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 $y > 1$.
Show that $\beta ( x , y ) = \int _ { 0 } ^ { + \infty } \frac { u ^ { x - 1 } } { ( 1 + u ) ^ { x + y } } \mathrm {~d} u$.
One may use the change of variable $t = \frac { u } { 1 + u }$.

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 $x > 0 , \psi ( x + 1 ) - \psi ( x ) = \frac { 1 } { x }$.

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 $f$.
We fix $\delta > 0$ and $\alpha \in \mathbb { R }$. Show that for all $x > 0$, the function $t \mapsto e ^ { - t / x } t ^ { \alpha }$ is integrable on $[ \delta , + \infty [$ and that for all $n \in \mathbb { N }$, we have: $$\int _ { \delta } ^ { + \infty } e ^ { - t / x } t ^ { \alpha } d t = o \left( x ^ { n } \right) \quad \text { when } x \rightarrow 0 ^ { + }$$ Deduce that for all $n \in \mathbb { N }$, $$\int _ { \delta } ^ { + \infty } e ^ { - t / x } \rho _ { N } ( t ) d t = o \left( x ^ { n } \right) \quad \text { when } x \rightarrow 0 ^ { + }$$

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 $f$.
We fix $\varepsilon > 0$. Show the existence of $\delta > 0$ and a constant $C ^ { \prime }$ independent of $\varepsilon$ and $\delta$ such that $$\forall x > 0 , \quad \left| \int _ { 0 } ^ { \delta } e ^ { - t / x } \rho _ { N } ( t ) d t \right| \leqslant C ^ { \prime } \varepsilon x ^ { ( N + \lambda ) / \mu }$$

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 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: $$F ( x ) = \sum _ { k = 0 } ^ { N } a _ { k } \Gamma \left( \frac { k + \lambda } { \mu } \right) x ^ { ( k + \lambda ) / \mu } + o \left( x ^ { ( N + \lambda ) / \mu } \right) \quad \text { when } x \rightarrow 0 ^ { + } .$$

Justify that, if $\varphi$ is a $\mathcal{C}^1$ application from $[0, \pi]$ to $\mathbf{R}$, then $$\lim_{x \to +\infty} \int_0^{\pi} \varphi(t) \sin(xt) \mathrm{d}t = 0$$ and conclude that $$\sigma(1) = \frac{\pi^2}{6}$$

Let $f : \mathbb { R } _ { + } \rightarrow \mathbb { R }$, be a piecewise continuous function, strictly positive and integrable. For all $x > 0$, we define
$$g ( x ) = \frac { 1 } { x } \int _ { 0 } ^ { x } t f ( t ) \mathrm { d } t \quad \text { and } \quad h ( x ) = \frac { 1 } { x } g ( x ) = \frac { 1 } { x ^ { 2 } } \int _ { 0 } ^ { x } t f ( t ) \mathrm { d } t .$$
Deduce that the integral $\int _ { 0 } ^ { + \infty } h ( x ) \mathrm { d } x$ converges and that
$$\int _ { 0 } ^ { + \infty } f ( x ) \mathrm { d } x = \int _ { 0 } ^ { + \infty } h ( x ) \mathrm { d } x .$$
You may use integration by parts.

Let $f$ be a piecewise continuous function, strictly positive, integrable on $\mathbb { R } _ { + }$. Prove that, for all $x > 0$,
$$\exp \left( \frac { 1 } { x } \int _ { 0 } ^ { x } \ln ( f ( t ) ) \mathrm { d } t \right) \leqslant \frac { \mathrm { e } } { x ^ { 2 } } \int _ { 0 } ^ { x } t f ( t ) \mathrm { d } t$$
You may note that $\ln ( f ( t ) ) = \ln ( t f ( t ) ) - \ln ( t )$.

Let $f$ be a piecewise continuous function, strictly positive, integrable on $\mathbb { R } _ { + }$. Deduce that $x \mapsto \exp \left( \frac { 1 } { x } \int _ { 0 } ^ { x } \ln ( f ( t ) ) \mathrm { d } t \right)$ is integrable on $\mathbb { R } _ { + } ^ { * }$ and that
$$\int _ { 0 } ^ { + \infty } \exp \left( \frac { 1 } { x } \int _ { 0 } ^ { x } \ln ( f ( t ) ) \mathrm { d } t \right) \mathrm { d } x \leqslant \mathrm { e } \int _ { 0 } ^ { + \infty } f ( x ) \mathrm { d } x$$

We assume that $\left( a _ { n } \right) _ { n \in \mathbb { N } ^ { * } }$ is a decreasing sequence of strictly positive real numbers. We denote by $f$ the step function which, for all $k \in \mathbb { N } ^ { * }$, equals $a _ { k }$ on the interval $[ k - 1 , k [$.
Prove that, for all $k$ in $\mathbb { N } ^ { * }$,
$$\int _ { k - 1 } ^ { k } \exp \left( \frac { 1 } { x } \int _ { 0 } ^ { x } \ln ( f ( t ) ) \mathrm { d } t \right) \mathrm { d } x \geqslant \exp \left( \frac { 1 } { k } \sum _ { i = 1 } ^ { k } \ln \left( a _ { i } \right) \right)$$
You may use the previous question.

We assume that $\left( a _ { n } \right) _ { n \in \mathbb { N } ^ { * } }$ is a decreasing sequence of strictly positive real numbers. We denote by $f$ the step function which, for all $k \in \mathbb { N } ^ { * }$, equals $a _ { k }$ on the interval $[ k - 1 , k [$.
Deduce Carleman's inequality in the case where $\left( a _ { n } \right) _ { n \in \mathbb { N } ^ { * } }$ is a decreasing sequence.

Explain how one can remove the decreasing hypothesis in Carleman's inequality.

Let $n$ be in $\mathbb { N } ^ { * }$. We denote by $U _ { n }$ the open set $\left( \mathbb { R } _ { + } ^ { * } \right) ^ { n }$. Its closure, denoted $\overline { U _ { n } }$, is $\left( \mathbb { R } _ { + } \right) ^ { n }$.
Let $s > 0$. We define the functions $f$ and $g _ { s }$ on $\overline { U _ { n } }$ by setting, for all $x = \left( x _ { 1 } , \ldots , x _ { n } \right) \in \overline { U _ { n } }$,
$$f ( x ) = \prod _ { k = 1 } ^ { n } x _ { k } \quad \text { and } \quad g _ { s } ( x ) = \left( \sum _ { k = 1 } ^ { n } x _ { k } \right) - s .$$
We denote by $X _ { s }$ the subset of $\overline { U _ { n } }$ consisting of the zeros of $g _ { s } : X _ { s } = \left\{ x \in \overline { U _ { n } } \mid g _ { s } ( x ) = 0 \right\}$.
We admit that $f$ and $g _ { s }$ are of class $\mathcal { C } ^ { 1 }$ on $U _ { n }$. Give the expression of their gradient at a point $x = \left( x _ { 1 } , \ldots , x _ { n } \right)$ of $U _ { n }$.

Let $n$ be in $\mathbb { N } ^ { * }$. We denote by $U _ { n }$ the open set $\left( \mathbb { R } _ { + } ^ { * } \right) ^ { n }$. Its closure, denoted $\overline { U _ { n } }$, is $\left( \mathbb { R } _ { + } \right) ^ { n }$.
Let $s > 0$. We define the functions $f$ and $g _ { s }$ on $\overline { U _ { n } }$ by setting, for all $x = \left( x _ { 1 } , \ldots , x _ { n } \right) \in \overline { U _ { n } }$,
$$f ( x ) = \prod _ { k = 1 } ^ { n } x _ { k } \quad \text { and } \quad g _ { s } ( x ) = \left( \sum _ { k = 1 } ^ { n } x _ { k } \right) - s .$$
We denote by $X _ { s }$ the subset of $\overline { U _ { n } }$ consisting of the zeros of $g _ { s } : X _ { s } = \left\{ x \in \overline { U _ { n } } \mid g _ { s } ( x ) = 0 \right\}$.
Prove that the restriction of $f$ to $X _ { s }$ admits a maximum on $X _ { s }$ and that this maximum is in fact attained on $X _ { s } \cap U _ { n }$.
You may verify that $f$ is strictly positive at certain points of $X _ { s } \cap U _ { n }$.

Let $n$ be in $\mathbb { N } ^ { * }$. We denote by $U _ { n }$ the open set $\left( \mathbb { R } _ { + } ^ { * } \right) ^ { n }$. Its closure, denoted $\overline { U _ { n } }$, is $\left( \mathbb { R } _ { + } \right) ^ { n }$.
Let $s > 0$. We define the functions $f$ and $g _ { s }$ on $\overline { U _ { n } }$ by setting, for all $x = \left( x _ { 1 } , \ldots , x _ { n } \right) \in \overline { U _ { n } }$,
$$f ( x ) = \prod _ { k = 1 } ^ { n } x _ { k } \quad \text { and } \quad g _ { s } ( x ) = \left( \sum _ { k = 1 } ^ { n } x _ { k } \right) - s .$$
We denote by $X _ { s }$ the subset of $\overline { U _ { n } }$ consisting of the zeros of $g _ { s } : X _ { s } = \left\{ x \in \overline { U _ { n } } \mid g _ { s } ( x ) = 0 \right\}$.
We denote by $a = \left( a _ { 1 } , \ldots , a _ { n } \right)$ an element of $X _ { s } \cap U _ { n }$ at which the restriction of $f$ to $X _ { s }$ attains its maximum.
Prove that there exists a real number $\lambda > 0$ such that, for all $k$ in $\llbracket 1 , n \rrbracket , a _ { k } = \frac { f ( a ) } { \lambda }$.

Let $n$ be in $\mathbb { N } ^ { * }$. We denote by $U _ { n }$ the open set $\left( \mathbb { R } _ { + } ^ { * } \right) ^ { n }$. Its closure, denoted $\overline { U _ { n } }$, is $\left( \mathbb { R } _ { + } \right) ^ { n }$.
Let $s > 0$. We define the functions $f$ and $g _ { s }$ on $\overline { U _ { n } }$ by setting, for all $x = \left( x _ { 1 } , \ldots , x _ { n } \right) \in \overline { U _ { n } }$,
$$f ( x ) = \prod _ { k = 1 } ^ { n } x _ { k } \quad \text { and } \quad g _ { s } ( x ) = \left( \sum _ { k = 1 } ^ { n } x _ { k } \right) - s .$$
We denote by $X _ { s }$ the subset of $\overline { U _ { n } }$ consisting of the zeros of $g _ { s } : X _ { s } = \left\{ x \in \overline { U _ { n } } \mid g _ { s } ( x ) = 0 \right\}$.
Prove that, for all $\left( x _ { 1 } , \ldots , x _ { n } \right) \in U _ { n } \cap X _ { s } , \left( \prod _ { i = 1 } ^ { n } x _ { i } \right) ^ { 1 / n } \leqslant \frac { 1 } { n } \sum _ { i = 1 } ^ { n } x _ { i }$ and deduce the arithmetic-geometric inequality
$$\forall \left( x _ { 1 } , \ldots , x _ { n } \right) \in \left( \mathbb { R } _ { + } \right) ^ { n } , \quad \left( \prod _ { i = 1 } ^ { n } x _ { i } \right) ^ { 1 / n } \leqslant \frac { 1 } { n } \sum _ { i = 1 } ^ { n } x _ { i }$$