \prime } ( a )$ as a function of $\Gamma ( x ) , e ^ { - a }$ and $a ^ { y - 1 }$.

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 $(\mathcal{R})$.

We define the function $\psi$ (called the digamma function) on $\mathbb { R } ^ { + * }$ as the derivative of $x \mapsto \ln ( \Gamma ( x ) )$. For every real $x > 0 , \psi ( x ) = \frac { \Gamma ^ { \prime } ( x ) } { \Gamma ( x ) }$. We admit that $\Gamma$ satisfies, for every real $x > 0$, the relation $\Gamma ( x + 1 ) = x \Gamma ( x )$.
Show that for every real $x > 0 , \psi ( x + 1 ) - \psi ( x ) = \frac { 1 } { x }$.

For $( x , y )$ in $\left( \mathbb { R } ^ { + * } \right) ^ { 2 }$, we define $\beta ( x , y ) = \int _ { 0 } ^ { 1 } t ^ { x - 1 } ( 1 - t ) ^ { y - 1 } \mathrm {~d} t$. We have the relation $(\mathcal{R})$: $\beta ( x , y ) = \frac { \Gamma ( x ) \Gamma ( y ) } { \Gamma ( x + y ) }$. We define $\psi ( x ) = \frac { \Gamma ^ { \prime } ( x ) } { \Gamma ( x ) }$.
From the relation $(\mathcal{R})$, justify that $\frac { \partial \beta } { \partial y }$ is defined on $\left( \mathbb { R } ^ { + * } \right) ^ { 2 }$.
Establish that for all real $x > 0$ and $y > 0 , \frac { \partial \beta } { \partial y } ( x , y ) = \beta ( x , y ) ( \psi ( y ) - \psi ( x + y ) )$.

For $( x , y )$ in $\left( \mathbb { R } ^ { + * } \right) ^ { 2 }$, we define $\beta ( x , y ) = \int _ { 0 } ^ { 1 } t ^ { x - 1 } ( 1 - t ) ^ { y - 1 } \mathrm {~d} t$.
Let $x > 0$ be fixed. What is the monotonicity on $\mathbb { R } ^ { + * }$ of the function $y \mapsto \beta ( x , y )$?

We define $\psi ( x ) = \frac { \Gamma ^ { \prime } ( x ) } { \Gamma ( x ) }$ on $\mathbb{R}^{+*}$.
Show that the function $\psi$ is increasing on $\mathbb { R } ^ { + * }$.

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 $n \geqslant 1$ $$\psi ( 1 + x ) - \psi ( 1 ) = \psi ( n + x + 1 ) - \psi ( n + 1 ) + \sum _ { k = 1 } ^ { n } \left( \frac { 1 } { k } - \frac { 1 } { k + x } \right)$$

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 $\mathbb{R}^{+*}$, and $\psi$ is increasing on $\mathbb{R}^{+*}$.
Let $n$ be an integer $\geqslant 2$ and $x$ a real $> - 1$. We set $p = E ( x ) + 1$, where $E ( x )$ denotes the integer part of $x$. Prove that $$0 \leqslant \psi ( n + x + 1 ) - \psi ( n ) \leqslant H _ { n + p } - H _ { n - 1 } \leqslant \frac { p + 1 } { n }$$

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 $B$ the function defined on $\mathbb { R } ^ { + * }$ by $B ( x ) = \frac { \partial ^ { 2 } \beta } { \partial y ^ { 2 } } ( x , 1 )$, where $\beta ( x , y ) = \int _ { 0 } ^ { 1 } t ^ { x - 1 } ( 1 - t ) ^ { y - 1 } \mathrm {~d} t$ and $\psi ( x ) = \frac { \Gamma ^ { \prime } ( x ) } { \Gamma ( x ) }$. We have $\frac { \partial \beta } { \partial y } ( x , y ) = \beta ( x , y ) ( \psi ( y ) - \psi ( x + y ) )$.
Justify that $B$ is defined on $\mathbb { R } ^ { + * }$.
Using the relation found in III.B.1, establish that for every real $x > 0$ $$x B ( x ) = ( \psi ( 1 + x ) - \psi ( 1 ) ) ^ { 2 } + \left( \psi ^ { \prime } ( 1 ) - \psi ^ { \prime } ( 1 + x ) \right)$$
Deduce that $B$ is $\mathcal { C } ^ { \infty }$ on $\mathbb { R } ^ { + * }$.

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 $x > 0 , B ( x ) = \int _ { 0 } ^ { 1 } ( \ln ( 1 - t ) ) ^ { 2 } t ^ { x - 1 } \mathrm {~d} t$.