We fix two functions $f$ and $g$ in $E$. For $x > 0$, we set $F ( x ) = - U ( f ) ^ { \prime } ( x ) \mathrm { e } ^ { - x }$. It has been shown that $| U ( g ) ( x ) | \leqslant 4 \| g \| \frac { \sqrt { x } \mathrm { e } ^ { x / 2 } } { 1 + x }$ and $\left| U ( f ) ^ { \prime } ( x ) \right| \leqslant \| f \| \frac { \mathrm { e } ^ { x / 2 } } { \sqrt { x } }$. Show that for all $x > 0 , | F ( x ) U ( g ) ( x ) | \leqslant \frac { 4 \| f \| \| g \| } { 1 + x }$.

We fix two functions $f$ and $g$ in $E$. For $x > 0$, we set $F ( x ) = - U ( f ) ^ { \prime } ( x ) \mathrm { e } ^ { - x }$ where $U(f)^\prime(x) = \mathrm { e } ^ { x } \int _ { x } ^ { + \infty } f ( t ) \frac { \mathrm { e } ^ { - t } } { t } \mathrm {~d} t$. Show that for all $x \in ] 0,1] , | F ( x ) | \leqslant \| f \| \left( \mathrm { e } ^ { - 1 } - \ln ( x ) \right) ^ { 1 / 2 }$. One may use Question 19.

We fix two functions $f$ and $g$ in $E$. For $x > 0$, we set $F ( x ) = - U ( f ) ^ { \prime } ( x ) \mathrm { e } ^ { - x }$. Show the existence and calculate the values of the limits at $0$ and at $+ \infty$ of the function $t \mapsto F ( t ) U ( g ) ( t )$.

We fix two functions $f$ and $g$ in $E$. For $x > 0$, we set $F ( x ) = - U ( f ) ^ { \prime } ( x ) \mathrm { e } ^ { - x }$, which is an antiderivative of $x \mapsto f ( x ) \frac { \mathrm { e } ^ { - x } } { x }$. The limits of $t \mapsto F(t)U(g)(t)$ at $0$ and $+\infty$ are both $0$. Show that $$\langle f \mid U ( g ) \rangle = \int _ { 0 } ^ { + \infty } U ( f ) ^ { \prime } ( t ) U ( g ) ^ { \prime } ( t ) \mathrm { e } ^ { - t } \mathrm {~d} t.$$

We fix two functions $f$ and $g$ in $E$. Using the result of Question 28, deduce that $\langle f \mid U ( g ) \rangle = \langle U ( f ) \mid g \rangle$.

We consider two strictly positive real numbers $a$ and $b$, and we set $\rho = \frac{b-a}{b+a}$. We call $\Psi$ the application from $\mathbf{R}$ to $\mathbf{R}$ defined by: $$\forall x \in \mathbf{R}, \Psi(x) = \ln(a^2 \cos^2 x + b^2 \sin^2 x)$$
Deduce that for all $x \in \mathbf{R}$, $$\Psi(x) = 2\ln\left(\frac{a+b}{2}\right) - 2\sum_{k=1}^{+\infty} \frac{\cos(2kx)}{k}\rho^k$$

Let $g$ be the function defined by
$$\begin{aligned} g : ] - \pi ; \pi [ & \longrightarrow \mathbf { C } \\ \theta & \longmapsto e ^ { \mathrm { i } x \theta } \int _ { 0 } ^ { + \infty } \frac { t ^ { x - 1 } } { 1 + t e ^ { \mathrm { i } \theta } } \mathrm {~d} t \end{aligned}$$
where $x$ is a fixed element of $]0;1[$. Show that for all $\theta \in ] 0 ; \pi [$,
$$g ( \theta ) \sin ( x \theta ) = \frac { 1 } { 2 \mathrm { i } } \left( g ( - \theta ) e ^ { \mathrm { i } x \theta } - g ( \theta ) e ^ { - \mathrm { i } x \theta } \right) = \sin ( \theta ) \int _ { 0 } ^ { + \infty } \frac { t ^ { x } } { t ^ { 2 } + 2 t \cos ( \theta ) + 1 } \mathrm {~d} t$$

Let $g$ be the function defined by
$$\begin{aligned} g : ] - \pi ; \pi [ & \longrightarrow \mathbf { C } \\ \theta & \longmapsto e ^ { \mathrm { i } x \theta } \int _ { 0 } ^ { + \infty } \frac { t ^ { x - 1 } } { 1 + t e ^ { \mathrm { i } \theta } } \mathrm {~d} t \end{aligned}$$
where $x$ is a fixed element of $]0;1[$. Deduce that:
$$\forall \theta \in ] 0 ; \pi \left[ , \quad g ( \theta ) \sin ( \theta x ) = \int _ { \cot ( \theta ) } ^ { + \infty } \frac { ( u \sin ( \theta ) - \cos ( \theta ) ) ^ { x } } { 1 + u ^ { 2 } } \mathrm {~d} u , \right.$$
where $\cot ( \theta ) = \frac { \cos ( \theta ) } { \sin ( \theta ) }$.

Recall that $x$ is a fixed element of $]0;1[$. Show that:
$$\int _ { 0 } ^ { + \infty } \frac { t ^ { x - 1 } } { 1 + t } \mathrm {~d} t = \int _ { 0 } ^ { 1 } \left( \frac { t ^ { x - 1 } } { 1 + t } + \frac { t ^ { - x } } { 1 + t } \right) \mathrm { d } t$$

Show that the integral
$$\int _ { 0 } ^ { + \infty } \frac { 1 - ( \cos ( t ) ) ^ { 2 p + 1 } } { t ^ { 2 } } \mathrm {~d} t$$
converges and that:
$$\int _ { 0 } ^ { + \infty } \frac { 1 - ( \cos ( t ) ) ^ { 2 p + 1 } } { t ^ { 2 } } \mathrm {~d} t = ( 2 p + 1 ) \int _ { 0 } ^ { + \infty } ( \cos ( t ) ) ^ { 2 p } \frac { \sin ( t ) } { t } \mathrm {~d} t$$

Show that for all $n \in \mathbf { N } ^ { * }$:
$$\int _ { \frac { \pi } { 2 } + ( n - 1 ) \pi } ^ { \frac { \pi } { 2 } + n \pi } ( \cos ( t ) ) ^ { 2 p } \frac { \sin ( t ) } { t } \mathrm {~d} t = \int _ { 0 } ^ { \frac { \pi } { 2 } } ( \cos ( t ) ) ^ { 2 p } \frac { 2 ( - 1 ) ^ { n } t \sin ( t ) } { t ^ { 2 } - n ^ { 2 } \pi ^ { 2 } } \mathrm {~d} t$$

Deduce that:
$$\int _ { \frac { \pi } { 2 } } ^ { + \infty } ( \cos ( t ) ) ^ { 2 p } \frac { \sin ( t ) } { t } \mathrm {~d} t = \int _ { 0 } ^ { \frac { \pi } { 2 } } ( \cos ( t ) ) ^ { 2 p } \left( \sum _ { n = 1 } ^ { + \infty } \frac { 2 ( - 1 ) ^ { n } t \sin ( t ) } { t ^ { 2 } - n ^ { 2 } \pi ^ { 2 } } \right) \mathrm { d } t$$

Deduce that:
$$\int _ { 0 } ^ { + \infty } ( \cos ( t ) ) ^ { 2 p } \frac { \sin ( t ) } { t } \mathrm {~d} t = \int _ { 0 } ^ { \frac { \pi } { 2 } } ( \cos ( t ) ) ^ { 2 p } \mathrm {~d} t$$
In the case $p = 0$, this integral is commonly called the ``Dirichlet Integral''.

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

Show that for all $s \in \mathbf { R }$
$$\int _ { 0 } ^ { + \infty } \frac { 1 - \cos ( s t ) } { t ^ { 2 } } \mathrm {~d} t = \frac { \pi } { 2 } | s |$$

We fix $( p , q ) \in \left( \mathbf { N } ^ { * } \right) ^ { 2 }$ and set $\alpha _ { p , q } := \dfrac { p } { q }$. We define, for all $t \in \mathbf { R } _ { + }$, the application $I _ { p , q } : \mathbf { R } _ { + } \rightarrow \mathbf { R }$ by $$I _ { p , q } ( t ) := \int _ { 0 } ^ { 1 } \frac { x ^ { ( t + 1 ) \alpha _ { p , q } } } { 1 + x ^ { \alpha _ { p , q } } } d x$$ Show that, for all $( p , q ) \in \left( \mathrm { N } ^ { * } \right) ^ { 2 }$, $$S _ { p , q } = \int _ { 0 } ^ { 1 } \frac { t ^ { q - 1 } } { 1 + t ^ { p } } d t$$

We define $E _ { 1 } := \left\{ ( p , q ) \in \left( \mathbf { N } ^ { * } \right) ^ { 2 } : p = q \right\}$ and $S _ { p , q } = \int _ { 0 } ^ { 1 } \dfrac { t ^ { q - 1 } } { 1 + t ^ { p } } d t$.
Show that, for all $( p , q ) \in E _ { 1 }$, $$S _ { p , q } = \frac { \ln 2 } { p }$$

We define $E _ { 2 } := \left\{ ( p , q ) \in \left( \mathbf { N } ^ { * } \right) ^ { 2 } : p < q , p \mid q \right\}$ and $S _ { p , q } = \int _ { 0 } ^ { 1 } \dfrac { t ^ { q - 1 } } { 1 + t ^ { p } } d t$.
For all pairs $( p , q ) \in E _ { 2 }$, show that there exists a constant $\lambda := \lambda ( p , q )$ which one will determine, such that $$S _ { p , q } = \frac { ( - 1 ) ^ { \lambda - 1 } } { p } \left( \ln ( 2 ) - \sum _ { k = 1 } ^ { \lambda - 1 } \frac { ( - 1 ) ^ { k - 1 } } { k } \right)$$

We are given $f \in \mathcal{C}^1(\mathbb{R})$, convex, admitting a minimizer $x_* \in \mathbb{R}$, with $f'$ being $L$-Lipschitzian, and $0 < \tau < 2/L$. The sequence $(x_n)_{n \in \mathbb{N}}$ is defined by $x_{n+1} := x_n - \tau f'(x_n)$. a) Show that for all $x, y \in \mathbb{R}$ $$f(y) \geq f(x) + f'(x)(y-x)$$ Hint: consider a Taylor expansion of the convexity inequality when $t \rightarrow 0^+$. b) Show that for all $x, y \in \mathbb{R}$ $$f(y) \leq f(x) + f'(x)(y-x) + \frac{L}{2}(y-x)^2$$ c) Establish that for all $n \in \mathbb{N}$ $$f(x_{n+1}) \leq f(x_n) - \frac{\tau}{2}(2 - \tau L)\left|f'(x_n)\right|^2$$ Deduce that the sequence $\left(f(x_n)\right)_{n \in \mathbb{N}}$ is decreasing.

We fix a pair $( p , q ) \in E _ { 3 } := \left\{ ( p , q ) \in \left( \mathbf { N } ^ { * } \right) ^ { 2 } : p > q \right\}$, and set $\theta _ { k } := ( 2 k + 1 ) \dfrac { \pi } { p }$. We admit that for all $0 \leq k \leq \lfloor p / 2 \rfloor - 1$, $$\int _ { 0 } ^ { 1 } F _ { k } ( t ) d t = \cos \left( q \theta _ { k } \right) \ln \left( 2 \sin \left( \frac { \theta _ { k } } { 2 } \right) \right) - \frac { \pi } { 2 p } ( p - 1 - 2 k ) \sin \left( q \theta _ { k } \right)$$
Deduce from the previous questions that, for all $( p , q ) \in E _ { 3 }$, $$S _ { p , q } = \frac { 1 } { p } \left( \frac { \pi } { p } \sum _ { k = 0 } ^ { \lfloor p / 2 \rfloor - 1 } ( p - 1 - 2 k ) \sin \left( q \theta _ { k } \right) - 2 \sum _ { k = 0 } ^ { \lfloor p / 2 \rfloor - 1 } \cos \left( q \theta _ { k } \right) \ln \left( \sin \left( \frac { \theta _ { k } } { 2 } \right) \right) \right)$$

Using the formula established for $( p , q ) \in E _ { 3 }$: $$S _ { p , q } = \frac { 1 } { p } \left( \frac { \pi } { p } \sum _ { k = 0 } ^ { \lfloor p / 2 \rfloor - 1 } ( p - 1 - 2 k ) \sin \left( q \theta _ { k } \right) - 2 \sum _ { k = 0 } ^ { \lfloor p / 2 \rfloor - 1 } \cos \left( q \theta _ { k } \right) \ln \left( \sin \left( \frac { \theta _ { k } } { 2 } \right) \right) \right)$$ where $\theta_k := (2k+1)\dfrac{\pi}{p}$, deduce the exact values of $S _ { 2,1 }$ and $S _ { 3,1 }$.

For all $( p , q ) \in \left( \mathrm { N } ^ { * } \right) ^ { 2 }$, define $R _ { p , q } := \dfrac { 1 } { q } I _ { p , q }$ where $$I _ { p , q } ( t ) := \int _ { 0 } ^ { 1 } \frac { x ^ { ( t + 1 ) \alpha _ { p , q } } } { 1 + x ^ { \alpha _ { p , q } } } d x, \quad \alpha_{p,q} = \frac{p}{q}.$$
Using the change of variables $s = x ^ { n + 1 }$ in $I _ { p , q } ( n )$, prove that $$R _ { p , q } ( n ) \sim \frac { 1 } { 2 p n } \quad ( n \rightarrow + \infty )$$

For all $( p , q ) \in \left( \mathrm { N } ^ { * } \right) ^ { 2 }$, define $R _ { p , q } := \dfrac { 1 } { q } I _ { p , q }$ where $$I _ { p , q } ( t ) := \int _ { 0 } ^ { 1 } \frac { x ^ { ( t + 1 ) \alpha _ { p , q } } } { 1 + x ^ { \alpha _ { p , q } } } d x, \quad \alpha_{p,q} = \frac{p}{q},$$ and recall that $$\phi _ { p , q } ( n ) = \frac { 1 } { q } \left( \int _ { 0 } ^ { 1 } \frac { 1 } { 1 + x ^ { \alpha _ { p , q } } } d x + ( - 1 ) ^ { n } I _ { p , q } ( n ) \right).$$
Using the result $R _ { p , q } ( n ) \sim \dfrac { 1 } { 2 p n }$ as $n \to +\infty$, deduce the convergence rate of the alternating congruent-harmonic series $\sum u _ { k }$, that is, that of the sequence of partial sums $\left( \phi _ { p , q } ( n ) \right) _ { n \in \mathbf { N } }$.

Let $r$ and $s$ be two strictly positive natural integers such that $r \geqslant s$. We set
$$J _ { r , s } = \int _ { 0 } ^ { 1 } f _ { r , s } ( y ) \mathrm { d } y = \int _ { 0 } ^ { 1 } \int _ { 0 } ^ { 1 } \frac { x ^ { r } y ^ { s } } { 1 - x y } \mathrm {~d} x \mathrm {~d} y$$
Show that
$$J _ { r , s } = \sum _ { k = 0 } ^ { + \infty } \frac { 1 } { ( r + k + 1 ) ( s + k + 1 ) }$$

Show that $$\forall u \in \mathbb{R}, \quad \forall a \in \mathbb{R}_+^*, \quad \mathrm{e}^{\frac{au^2}{2}} = \int_{-\infty}^{+\infty} \mathrm{e}^{ut - \frac{t^2}{2a}} \frac{\mathrm{~d}t}{\sqrt{2\pi a}}$$