For every natural number $n$, we consider the following integrals:
$$I_n = \int_0^{\pi} \mathrm{e}^{-nx} \sin(x) \mathrm{d}x, \quad J_n = \int_0^{\pi} \mathrm{e}^{-nx} \cos(x) \mathrm{d}x$$

1. Calculate $I_0$.
2. a. Justify that, for every natural number $n$, we have $I_n \geqslant 0$. b. Show that, for every natural number $n$, we have $I_{n+1} - I_n \leqslant 0$. c. Deduce from the two previous questions that the sequence $(I_n)$ converges.
3. a. Show that, for every natural number $n$, we have: $$I_n \leqslant \int_0^{\pi} \mathrm{e}^{-nx} \mathrm{d}x$$ b. Show that, for every natural number $n \geqslant 1$, we have: $$\int_0^{\pi} \mathrm{e}^{-nx} \mathrm{d}x = \frac{1 - \mathrm{e}^{-n\pi}}{n}.$$ c. Deduce from the two previous questions the limit of the sequence $(I_n)$.
4. a. By integrating by parts the integral $I_n$ in two different ways, establish the two following relations, for every natural number $n \geqslant 1$: $$I_n = 1 + \mathrm{e}^{-n\pi} - nJ_n \quad \text{and} \quad I_n = \frac{1}{n}J_n$$ b. Deduce that, for every natural number $n \geqslant 1$, we have $$I_n = \frac{1 + \mathrm{e}^{-n\pi}}{n^2 + 1}$$
5. It is desired to obtain the rank $n$ from which the sequence $(I_n)$ becomes less than $0.1$. Copy and complete the fifth line of the Python script below with the appropriate command. \begin{verbatim} from math import * def seuil() : n = 0 I = 2 ... n=n+1 I =(1+exp(-n*pi))/(n*n+1) return n \end{verbatim}

We study the sequence $\left(u_{n}\right)_{n \in \mathbb{N}^{*}}$ defined by $$\forall n \in \mathbb{N}^{*}, \quad u_{n} = \int_{0}^{\infty} \frac{1 - (\cos t)^{n}}{t^{2}} \mathrm{~d}t$$ Show that $u_{1} = u_{2} = \frac{\pi}{2}$.

We admit the identity $\int_{-\infty}^{+\infty} \exp(-x^2) \mathrm{d}x = \sqrt{\pi}$.
(a) Show that for all integer $n \in \mathbb{N}$, we have $$n! = \int_0^{+\infty} e^{-t} t^n \mathrm{d}t$$
(b) Using the preceding results, recover Stirling's formula giving an asymptotic equivalent of $n!$.

We admit the identity $$\int _ { - \infty } ^ { + \infty } \exp \left( - x ^ { 2 } \right) \mathrm { d } x = \sqrt { \pi }$$
(a) Show that for all integer $n \in \mathbb { N }$, we have $$n ! = \int _ { 0 } ^ { + \infty } e ^ { - t } t ^ { n } \mathrm {~d} t$$
(b) Using the preceding results, recover Stirling's formula giving an asymptotic equivalent of $n !$.

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

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.

Let
$$a _ { n } = \int _ { 0 } ^ { 1 } x ^ { 2n } \sqrt { 1 - x ^ { 2 } } \, d x \quad ( n = 0,1,2 , \cdots )$$
We are to find the value of the limit $\lim _ { n \rightarrow \infty } \frac { a _ { n } } { a _ { n - 1 } }$.
(1) First, let us find $a _ { 0 }$ and $a _ { 1 }$. Since the area of a circle with the radius 1 is $\pi$, we see that
$$a _ { 0 } = \int _ { 0 } ^ { 1 } \sqrt { 1 - x ^ { 2 } } \, d x = \frac { \pi } { \mathbf { A } }$$
Next, the partial integral method applied to $a _ { 1 }$ gives
$$\begin{aligned} a _ { 1 } & = \int _ { 0 } ^ { 1 } x ^ { 2 } \sqrt { 1 - x ^ { 2 } } \, d x \\ & = - \frac { \mathbf { B } } { \mathbf { C } } \left[ x \left( 1 - x ^ { 2 } \right) ^ { \frac { \mathbf { D } } { \mathbf { E } } } \right] _ { 0 } ^ { 1 } + \frac { \mathbf { F } } { \mathbf { G } } \int _ { 0 } ^ { 1 } \left( 1 - x ^ { 2 } \right) ^ { \frac { \mathbf { H } } { \mathbf{I} } } d x \\ & = \frac { \mathbf { J } } { \mathbf { K } } \left\{ \int _ { 0 } ^ { 1 } \sqrt { 1 - x ^ { 2 } } \, d x - \int _ { 0 } ^ { 1 } x ^ { \mathbf { L } } \sqrt { 1 - x ^ { 2 } } \, d x \right\} \end{aligned}$$
Thus we have
$$a _ { 1 } = \frac { \pi } { \mathbf { M N } } .$$
(2) For $\mathbf { O } \sim \mathbf { U }$ in the following sentences, choose the correct answer from among choices (0) $\sim$ (9) below.
When the partial integral method is applied to $a _ { n }$ in the same way as for $a _ { 1 }$, we get
$$a _ { n } = \frac { \mathbf { O } } { \mathbf { P } } \left\{ \int _ { 0 } ^ { 1 } x ^ { \mathbf { Q } } \sqrt { 1 - x ^ { 2 } } \, d x - \int _ { 0 } ^ { 1 } x ^ { \mathbf { R } } \sqrt { 1 - x ^ { 2 } } \, d x \right\} \quad ( n = 1,2,3 , \cdots )$$
Hence we have
$$( \mathbf { S } ) a _ { n } = ( \mathbf { T } ) a _ { n - 1 } ,$$
and so
$$\lim _ { n \rightarrow \infty } \frac { a _ { n } } { a _ { n - 1 } } = \mathbf { U }$$
(0) 0
(1) 1
(2) 2
(3) 3
(4) 4
(5) $2 n - 2$ (6) $2 n - 1$ (7) $2 n$ (8) $2 n + 1$ (9) $2 n + 2$