Show that $$\lim _ { n \rightarrow \infty } \sqrt { n } I _ { n } = \int _ { 0 } ^ { + \infty } \mathrm { e } ^ { - u ^ { 2 } } \mathrm {~d} u$$ where $I _ { n } = \int _ { 0 } ^ { 1 } \frac { 1 } { \left( 1 + t ^ { 2 } \right) ^ { n } } \mathrm {~d} t$.

If $n \in \mathbf{N}$, we denote by $D_n$ the improper integral $\int_0^{\pi/2} (\ln(\sin(t)))^n \mathrm{~d}t$.
Justify that, if $n \in \mathbf{N}$, the improper integral $D_n$ is convergent, then show that $$D_1 = \int_0^{\pi/2} \ln(\cos(t)) \mathrm{d}t$$

Deduce the values of $$\int _ { 0 } ^ { + \infty } \mathrm { e } ^ { - u ^ { 2 } } \mathrm {~d} u \quad \text { then of } \quad \int _ { - \infty } ^ { + \infty } \mathrm { e } ^ { - u ^ { 2 } / 2 } \mathrm {~d} u .$$

Let $x > 0$. By writing that $\varphi ( t ) \leqslant \frac { t } { x } \varphi ( t )$ for all $t \geqslant x$, show that $$\int _ { x } ^ { + \infty } \varphi ( t ) \mathrm { d } t \leqslant \frac { \varphi ( x ) } { x }$$ where $\varphi ( x ) = \frac { 1 } { \sqrt { 2 \pi } } \mathrm { e } ^ { - x ^ { 2 } / 2 }$.

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 $g ( x ) = \int _ { 0 } ^ { + \infty } \frac { 1 } { x } t f ( t ) \mathbb { 1 } _ { [ 0 , x ] } ( t ) \mathrm { d } t$.

We denote by $\tilde{h}$ the restriction of the function $h : ]0,1[ \longrightarrow \mathbf{R},\; t \longmapsto \frac{1}{\sqrt{t(1-t)}}$ to the interval $\left]0, \frac{1}{2}\right]$. Verify that the function $\tilde{h}$ is decreasing on $]0, \frac{1}{2}[$, then show that the function $\tilde{h}$ belongs to $\mathscr{D}_{0, \frac{1}{2}}$.

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 the function $h$ is integrable on $]0,1[$ and that: $$\int_0^1 h(t)\, dt = 2\int_0^{\frac{1}{2}} \tilde{h}(t)\, dt.$$

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

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: $$\lim_{n \rightarrow +\infty} \sum_{k=1}^{2n} \frac{1}{2n+1} h\!\left(\frac{k}{2n+1}\right) = \int_0^1 h(t)\, dt.$$

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 $[ k - 1 , k [$.
Let $k$ be in $\mathbb { N } ^ { * }$. Prove that the function $v _ { k }$ defined on $[ k - 1 , k ]$ by
$$\left\{ \begin{array} { l } v _ { k } ( x ) = \frac { 1 } { x } \sum _ { i = 1 } ^ { k - 1 } \ln \left( a _ { i } \right) + \frac { 1 } { x } ( x - k + 1 ) \ln \left( a _ { k } \right) \quad \text { if } k \geqslant 2 \\ v _ { 1 } ( x ) = \ln \left( a _ { 1 } \right) \end{array} \right.$$
is minimal for $x = k$.

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 $\mathscr{D}_{0,1}$.

Consider the function $h : ]0,1[ \longrightarrow \mathbf{R},\; t \longmapsto \frac{1}{\sqrt{t(1-t)}}$. Show that: $$\int_0^1 h(t)\, dt = \pi.$$

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

Consider a sequence $(\varepsilon_n)_{n \in \mathbb{N}}$ of real numbers strictly greater than $-1$, convergent with limit zero. Show that: $$\lim_{n \rightarrow +\infty} \sum_{i=1}^{n-1} \frac{|\varepsilon_i|}{\sqrt{i(n-i)}} = 0.$$

By comparison with an integral, establish that $$\sum_{k=1}^{n} \ln(k) \underset{n \rightarrow +\infty}{=} n\ln(n) - n + O(\ln(n))$$

Problem 1: calculation of an integral
For $x \geqslant 0$ we define $$f ( x ) = \int _ { 0 } ^ { \infty } \frac { e ^ { - t x } } { 1 + t ^ { 2 } } \mathrm {~d} t \quad \text { and } \quad g ( x ) = \int _ { 0 } ^ { \infty } \frac { \sin t } { t + x } \mathrm {~d} t$$
Study of $f$. a. Show that $f ( x )$ is well defined for all $x \geqslant 0$. b. Show with precision that the function $f$ is of class $C ^ { 2 }$ on $] 0 , \infty [$, and also continuous at 0. c. Calculate $f + f ^ { \prime \prime }$ and deduce that $f$ is a solution of a linear second-order differential equation.

Problem 1: calculation of an integral
For $x \geqslant 0$ we define $$f ( x ) = \int _ { 0 } ^ { \infty } \frac { e ^ { - t x } } { 1 + t ^ { 2 } } \mathrm {~d} t \quad \text { and } \quad g ( x ) = \int _ { 0 } ^ { \infty } \frac { \sin t } { t + x } \mathrm {~d} t$$
Study of $g$. a. Show that $g ( x )$ is well defined for all $x \geqslant 0$. b. Show that $g$ is of class $C ^ { 2 }$ on $] 0 , \infty [$.
For this you may use the change of variable $u = t + x$ and express $g$ in terms of the functions $C : x \mapsto \int _ { x } ^ { \infty } \frac { \cos u } { u } \mathrm {~d} u$ and $S : x \mapsto \int _ { x } ^ { \infty } \frac { \sin u } { u } \mathrm {~d} u$. c. Determine a linear second-order differential equation satisfied by $f$.

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 $\mathbf { R } _ { + }$.

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 $] 0,1 [$.

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