The function $q$ associates to any real $x$ the real number $q ( x ) = x - \lfloor x \rfloor - \frac { 1 } { 2 }$, where $\lfloor x \rfloor$ denotes the integer part of $x$.
Show that for all integer $n > 1$,
$$\int _ { 1 } ^ { n } \frac { q ( u ) } { u } \mathrm {~d} u = \ln ( n ! ) + ( n - 1 ) - n \ln ( n ) - \frac { 1 } { 2 } \ln ( n ) = \ln \left( \frac { n ! e ^ { n } } { n ^ { n } \sqrt { n } } \right) - 1$$

Show that for all integer $n > 1$, $$\int_{1}^{n} \frac{q(u)}{u} \mathrm{~d}u = \ln(n!) + (n-1) - n\ln(n) - \frac{1}{2}\ln(n) = \ln\left(\frac{n! e^n}{n^n \sqrt{n}}\right) - 1.$$

The function $q$ associates to any real $x$ the real number $q ( x ) = x - \lfloor x \rfloor - \frac { 1 } { 2 }$, where $\lfloor x \rfloor$ denotes the integer part of $x$.
Show that $\int _ { \lfloor x \rfloor } ^ { x } \frac { q ( u ) } { u } \mathrm {~d} u$ tends to 0 as $x$ tends to $+ \infty$, and deduce the convergence of the integral $\int _ { 1 } ^ { + \infty } \frac { q ( u ) } { u } \mathrm {~d} u$, as well as the equality
$$\int _ { 1 } ^ { + \infty } \frac { q ( u ) } { u } \mathrm {~d} u = \frac { \ln ( 2 \pi ) } { 2 } - 1$$

Show that $\int_{\lfloor x \rfloor}^{x} \frac{q(u)}{u} \mathrm{~d}u$ tends to 0 when $x$ tends to $+\infty$, and deduce the convergence of the integral $\int_{1}^{+\infty} \frac{q(u)}{u} \mathrm{~d}u$, as well as the equality $$\int_{1}^{+\infty} \frac{q(u)}{u} \mathrm{~d}u = \frac{\ln(2\pi)}{2} - 1$$

Show that $$\int_{0}^{1} \ln\left(\frac{1-e^{-tu}}{t}\right) \mathrm{d}u \underset{t \rightarrow 0^+}{\longrightarrow} -1.$$ One may begin by establishing that $x \mapsto \frac{1-e^{-x}}{x}$ is decreasing on $\mathbf{R}_+$.

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 $q ( x ) = x - \lfloor x \rfloor - \frac { 1 } { 2 }$.
Show that $u _ { k }$ is continuous on $\mathbf { R } _ { + }$ for all $k \in \mathbf { N } ^ { * }$.

For $k \in \mathbf{N}^*$ and $t \in \mathbf{R}_+$, we set $$u_k(t) = \int_{k/2}^{(k+1)/2} \frac{tq(u)}{e^{tu}-1} \mathrm{du} \text{ if } t > 0 \text{, and } u_k(t) = \int_{k/2}^{(k+1)/2} \frac{q(u)}{u} \mathrm{du} \text{ if } t = 0.$$ Show that $u_k$ is continuous on $\mathbf{R}$, for all $k \in \mathbf{N}^*$.

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 $$\forall n \in \mathbf{N}^*, \left|\sum_{k=n}^{+\infty} u_k(t)\right| \leq \frac{1}{2n}.$$

For any real $\alpha > 0$, consider the function $h _ { \alpha } : t \mapsto \ln \left( \frac { 1 - t ^ { 2 } } { \alpha ^ { 2 } + t ^ { 2 } } \right)$ and set $J _ { \alpha } = \int _ { 0 } ^ { 1 } h _ { \alpha } ( t ) \, \mathrm { d } t$. Deduce that $$J _ { \alpha } = 2 \ln ( 2 ) - \ln \left( 1 + \alpha ^ { 2 } \right) - 2 \alpha \arctan \left( \frac { 1 } { \alpha } \right).$$

For all $n \in \mathbb { N } ^ { * }$, consider in $]0,1[$ the points $a _ { k , n }$ given, for $k \in \llbracket 0 , n - 1 \rrbracket$, by $a _ { k , n } = \frac { 2 k + 1 } { 2 n }$ and $$S _ { n } \left( h _ { \alpha } \right) = \frac { 1 } { n } \sum _ { k = 0 } ^ { n - 1 } h _ { \alpha } \left( a _ { k , n } \right).$$ For all $n \in \mathbb { N } ^ { * }$, show that $$\int _ { 1 / 2 n } ^ { ( 2 n - 1 ) / 2 n } h _ { \alpha } ( t ) \, \mathrm { d } t + \frac { 1 } { n } h _ { \alpha } \left( \frac { 2 n - 1 } { 2 n } \right) \leqslant S _ { n } \left( h _ { \alpha } \right) \leqslant \frac { 1 } { n } h _ { \alpha } \left( \frac { 1 } { 2 n } \right) + \int _ { 1 / 2 n } ^ { ( 2 n - 1 ) / 2 n } h _ { \alpha } ( t ) \, \mathrm { d } t.$$

For all $n \in \mathbb { N } ^ { * }$, consider in $]0,1[$ the points $a _ { k , n } = \frac { 2 k + 1 } { 2 n }$ for $k \in \llbracket 0 , n - 1 \rrbracket$, and $S _ { n } \left( h _ { \alpha } \right) = \frac { 1 } { n } \sum _ { k = 0 } ^ { n - 1 } h _ { \alpha } \left( a _ { k , n } \right)$, where $J_\alpha = \int_0^1 h_\alpha(t)\,\mathrm{d}t$. Deduce that the sequence $\left( S _ { n } \left( h _ { \alpha } \right) \right) _ { n \in \mathbb { N } ^ { * } }$ converges to $J _ { \alpha }$.

For all $n \in \mathbb { N } ^ { * }$, consider the points $a _ { k , n } = \frac { 2 k + 1 } { 2 n }$ for $k \in \llbracket 0 , n - 1 \rrbracket$. Let $\gamma > 0$ be such that $J_\alpha > 0$ for all $\alpha \in ]0, \gamma[$. Show that, for $\alpha \in ] 0 , \gamma [$, the sequence $\left( \left| \prod _ { k = 0 } ^ { n - 1 } \frac { 1 - a _ { k , n } ^ { 2 } } { \alpha ^ { 2 } + a _ { k , n } ^ { 2 } } \right| \right) _ { n \in \mathbb { N } ^ { * } }$ diverges to $+ \infty$.

Let $I = [ - 1,1 ]$, $\alpha > 0$, and $$f _ { \alpha } : \begin{array}{ccc} {[-1,1]} & \rightarrow & \mathbb{R} \\ x & \mapsto & \dfrac{1}{\alpha^2 + x^2} \end{array}.$$ For $n \in \mathbb{N}^*$, let $a_{k,n} = \frac{2k+1}{2n}$ for $k \in \llbracket 0, n-1 \rrbracket$, and let $R_n \in \mathbb{R}_{2n-1}[X]$ be the interpolating polynomial of $f_\alpha$ at the $2n$ real numbers $\{\pm a_{k,n} \mid k \in \llbracket 0, n-1 \rrbracket\}$. Set $Q_n(X) = 1 - (X^2 + \alpha^2) R_n(X)$. Show that $R_n$ is an even polynomial and determine $Q_n(\alpha \mathrm{i})$.

Let $I = [-1,1]$, $\alpha > 0$, $a_{k,n} = \frac{2k+1}{2n}$ for $k \in \llbracket 0, n-1 \rrbracket$, $R_n \in \mathbb{R}_{2n-1}[X]$ the interpolating polynomial of $f_\alpha(x) = \frac{1}{\alpha^2+x^2}$ at $\{\pm a_{k,n} \mid k \in \llbracket 0,n-1\rrbracket\}$, and $Q_n(X) = 1 - (X^2 + \alpha^2)R_n(X)$. Show that there exists $\lambda_n \in \mathbb{R}$ such that $$\forall x \in [ - 1,1 ] , \quad Q _ { n } ( x ) = \lambda _ { n } \prod _ { k = 0 } ^ { n - 1 } \left( x ^ { 2 } - a _ { k , n } ^ { 2 } \right).$$

Let $I = [-1,1]$, $\alpha > 0$, $a_{k,n} = \frac{2k+1}{2n}$ for $k \in \llbracket 0, n-1 \rrbracket$, $R_n \in \mathbb{R}_{2n-1}[X]$ the interpolating polynomial of $f_\alpha(x) = \frac{1}{\alpha^2+x^2}$ at $\{\pm a_{k,n} \mid k \in \llbracket 0,n-1\rrbracket\}$, and $Q_n(X) = \lambda_n \prod_{k=0}^{n-1}(x^2 - a_{k,n}^2)$. Deduce that, for all $x \in [ - 1,1 ]$, $$f _ { \alpha } ( x ) - R _ { n } ( x ) = \frac { ( - 1 ) ^ { n } } { x ^ { 2 } + \alpha ^ { 2 } } \prod _ { k = 0 } ^ { n - 1 } \frac { x ^ { 2 } - a _ { k , n } ^ { 2 } } { \alpha ^ { 2 } + a _ { k , n } ^ { 2 } }.$$

Let $I = [-1,1]$, $\alpha > 0$, $a_{k,n} = \frac{2k+1}{2n}$ for $k \in \llbracket 0, n-1 \rrbracket$, $R_n \in \mathbb{R}_{2n-1}[X]$ the interpolating polynomial of $f_\alpha(x) = \frac{1}{\alpha^2+x^2}$ at $\{\pm a_{k,n} \mid k \in \llbracket 0,n-1\rrbracket\}$, and $\gamma > 0$ such that $J_\alpha > 0$ for all $\alpha \in ]0,\gamma[$. Suppose that $\alpha < \gamma$. Show that $$\lim _ { n \rightarrow + \infty } \left| f _ { \alpha } ( 1 ) - R _ { n } ( 1 ) \right| = + \infty.$$

Show that $$I _ { n } \geqslant \frac { 1 } { 2 ^ { n } }.$$ where $I _ { n } = \int _ { 0 } ^ { 1 } \frac { 1 } { \left( 1 + t ^ { 2 } \right) ^ { n } } \mathrm {~d} t$.

Justify the existence of $K _ { n } = \int _ { 0 } ^ { + \infty } \frac { 1 } { \left( 1 + t ^ { 2 } \right) ^ { n } } \mathrm {~d} t$ and give the exact value of $K _ { 1 }$.

Show that $$\int _ { 1 } ^ { + \infty } \frac { 1 } { \left( 1 + t ^ { 2 } \right) ^ { n } } \mathrm {~d} t = O \left( \frac { 1 } { n 2 ^ { n } } \right) .$$ One may lower bound $1 + t ^ { 2 }$ by a polynomial of degree 1.

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

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: $$f(x) \underset{x \to +\infty}{\sim} \sqrt{\frac{\pi}{2x}}$$

Justify that $$\sqrt { n } I _ { n } = \int _ { 0 } ^ { \sqrt { n } } \frac { 1 } { \left( 1 + u ^ { 2 } / n \right) ^ { n } } \mathrm {~d} u$$ where $I _ { n } = \int _ { 0 } ^ { 1 } \frac { 1 } { \left( 1 + t ^ { 2 } \right) ^ { n } } \mathrm {~d} t$.