For every natural integer $k$ we set $$m_{k} = \frac{1}{2\pi} \int_{-2}^{2} x^{k} \sqrt{4 - x^{2}} \, \mathrm{d}x$$ Using the change of variable $x = 2\sin t$, calculate $m_{0}$.

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

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

Throughout this problem, $I = ]-1, +\infty[$, and $f(x) = \int_0^{\pi/2} (\sin(t))^x \mathrm{~d}t$. If $n \in \mathbf{N}$, we denote by $D_n$ the improper integral $\int_0^{\pi/2} (\ln(\sin(t)))^n \mathrm{~d}t$.
Calculate $f'(0)$ and $f'(1)$.

Let $r$ be the function defined by
$$\begin{aligned} r : ] - \pi ; \pi [ & \longrightarrow \mathbf { C } \\ \theta & \longmapsto \int _ { 0 } ^ { + \infty } \frac { t ^ { x - 1 } } { 1 + t e ^ { \mathrm { i } \theta } } \mathrm {~d} t \end{aligned}$$
where $x$ is a fixed element of $]0;1[$. Show that the function $r$ is of class $\mathcal { C } ^ { 1 }$ on $] - \pi ; \pi [$ and that:
$$\forall \theta \in ] - \pi ; \pi \left[ , \quad r ^ { \prime } ( \theta ) = - \mathrm { i } e ^ { \mathrm { i } \theta } \cdot \int _ { 0 } ^ { + \infty } \frac { t ^ { x } } { \left( 1 + t \mathrm { e } ^ { \mathrm { i } \theta } \right) ^ { 2 } } \mathrm {~d} t . \right.$$
Hint: let $\beta \in ] 0 ; \pi [$, show that for all $\theta \in [ - \beta ; \beta ]$ and $t \in [ 0 , + \infty [$, $\left| 1 + t e ^ { i \theta } \right| ^ { 2 } \geq \left| 1 + t e ^ { i \beta } \right| ^ { 2 } = ( t + \cos ( \beta ) ) ^ { 2 } + ( \sin ( \beta ) ) ^ { 2 }$.

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 $$I _ { p , q } ( t ) := \int _ { 0 } ^ { 1 } \frac { x ^ { ( t + 1 ) \alpha _ { p , q } } } { 1 + x ^ { \alpha _ { p , q } } } d x$$ Show that, for all $( p , q ) \in \left( \mathrm { N } ^ { * } \right) ^ { 2 }$, $$S _ { p , q } = \int _ { 0 } ^ { 1 } \frac { t ^ { q - 1 } } { 1 + t ^ { p } } d t$$

Show that $$\forall u \in \mathbb{R}, \quad \forall a \in \mathbb{R}_+^*, \quad \mathrm{e}^{\frac{au^2}{2}} = \int_{-\infty}^{+\infty} \mathrm{e}^{ut - \frac{t^2}{2a}} \frac{\mathrm{~d}t}{\sqrt{2\pi a}}$$

In this subsection, we assume that $J_n = J_n^{(\mathrm{C})}$, the matrix introduced in subsection A-II.
Deduce that $$Z_n(h) = \sqrt{\frac{n}{2\mathrm{e}^{\beta}\pi\beta}} \int_{-\infty}^{+\infty} \left(\sum_{x \in \Lambda_n} \mathrm{e}^{(t+h)s_n(x)}\right) \mathrm{e}^{-\frac{nt^2}{2\beta}} \mathrm{~d}t$$

In this subsection, we assume that $J_n = J_n^{(\mathrm{C})}$, the matrix introduced in subsection A-II.
We set $G_h : x \longmapsto \frac{(x-h)^2}{2\beta} - \ln(2\operatorname{ch}(x))$.
Show that $$Z_n(h) = \sqrt{\frac{n}{2\mathrm{e}^{\beta}\pi\beta}} \int_{-\infty}^{+\infty} \mathrm{e}^{-nG_h(x)} \mathrm{d}x$$

In this subsection, we assume that $J_n = J_n^{(\mathrm{C})}$, the matrix introduced in subsection A-II.
We set $G_h : x \longmapsto \frac{(x-h)^2}{2\beta} - \ln(2\operatorname{ch}(x))$. We now assume that $h > 0$.
We denote $\widehat{G}_h : x \longmapsto G_h(x + u_h) - \min G_h$. Show that $$\psi_n(h) = -G_h(u_h) - \frac{1}{2n}\ln\left(2\mathrm{e}^{\beta}\pi\beta\right) + \frac{1}{n}\ln\left(\int_{-\infty}^{+\infty} \mathrm{e}^{-n\widehat{G}_h\left(\frac{t}{\sqrt{n}}\right)} \mathrm{d}t\right)$$

If $b = \int _ { 0 } ^ { 1 } \frac { e ^ { t } } { t + 1 } d t$ then $\int _ { a - 1 } ^ { a } \frac { e ^ { - t } } { t - a - 1 } d t$ is
(A) $b e ^ { a }$
(B) $b e ^ { - a }$
(C) $- b e ^ { - a }$
(D) $- b e ^ { a }$

Evaluate $\displaystyle I = \int_{1/2014}^{2014} \frac{\tan^{-1} x}{x}\, dx$.
(A) $\dfrac{\pi}{2} \log(2014)$ (B) $\pi \log(2014)$ (C) $2\pi \log(2014)$ (D) $\dfrac{\pi}{4} \log(2014)$

If $b = \int _ { 0 } ^ { 1 } \frac { e ^ { t } } { t + 1 } d t$ then $\int _ { a - 1 } ^ { a } \frac { e ^ { - t } } { t - a - 1 } d t$ is
(A) $b e ^ { a }$
(B) $b e ^ { - a }$
(C) $- b e ^ { - a }$
(D) $- b e ^ { a }$

If $b = \int _ { 0 } ^ { 1 } \frac { e ^ { t } } { t + 1 } d t$ then $\int _ { a - 1 } ^ { a } \frac { e ^ { - t } } { t - a - 1 } d t$ is
(A) $b e ^ { a }$
(B) $b e ^ { - a }$
(C) $- b e ^ { - a }$
(D) $- b e ^ { a }$

The value of the integral $$\int _ { \pi / 2 } ^ { 5 \pi / 2 } \frac { e ^ { \tan ^ { - 1 } ( \sin x ) } } { e ^ { \tan ^ { - 1 } ( \sin x ) } + e ^ { \tan ^ { - 1 } ( \cos x ) } } d x$$ equals
(A) 1
(B) $\pi$
(C) $e$
(D) none of these

The value of the integral $$\int _ { \pi / 2 } ^ { 5 \pi / 2 } \frac { e ^ { \tan ^ { - 1 } ( \sin x ) } } { e ^ { \tan ^ { - 1 } ( \sin x ) } + e ^ { \tan ^ { - 1 } ( \cos x ) } } d x$$ equals
(A) 1
(B) $\pi$
(C) $e$
(D) none of these

The following integral
$$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} (2\operatorname{cosec} x)^{17}\, dx$$
is equal to
(A) $\int_{0}^{\log(1+\sqrt{2})} 2\left(e^{u} + e^{-u}\right)^{16} du$
(B) $\int_{0}^{\log(1+\sqrt{2})} \left(e^{u} + e^{-u}\right)^{17} du$
(C) $\int_{0}^{\log(1+\sqrt{2})} \left(e^{u} - e^{-u}\right)^{17} du$
(D) $\int_{0}^{\log(1+\sqrt{2})} 2\left(e^{u} - e^{-u}\right)^{16} du$

If
$$\alpha = \int _ { 0 } ^ { 1 } \left( e ^ { 9 x + 3 \tan ^ { - 1 } x } \right) \left( \frac { 12 + 9 x ^ { 2 } } { 1 + x ^ { 2 } } \right) d x$$
where $\tan ^ { - 1 } x$ takes only principal values, then the value of $\left( \log _ { e } | 1 + \alpha | - \frac { 3 \pi } { 4 } \right)$ is

If $I = \sum _ { k = 1 } ^ { 98 } \int _ { k } ^ { k + 1 } \frac { k + 1 } { x ( x + 1 ) } d x$, then
[A] $I > \log _ { e } 99$
[B] $I < \log _ { e } 99$
[C] $I < \frac { 49 } { 50 }$
[D] $I > \frac { 49 } { 50 }$

The value of the integral
$$\int _ { 0 } ^ { \frac { 1 } { 2 } } \frac { 1 + \sqrt { 3 } } { \left( ( x + 1 ) ^ { 2 } ( 1 - x ) ^ { 6 } \right) ^ { \frac { 1 } { 4 } } } d x$$
is $\_\_\_\_$ .

Consider the equation
$$\int _ { 1 } ^ { e } \frac { \left( \log _ { \mathrm { e } } x \right) ^ { 1 / 2 } } { x \left( a - \left( \log _ { \mathrm { e } } x \right) ^ { 3 / 2 } \right) ^ { 2 } } d x = 1 , \quad a \in ( - \infty , 0 ) \cup ( 1 , \infty )$$
Which of the following statements is/are TRUE ?
(A) No $a$ satisfies the above equation
(B) An integer $a$ satisfies the above equation
(C) An irrational number $a$ satisfies the above equation
(D) More than one $a$ satisfy the above equation

If
$$\alpha = \int _ { \frac { 1 } { 2 } } ^ { 2 } \frac { \tan ^ { - 1 } x } { 2 x ^ { 2 } - 3 x + 2 } d x$$
then the value of $\sqrt { 7 } \tan \left( \frac { 2 \alpha \sqrt { 7 } } { \pi } \right)$ is $\_\_\_\_$. (Here, the inverse trigonometric function $\tan ^ { - 1 } x$ assumes values in $\left( - \frac { \pi } { 2 } , \frac { \pi } { 2 } \right)$.)

$\int \frac { d x } { \cos x + \sqrt { 3 } \sin x }$ equals
(1) $\frac { 1 } { 2 } \log \tan \left( \frac { x } { 2 } + \frac { \pi } { 12 } \right) + c$
(2) $\frac { 1 } { 2 } \log \tan \left( \frac { x } { 2 } - \frac { \pi } { 12 } \right) + c$
(3) $\log \tan \left( \frac { x } { 2 } + \frac { \pi } { 12 } \right) + c$
(4) $\log \tan \left( \frac { x } { 2 } - \frac { \pi } { 12 } \right) + c$