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 $\beta ( x + 1 , y ) = \frac { x } { x + y } \beta ( x , y )$.

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 denote $B$ the function defined on $\mathbb { R } ^ { + * }$ by $B ( x ) = \int _ { 0 } ^ { 1 } ( \ln ( 1 - t ) ) ^ { 2 } t ^ { x - 1 } \mathrm {~d} t$.
Give without justification an expression, using an integral, of $B ^ { ( p ) } ( x )$, for every natural integer $p$ and every real $x > 0$.

We denote $S _ { r } = \sum _ { n = 1 } ^ { + \infty } \frac { H _ { n } } { ( n + 1 ) ^ { r } }$ for $r \geqslant 2$, and $B ( x ) = \int _ { 0 } ^ { 1 } ( \ln ( 1 - t ) ) ^ { 2 } t ^ { x - 1 } \mathrm {~d} t$. We have shown that $S _ { r } = \frac { ( - 1 ) ^ { r } } { 2 ( r - 2 ) ! } \int _ { 0 } ^ { 1 } \frac { ( \ln t ) ^ { r - 2 } ( \ln ( 1 - t ) ) ^ { 2 } } { t } \mathrm {~d} t$.
Deduce that for every integer $r \geqslant 2 , S _ { r } = \frac { ( - 1 ) ^ { r } } { 2 ( r - 2 ) ! } \lim _ { x \rightarrow 0 ^ { + } } B ^ { ( r - 2 ) } ( x )$.

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 define $\varphi$ the function defined on $] - 1 , + \infty [$ by $\varphi ( x ) = ( \psi ( 1 + x ) - \psi ( 1 ) ) ^ { 2 } + \left( \psi ^ { \prime } ( 1 ) - \psi ^ { \prime } ( 1 + x ) \right)$.
Show that $\varphi$ is $\mathcal { C } ^ { \infty }$ on its domain of definition and give for every natural integer $n \geqslant 2$ the value of $\varphi ^ { ( n ) } ( 0 )$ as a function of the successive derivatives of $\psi$ at the point 1.

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 $\varphi ( x ) = ( \psi ( 1 + x ) - \psi ( 1 ) ) ^ { 2 } + \left( \psi ^ { \prime } ( 1 ) - \psi ^ { \prime } ( 1 + x ) \right)$. We have $x B(x) = \varphi(x)$ for $x > 0$, and $S _ { r } = \frac { ( - 1 ) ^ { r } } { 2 ( r - 2 ) ! } \lim _ { x \rightarrow 0 ^ { + } } B ^ { ( r - 2 ) } ( x )$. We have also shown that $\psi ( 1 + x ) = \psi ( 1 ) + \sum _ { n = 1 } ^ { + \infty } ( - 1 ) ^ { n + 1 } \zeta ( n + 1 ) x ^ { n }$ for $x \in ] -1, 1[$.
Conclude that, for every integer $r \geqslant 3$, $$2 S _ { r } = r \zeta ( r + 1 ) - \sum _ { k = 1 } ^ { r - 2 } \zeta ( k + 1 ) \zeta ( r - k )$$

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

We study the sequence $\left(u_{n}\right)_{n \in \mathbb{N}^{*}}$ defined by $$\forall n \in \mathbb{N}^{*}, \quad u_{n} = \int_{0}^{\infty} \frac{1 - (\cos t)^{n}}{t^{2}} \mathrm{~d}t$$ Show that $$\left. \left. \forall (n, u) \in \mathbb{N}^{*} \times \right] 0, 1 \right], \quad \left| 1 - (\cos(\sqrt{2u/n}))^{n} \right| \leqslant u$$

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

Using the change of variable $t = \frac { \ln x } { 2 \pi }$, demonstrate that $$\forall p \in \mathbb { N } \quad I _ { p } = \frac { e ^ { - p ^ { 2 } \pi ^ { 2 } } } { 2 \pi } \int _ { 0 } ^ { + \infty } x ^ { p - 1 } \exp \left( - \frac { \ln ^ { 2 } x } { 4 \pi ^ { 2 } } \right) \sin \left( \frac { \ln x } { 2 \pi } \right) \mathrm { d } x$$

For every natural integer $k$ we set $$m_{k} = \frac{1}{2\pi} \int_{-2}^{2} x^{k} \sqrt{4 - x^{2}} \, \mathrm{d}x$$ Using the change of variable $x = 2\sin t$, calculate $m_{0}$.

Let $P \in \mathbb { R } [ X ]$ and $Q \in \mathbb { R } [ X ]$. Show that the function $x \mapsto P ( 4 x ) Q ( 4 x ) \frac { \sqrt { 1 - x } } { \sqrt { x } }$ is integrable on $\left. ] 0,1 \right]$.

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 $q ( x ) = x - \lfloor x \rfloor - \frac { 1 } { 2 }$.
Let $t \in \mathbf { R } _ { + } ^ { * }$. Show successively that $\left| u _ { k } ( t ) \right| = \int _ { k/2 } ^ { ( k + 1 ) / 2 } \frac { t | q ( u ) | } { e ^ { t u } - 1 } \mathrm {~d} u$, then $u _ { k } ( t ) = ( - 1 ) ^ { k } \left| u _ { k } ( t ) \right|$ for all integer $k \geq 1$, and finally establish that
$$\forall n \in \mathbf { N } ^ { * } , \left| \sum _ { k = n } ^ { + \infty } u _ { k } ( t ) \right| \leq \frac { 1 } { 2 n } .$$
We admit in what follows that this bound also holds for $t = 0$.

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 $$\int_{1}^{+\infty} \frac{tq(u)}{e^{tu}-1} \mathrm{~d}u \underset{t \rightarrow 0^+}{\longrightarrow} \frac{\ln(2\pi)}{2} - 1$$

The function $q$ associates to any real $x$ the real number $q ( x ) = x - \lfloor x \rfloor - \frac { 1 } { 2 }$.
Show, for all real $t > 0$, the identity
$$\int _ { 1 } ^ { + \infty } \frac { t q ( u ) } { e ^ { t u } - 1 } \mathrm {~d} u = - \frac { 1 } { 2 } \ln \left( 1 - e ^ { - t } \right) - \ln P \left( e ^ { - t } \right) - \int _ { 1 } ^ { + \infty } \ln \left( 1 - e ^ { - t u } \right) \mathrm { d } u$$

Show, for all real $t > 0$, the identity $$\int_{1}^{+\infty} \frac{tq(u)}{e^{tu}-1} \mathrm{~d}u = -\frac{1}{2}\ln(1-e^{-t}) - \ln P(e^{-t}) - \int_{1}^{+\infty} \ln(1-e^{-tu}) \mathrm{d}u$$

Conclude that $$\ln P(e^{-t}) = \frac{\pi^2}{6t} + \frac{\ln(t)}{2} - \frac{\ln(2\pi)}{2} + o(1) \text{ when } t \text{ tends to } 0^+.$$

We have $P ( z ) = \sum _ { n = 0 } ^ { + \infty } p _ { n } z ^ { n }$ for all $z \in D$, where $p_n$ denotes the number of partitions of $n$.
Let $n \in \mathbf { N }$. Show that for all real $t > 0$,
$$p _ { n } = \frac { e ^ { n t } P \left( e ^ { - t } \right) } { 2 \pi } \int _ { - \pi } ^ { \pi } e ^ { - i n \theta } \frac { P \left( e ^ { - t } e ^ { i \theta } \right) } { P \left( e ^ { - t } \right) } \mathrm { d } \theta \tag{1}$$

Let $x \in [ 0,1 [$ and $\theta \in \mathbf { R }$. Using the function $L$, show that
$$\left| \frac { 1 - x } { 1 - x e ^ { i \theta } } \right| \leq \exp ( - ( 1 - \cos \theta ) x ) .$$
Deduce that for all $x \in [ 0,1 [$ and all real $\theta$,
$$\left| \frac { P \left( x e ^ { i \theta } \right) } { P ( x ) } \right| \leq \exp \left( - \frac { 1 } { 1 - x } + \operatorname { Re } \left( \frac { 1 } { 1 - x e ^ { i \theta } } \right) \right)$$

Let $x \in [ 0,1 [$ and $\theta$ a real number. Show that
$$\frac { 1 } { 1 - x } - \operatorname { Re } \left( \frac { 1 } { 1 - x e ^ { i \theta } } \right) \geq \frac { x ( 1 - \cos \theta ) } { ( 1 - x ) \left( ( 1 - x ) ^ { 2 } + 2 x ( 1 - \cos \theta ) \right) }$$
Deduce that if $x \geq \frac { 1 } { 2 }$ then
$$\left| \frac { P \left( x e ^ { i \theta } \right) } { P ( x ) } \right| \leq \exp \left( - \frac { 1 - \cos \theta } { 6 ( 1 - x ) ^ { 3 } } \right) \quad \text { or } \quad \left| \frac { P \left( x e ^ { i \theta } \right) } { P ( x ) } \right| \leq \exp \left( - \frac { 1 } { 3 ( 1 - x ) } \right)$$
For this last result, distinguish two cases according to the relative values of $x ( 1 - \cos \theta )$ and $( 1 - x ) ^ { 2 }$.

Show that there exists a real $\alpha > 0$ such that
$$\forall \theta \in [ - \pi , \pi ] , 1 - \cos \theta \geq \alpha \theta ^ { 2 }$$
Deduce that there exist three reals $t _ { 0 } > 0 , \beta > 0$ and $\gamma > 0$ such that, for all $t \in ] 0 , t _ { 0 } ]$ and all $\theta \in [ - \pi , \pi ]$,
$$\left| \frac { P \left( e ^ { - t } e ^ { i \theta } \right) } { P \left( e ^ { - t } \right) } \right| \leq e ^ { - \beta \left( t ^ { - 3 / 2 } \theta \right) ^ { 2 } } \quad \text { or } \quad \left| \frac { P \left( e ^ { - t } e ^ { i \theta } \right) } { P \left( e ^ { - t } \right) } \right| \leq e ^ { - \gamma \left( t ^ { - 3 / 2 } | \theta | \right) ^ { 2 / 3 } }$$

We fix two functions $f$ and $g$ in $E$. For $x > 0$, we set $$F ( x ) = - U ( f ) ^ { \prime } ( x ) \mathrm { e } ^ { - x }$$ where $U(f)^\prime(x) = \mathrm { e } ^ { x } \int _ { x } ^ { + \infty } f ( t ) \frac { \mathrm { e } ^ { - t } } { t } \mathrm {~d} t$. Verify that $F$ is an antiderivative of $x \mapsto f ( x ) \frac { \mathrm { e } ^ { - x } } { x }$ on the interval $\mathbb { R } _ { + } ^ { * }$.

Deduce that
$$\int _ { - \pi } ^ { \pi } e ^ { - i \frac { \pi ^ { 2 } \theta } { 6 t ^ { 2 } } } \frac { P \left( e ^ { - t } e ^ { i \theta } \right) } { P \left( e ^ { - t } \right) } \mathrm { d } \theta = O \left( t ^ { 3 / 2 } \right) \quad \text { when } t \text { tends to } 0 ^ { + }$$