For $\mu > 0$ and $\varphi \in \mathcal{C}_{c}(\mathbb{R})$, we define $T_{\mu} : \varphi \mapsto T_{\mu}\varphi$, where for all $x \in \mathbb{R}$,
$$T_{\mu}\varphi(x) = \frac{1}{2\mu} \int_{x-\mu}^{x+\mu} \varphi(t)\, dt$$
For $k \geqslant 1$, show that if $\varphi \in \mathcal{C}_{c}^{k+1}(\mathbb{R})$, we have
$$\left\|(T_{\mu}\varphi)^{(k)} - \varphi^{(k)}\right\|_{\infty} \leqslant \frac{\mu}{2} \left\|\varphi^{(k+1)}\right\|_{\infty}.$$

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: $$\forall x \in \mathbf{R}, \Psi(x) = \ln(a^2 \cos^2 x + b^2 \sin^2 x)$$ and $\sigma(x) = \sum_{k=1}^{+\infty} \frac{x^k}{k^2}$.
Conclude that $$\int_0^{\pi} \Psi(x)^2 \mathrm{d}x = 4\pi\left(\ln\left(\frac{a+b}{2}\right)\right)^2 + 2\pi\sigma(\rho^2)$$

For $\mu > 0$ and $\varphi \in \mathcal{C}_{c}(\mathbb{R})$, we define $T_{\mu} : \varphi \mapsto T_{\mu}\varphi$, where for all $x \in \mathbb{R}$,
$$T_{\mu}\varphi(x) = \frac{1}{2\mu} \int_{x-\mu}^{x+\mu} \varphi(t)\, dt$$
We assume that $(\mu_{n})_{n \geqslant 1}$ is a sequence of strictly positive real numbers such that $\sum_{n \geqslant 1} \mu_{n}$ converges. We fix $\psi_{0} \in \mathcal{C}_{c}(\mathbb{R})$ and we define by recursion the sequence $(\psi_{n})_{n \geqslant 0}$ by
$$\forall n \geqslant 0,\quad \psi_{n+1} = T_{\mu_{n+1}} \psi_{n}$$
Show that for all $n \geqslant k$, $\psi_{n}$ is of class $C^{k}$.

Let $I$ be an interval of $\mathbb { R }$ and $\left( f _ { n } \right) _ { n \in \mathbb { N } ^ { * } }$ a sequence of functions piecewise continuous on $I$ that converges uniformly on $I$ to a function $f$ also piecewise continuous on $I$.
If $\left( u _ { n } \right) _ { n \in \mathbb { N } ^ { * } }$ (respectively $\left( v _ { n } \right) _ { n \in \mathbb { N } ^ { * } }$) is a sequence of real numbers belonging to $I$ that converges to $u \in I$ (respectively $v \in I$), show that $$\lim _ { n \rightarrow + \infty } \left( \int _ { u _ { n } } ^ { v _ { n } } f _ { n } ( x ) \mathrm { d } x \right) = \int _ { u } ^ { v } f ( x ) \mathrm { d } x$$

Let $f : [ a , b ] \rightarrow \mathbb { R }$ be a piecewise continuous function taking values in an interval $J$. Let $\varphi$ be a continuous and convex function on $J$. Prove that
$$\varphi \left( \frac { 1 } { b - a } \int _ { a } ^ { b } f ( t ) \mathrm { d } t \right) \leqslant \frac { 1 } { b - a } \int _ { a } ^ { b } \varphi \circ f ( t ) \mathrm { d } t .$$
You may use Riemann sums.

Let $f : \mathbb { R } _ { + } \rightarrow \mathbb { R }$, be a piecewise continuous function, strictly positive and integrable. For all $x > 0$, we define
$$g ( x ) = \frac { 1 } { x } \int _ { 0 } ^ { x } t f ( t ) \mathrm { d } t \quad \text { and } \quad h ( x ) = \frac { 1 } { x } g ( x ) = \frac { 1 } { x ^ { 2 } } \int _ { 0 } ^ { x } t f ( t ) \mathrm { d } t .$$
Determine the limit of $g ( x )$ as $x$ tends to 0.

Show that the function $\varphi : t \longmapsto \frac{1}{\sqrt{t}}$ is integrable on $]0,1[$, then show that the function $\varphi$ belongs to $\mathscr{D}_{0,1}$.

Let $f : \mathbb { R } _ { + } \rightarrow \mathbb { R }$, be a piecewise continuous function, strictly positive and integrable. For all $x > 0$, we define
$$g ( x ) = \frac { 1 } { x } \int _ { 0 } ^ { x } t f ( t ) \mathrm { d } t \quad \text { and } \quad h ( x ) = \frac { 1 } { x } g ( x ) = \frac { 1 } { x ^ { 2 } } \int _ { 0 } ^ { x } t f ( t ) \mathrm { d } t .$$
Determine the limit of $g ( x )$ as $x$ tends to $+ \infty$. Denoting by $\mathbb { 1 } _ { [ 0 , x ] }$ the indicator function of $[ 0 , x ]$, you may note that $g ( x ) = \int _ { 0 } ^ { + \infty } \frac { 1 } { x } t f ( t ) \mathbb { 1 } _ { [ 0 , x ] } ( t ) \mathrm { d } t$.

Let $f \in \mathcal { C } ^ { 0 } ( [ 0 , + \infty [ )$ and $\ell \in \mathbb { R }$. Prove that $$\left( \lim _ { x \rightarrow + \infty } f ( x ) = \ell \right) \Rightarrow \left( \lim _ { x \rightarrow + \infty } \frac { 1 } { x } \int _ { 0 } ^ { x } f ( t ) d t = \ell \right)$$

Using a counterexample, prove that the converse of the result in question 3.1 is false, i.e. that $$\left( \lim _ { x \rightarrow + \infty } \frac { 1 } { x } \int _ { 0 } ^ { x } f ( t ) d t = \ell \right) \not\Rightarrow \left( \lim _ { x \rightarrow + \infty } f ( x ) = \ell \right)$$ for $f \in \mathcal{C}^0([0,+\infty[)$.

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 $g$ 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, using the dominated convergence theorem, that:
$$\lim _ { \theta \rightarrow \pi ^ { - } } g ( \theta ) \sin ( x \theta ) = \int _ { - \infty } ^ { + \infty } \frac { \mathrm { d } u } { 1 + u ^ { 2 } }$$

Consider the function $h : ]0,1[ \longrightarrow \mathbf{R},\; t \longmapsto \frac{1}{\sqrt{t(1-t)}}$. Prove that: $$\lim_{n \longrightarrow +\infty} \sum_{k=1}^{2n-1} \frac{1}{2n} h\!\left(\frac{k}{2n}\right) = \int_0^1 h(t)\, dt.$$

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
$$\int _ { 0 } ^ { + \infty } \exp \left( \frac { 1 } { x } \int _ { 0 } ^ { x } \ln ( f ( t ) ) \mathrm { d } t \right) \mathrm { d } x \leqslant \mathrm { e } \int _ { 0 } ^ { + \infty } f ( x ) \mathrm { d } x$$

Consider the function $h : ]0,1[ \longrightarrow \mathbf{R},\; t \longmapsto \frac{1}{\sqrt{t(1-t)}}$, and let $\tilde{h}$ denote its restriction to $\left]0, \frac{1}{2}\right]$. Show that: $$\lim_{n \rightarrow +\infty} \sum_{k=1}^{n} \frac{1}{2n+1} h\!\left(\frac{k}{2n+1}\right) = \int_0^{\frac{1}{2}} h(t)\, dt.$$ Deduce that: $$\lim_{n \rightarrow +\infty} \sum_{k=1}^{2n} \frac{1}{2n+1} h\!\left(\frac{k}{2n+1}\right) = \int_0^1 h(t)\, dt.$$

Consider the function $h : ]0,1[ \longrightarrow \mathbf{R},\; t \longmapsto \frac{1}{\sqrt{t(1-t)}}$. Deduce from the previous questions that the function $h$ belongs to $\mathscr{D}_{0,1}$.

Study the variations of the function $t \mapsto t \ln(t)$ on $\mathbf{R}_{+}^{*}$. Verify that the function can be extended by continuity at 0 and verify that the quantity $\operatorname{Ent}_{\varphi}(g)$ is well defined for all $g \in C^{0}(\mathbf{R}) \cap CL(\mathbf{R})$ with strictly positive values such that $\int_{-\infty}^{+\infty} g(x) \varphi(x) \mathrm{d}x = 1$.
Recall that for $f \in C^{0}(\mathbf{R}) \cap CL(\mathbf{R})$ with strictly positive values such that $\int_{-\infty}^{+\infty} f(x) \varphi(x) \mathrm{d}x = 1$, the entropy of $f$ with respect to $\varphi$ is defined by: $$\operatorname{Ent}_{\varphi}(f) = \int_{-\infty}^{+\infty} \ln(f(x)) f(x) \varphi(x) \mathrm{d}x$$

Deduce the limit: $$\lim_{n \rightarrow +\infty} \sum_{i=1}^{n-1} \frac{1}{\sqrt{i(n-i)}}.$$

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 $$I _ { p , q } ( t ) := \int _ { 0 } ^ { 1 } \frac { x ^ { ( t + 1 ) \alpha _ { p , q } } } { 1 + x ^ { \alpha _ { p , q } } } d x$$ Prove that the application $I _ { p , q }$ is well-defined and continuous on $\mathbf { R } _ { + }$.

Problem 1: calculation of an integral
For $x \geqslant 0$ we define $$f ( x ) = \int _ { 0 } ^ { \infty } \frac { e ^ { - t x } } { 1 + t ^ { 2 } } \mathrm {~d} t \quad \text { and } \quad g ( x ) = \int _ { 0 } ^ { \infty } \frac { \sin t } { t + x } \mathrm {~d} t$$
Deduce the value of $\int _ { 0 } ^ { \infty } \frac { \sin t } { t } \mathrm {~d} t$.

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 $$I _ { p , q } ( t ) := \int _ { 0 } ^ { 1 } \frac { x ^ { ( t + 1 ) \alpha _ { p , q } } } { 1 + x ^ { \alpha _ { p , q } } } d x$$ Determine $$\lim _ { n \rightarrow + \infty } I _ { p , q } ( n ) = 0$$

Let $r$ and $s$ be two strictly positive natural integers such that $r \geqslant s$.
Let $y \in ] 0,1 [$. Justify that the function
$$x \mapsto \frac { x ^ { r } y ^ { s } } { 1 - x y }$$
is integrable on $[ 0,1 ]$.

Let $r$ and $s$ be two strictly positive natural integers such that $r \geqslant s$. For $y \in ] 0,1 [$, we set
$$f _ { r , s } ( y ) = \int _ { 0 } ^ { 1 } \frac { x ^ { r } y ^ { s } } { 1 - x y } \mathrm {~d} x$$
Show that $f _ { r , s }$ is continuous and integrable on $] 0,1 [$.

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 $\int_{-\infty}^{+\infty} \mathrm{e}^{-\frac{x^2}{2}} \mathrm{~d}x$ converges. We admit that its value is $\sqrt{2\pi}$.