From now on, $f$ denotes an infinitely differentiable function from $[ 0,1 ]$ to $\mathbb { R }$. We assume that there exists a unique point $x _ { 0 } \in \left[ 0,1 \left[ \right. \right.$ where $f ^ { \prime }$ vanishes. We also assume that $f ^ { \prime \prime } \left( x _ { 0 } \right) > 0$. We are also given an infinitely differentiable function $g : [ 0,1 ] \rightarrow \mathbb { R }$.
Show that we have, as $t \rightarrow + \infty$, $$\int _ { x _ { 0 } } ^ { 1 } g ( x ) \sin ( t f ( x ) ) \mathrm { d } x = g \left( x _ { 0 } \right) \int _ { x _ { 0 } } ^ { 1 } \sin ( t f ( x ) ) \mathrm { d } x + O \left( \frac { 1 } { t } \right)$$

We define, for all $x \in \mathbb{R}^{+*}$, $$\Gamma(x) = \int_0^{+\infty} t^{x-1} \mathrm{e}^{-t} \, \mathrm{d}t$$ Show that, for all $x \in \mathbb{R}^{+*}$, $$\Gamma(x+1) = x\Gamma(x)$$

We define, for all $x \in \mathbb{R}^{+*}$, $$\Gamma(x) = \int_0^{+\infty} t^{x-1} \mathrm{e}^{-t} \, \mathrm{d}t$$ Determine the value of $\Gamma(n)$, for $n \in \mathbb{N}^*$.

Let $\Gamma$ be the Gamma function defined as the pointwise limit on $]0, +\infty[$ of $\Gamma_n(x) = \frac{1}{x} e^{-\gamma x} \prod_{k=1}^{n} \frac{e^{x k^{-1}}}{1 + x k^{-1}}$.
Show that for all $a \in ]0, +\infty[$ and $x \in ]0, +\infty[$: $$\int_0^{+\infty} \frac{t^{x-1}}{(1+t)^{x+a}} dt = \frac{\Gamma(x)\Gamma(a)}{\Gamma(x+a)}.$$ Hint: you may set, for $x \in ]0, +\infty[$, $f(x) = \frac{\Gamma(x+a)}{\Gamma(a)} \int_0^{+\infty} \frac{t^{x-1}}{(1+t)^{x+a}} dt$.

Let $\Gamma$ be the Gamma function. Show that for all $x \in ]0,1[$: $$\int_0^{+\infty} \frac{t^{x-1}}{1+t} dt = \frac{\pi}{\sin(\pi x)}.$$

For all $n \in \mathbb { N }$, set $\mu _ { n } = \left( X ^ { n } \mid 1 \right)$ with the inner product $$( P \mid Q ) = \frac { 2 } { \pi } \int _ { 0 } ^ { 1 } P ( 4 x ) Q ( 4 x ) \frac { \sqrt { 1 - x } } { \sqrt { x } } \mathrm { ~d} x .$$ Using integration by parts, show $$\forall n \in \mathbb { N } ^ { * } , \quad 4 \mu _ { n - 1 } - \mu _ { n } = \frac { 2 \times 4 ^ { n } } { \pi } \int _ { 0 } ^ { 1 } x ^ { n - 3 / 2 } ( 1 - x ) ^ { 3 / 2 } \mathrm { d } x = \frac { 3 } { 2 n - 1 } \mu _ { n } .$$

Prove, without using what precedes, that
$$\int _ { 0 } ^ { + \infty } \frac { x e ^ { - x } } { 1 - e ^ { - x } } \mathrm {~d} x = \frac { \pi ^ { 2 } } { 6 }$$

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 $$u_k(t) = \int_{k/2}^{(k+1)/2} \frac{tq(u)}{e^{tu}-1} \mathrm{du} \text{ if } t > 0 \text{, and } u_k(t) = \int_{k/2}^{(k+1)/2} \frac{q(u)}{u} \mathrm{du} \text{ if } t = 0.$$ Let $t \in \mathbf{R}_+$. Show successively that $|u_k(t)| = \int_{k/2}^{(k+1)/2} \frac{t|q(u)|}{e^{tu}-1} du$ then $u_k(t) = (-1)^k |u_k(t)|$ for all integer $k \geq 1$, and finally establish that $$\forall n \in \mathbf{N}^*, \left|\sum_{k=n}^{+\infty} u_k(t)\right| \leq \frac{1}{2n}.$$

Show that, for all $k \in \mathbb { N } , \int _ { 0 } ^ { + \infty } t ^ { k } \mathrm { e } ^ { - t } \mathrm {~d} t = k !$.

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

To each function $f \in E$, we associate the function $U ( f )$ defined for all $x > 0$ by $U ( f ) ( x ) = \int _ { 0 } ^ { + \infty } \left( \mathrm { e } ^ { \min ( x , t ) } - 1 \right) f ( t ) \frac { \mathrm { e } ^ { - t } } { t } \mathrm {~d} t$. It has been shown that $\left| U ( f ) ^ { \prime } ( x ) \right| \leqslant \| f \| \frac { \mathrm { e } ^ { x / 2 } } { \sqrt { x } }$ and $\lim_{x\to 0} U(f)(x) = 0$. Deduce from the above that $U$ is an endomorphism of $E$ and that, for all $f \in E$ and all $x > 0$, $$| U ( f ) ( x ) | \leqslant 4 \| f \| \frac { \sqrt { x } \mathrm { e } ^ { x / 2 } } { 1 + x }$$

1. Let $a$ and $b$ be two real numbers with $a > 0$. Choose without justification the correct expression for $a ^ { b }$: $(A) : \mathrm { e } ^ { b \ln ( a ) }$ $(B) : \mathrm { e } ^ { a \ln ( b ) }$ $(C) : \mathrm { e } ^ { \ln ( a ) \ln ( b ) }$.
2. Let $x$ and $y$ be two real numbers such that $x < y$ and $t$ a real number in $]0,1[$. Compare $t ^ { x }$ and $t ^ { y }$.
3. Give, without proof, the power series expansion of the real exponential function and give its domain of validity.
4. We consider the function $\Gamma$ defined by $\Gamma ( x ) = \int _ { 0 } ^ { + \infty } t ^ { x - 1 } e ^ { - t } \mathrm {~d} t$. We admit that this function is defined on $]0 , + \infty[$ and that, for all strictly positive real $x$: $$\Gamma ( x + 1 ) = x \Gamma ( x )$$ Calculate $\Gamma ( 1 )$ and deduce, by using a proof by induction, the value of $\Gamma ( n + 1 )$ for $n \in \mathbb { N }$.
5. For $x \in \mathbb { R }$, we denote, when it makes sense: $$F ( x ) = \int _ { 0 } ^ { 1 } t ^ { t ^ { x } } \mathrm { d } t$$ where, as is customary, $t ^ { t ^ { x } } = t^{(t^{x})}$.
   1. [5.1.] Determine the domain of definition of $F$.
   2. [5.2.] Determine the monotonicity of $F$.
   3. [5.3.] Prove that for all non-negative real $x$, we have: $F ( x ) \geqslant \frac { 1 } { 2 }$.
   4. [5.4.] Prove that $F$ is continuous on its domain of definition.
   5. [5.5.] Determine $\lim _ { x \rightarrow + \infty } F ( x )$ and $\lim _ { x \rightarrow - \infty } F ( x )$. The theorems used will be cited with precision and we will ensure that their hypotheses are well verified.
   6. [5.6.] Then carefully draw up the table of variations of $F$ and give a general sketch of its representative curve in an orthonormal coordinate system. We will admit that $F ^ { \prime } ( 0 ) = \frac { 1 } { 4 }$ and we will draw the tangent line at the point with abscissa $x = 0$.
6. Let $x$ be a strictly positive real number. For every natural number $n$, we denote by $g _ { n }$ the function defined on $]0,1]$ by $g _ { n } ( t ) = \frac { t ^ { n x } \ln ^ { n } ( t ) } { n ! }$.
   1. [6.1.] Prove that the series of functions $\sum _ { n \in \mathbb { N } } g _ { n }$ converges pointwise on $]0,1]$ and give its sum.
   2. [6.2.] Prove that, for every natural number $n , \int _ { 0 } ^ { 1 } \left| g _ { n } ( t ) \right| \mathrm { d } t = \frac { 1 } { n ! } \frac { \Gamma ( n + 1 ) } { ( n x + 1 ) ^ { n + 1 } }$.
   3. [6.3.] Finally establish that we have: $$F ( x ) = \sum _ { n = 0 } ^ { + \infty } \frac { ( - 1 ) ^ { n } } { ( 1 + n x ) ^ { n + 1 } }$$

To every $p \in \mathbb{K}[X]$, we associate the function $L(p) = Lp$ from $\mathbb{K}$ to $\mathbb{K}$ defined by $$\forall x \in \mathbb{K}, \quad L(p)(x) = Lp(x) = -\int_0^{+\infty} \mathrm{e}^{-t} p'(x+t)\,\mathrm{d}t$$
Show that $\int_0^{+\infty} \mathrm{e}^{-t} t^k\,\mathrm{d}t$ exists for all $k \in \mathbb{N}$ and calculate its value.

Throughout this problem, $I = ]-1, +\infty[$, and $f(x) = \int_0^{\pi/2} (\sin(t))^x \mathrm{~d}t$.
Determine the domain of definition of $f$ and verify that $$\forall x \in I, (x+1)f(x) = (x+2)f(x+2)$$

Deduce that, as $n$ tends to $+ \infty$, $$I _ { n } \sim K _ { n } .$$ where $I _ { n } = \int _ { 0 } ^ { 1 } \frac { 1 } { \left( 1 + t ^ { 2 } \right) ^ { n } } \mathrm {~d} t$ and $K _ { n } = \int _ { 0 } ^ { + \infty } \frac { 1 } { \left( 1 + t ^ { 2 } \right) ^ { n } } \mathrm {~d} t$.

Throughout this problem, $I = ]-1, +\infty[$, and $f(x) = \int_0^{\pi/2} (\sin(t))^x \mathrm{~d}t$.
Justify that $f$ is of class $\mathcal{C}^2$, decreasing and convex on $I$.

Establish the recurrence relation $K _ { n } = K _ { n + 1 } + \frac { 1 } { 2 n } K _ { n }$, where $K _ { n } = \int _ { 0 } ^ { + \infty } \frac { 1 } { \left( 1 + t ^ { 2 } \right) ^ { n } } \mathrm {~d} t$.

Using the recurrence relation $K _ { n } = K _ { n + 1 } + \frac { 1 } { 2 n } K _ { n }$ and the fact that $I_n \sim K_n$, deduce a simple equivalent of $I _ { n }$ as $n$ tends to $+ \infty$, where $I _ { n } = \int _ { 0 } ^ { 1 } \frac { 1 } { \left( 1 + t ^ { 2 } \right) ^ { n } } \mathrm {~d} t$.

Throughout this problem, $I = ]-1, +\infty[$, and $f(x) = \int_0^{\pi/2} (\sin(t))^x \mathrm{~d}t$.
Show that for every natural number $n$, $$f(n)f(n+1) = \frac{\pi}{2(n+1)}$$ then that: $$f(x) \underset{x \to +\infty}{\sim} \sqrt{\frac{\pi}{2x}}$$

If $n \in \mathbf{N}$, we denote by $D_n$ the improper integral $\int_0^{\pi/2} (\ln(\sin(t)))^n \mathrm{~d}t$.
Verify that if $n \in \mathbf{N}^*$, then $$(-1)^n D_n = \int_0^{+\infty} \frac{u^n}{\sqrt{\mathrm{e}^{2u} - 1}} \mathrm{~d}u$$ then that $$D_n \underset{n \to +\infty}{\sim} (-1)^n n!$$

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 the function $g$ is of class $\mathcal { C } ^ { 1 }$ on $] - \pi ; \pi [$ and that for all $\theta \in ] - \pi ; \pi [$,
$$g ^ { \prime } ( \theta ) = \mathrm { i } e ^ { \mathrm { i } x \theta } \int _ { 0 } ^ { + \infty } h ^ { \prime } ( t ) \mathrm { d } t$$
where $h$ is the function defined by
$$\begin{aligned} h : ] 0 ; + \infty [ & \longrightarrow \mathbf { C } \\ t & \longmapsto \frac { t ^ { x } } { 1 + t e ^ { \mathrm { i } \theta } } . \end{aligned}$$
Calculate $h ( 0 )$ and
$$\lim _ { t \rightarrow + \infty } h ( t ) .$$
Deduce that the function $g$ is constant on $] - \pi ; \pi [$.

Let $f \in \mathcal { C } _ { b } ^ { 0 } ( [ 0 , + \infty [ )$. We define the Laplace transform of $f$ by the function $$\mathcal { L } ( f ) : t \in ] 0 , + \infty \left[ \mapsto \int _ { 0 } ^ { + \infty } e ^ { - t x } f ( x ) d x \right.$$ Prove that $\mathcal { L } ( f )$ is well-defined and of class $\mathcal { C } ^ { 1 }$ on $] 0 , + \infty [$, and express its derivative.

Let $f \in \mathcal { C } _ { b } ^ { 0 } ( [ 0 , + \infty [ )$. The purpose of this question is to prove that $$\left( \int _ { 0 } ^ { + \infty } f ( x ) d x \text { converges } \right) \Rightarrow \left( \lim _ { t \rightarrow 0 ^ { + } } \mathcal { L } ( f ) ( t ) = \int _ { 0 } ^ { + \infty } f ( x ) d x \right)$$
We assume that $\int _ { 0 } ^ { + \infty } f ( x ) d x$ converges.
(a) Prove that the function $$F : x \in \left[ 0 , + \infty \left[ \mapsto \int _ { x } ^ { + \infty } f ( t ) d t \right. \right.$$ is well-defined, continuous and bounded on $\left[ 0 , + \infty \right[$, of class $\mathcal { C } ^ { 1 }$ on $] 0 , + \infty [$ and satisfies $F ^ { \prime } = - f$.
(b) Prove that $\lim _ { t \rightarrow 0 ^ { + } } \mathcal { L } ( f ) ( t ) = \int _ { 0 } ^ { + \infty } f ( x ) d x$ by integration by parts.