Prove that $$\int _ { 0 } ^ { + \infty } \frac { \sin ( x ) } { x } d x = \frac { \pi } { 2 }$$

Let $f \in \mathcal { C } _ { b } ^ { 0 } ( [ 0 , + \infty [ )$ and $S \in \mathbb { R }$. Prove that $$\left( \lim _ { t \rightarrow 0 ^ { + } } \mathcal { L } ( f ) ( t ) = S \text { and } f ( t ) \underset { t \rightarrow + \infty } { = } O \left( \frac { 1 } { t } \right) \right) \Rightarrow \left( \int _ { 0 } ^ { + \infty } f ( x ) d x \text { converges and } \int _ { 0 } ^ { + \infty } f ( x ) d x = S \right)$$
For this, using the notations of question 5 of section 2, one can prove that there exist $M > 0$ and $A > 0$ such that for all $t > 0$ $$\begin{aligned} \left| \int _ { A } ^ { + \infty } f ( x ) g \left( e ^ { - t x } \right) d x - \int _ { A } ^ { + \infty } f ( x ) P _ { 1 } \left( e ^ { - t x } \right) d x \right| & \leqslant M \int _ { A } ^ { + \infty } Q \left( e ^ { - t x } \right) e ^ { - t x } \frac { 1 - e ^ { - t x } } { x } d x \\ & \leqslant M \int _ { 0 } ^ { 1 } Q ( u ) d u \end{aligned}$$

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 for all $\theta \in ] 0 ; \pi [$,
$$g ( \theta ) \sin ( x \theta ) = \frac { 1 } { 2 \mathrm { i } } \left( g ( - \theta ) e ^ { \mathrm { i } x \theta } - g ( \theta ) e ^ { - \mathrm { i } x \theta } \right) = \sin ( \theta ) \int _ { 0 } ^ { + \infty } \frac { t ^ { x } } { t ^ { 2 } + 2 t \cos ( \theta ) + 1 } \mathrm {~d} t$$

Let $f : \mathbb { R } _ { + } \rightarrow \mathbb { R }$, be a piecewise continuous function, strictly positive and integrable. For all $x > 0$, we define
$$g ( x ) = \frac { 1 } { x } \int _ { 0 } ^ { x } t f ( t ) \mathrm { d } t \quad \text { and } \quad h ( x ) = \frac { 1 } { x } g ( x ) = \frac { 1 } { x ^ { 2 } } \int _ { 0 } ^ { x } t f ( t ) \mathrm { d } t .$$
Deduce that the integral $\int _ { 0 } ^ { + \infty } h ( x ) \mathrm { d } x$ converges and that
$$\int _ { 0 } ^ { + \infty } f ( x ) \mathrm { d } x = \int _ { 0 } ^ { + \infty } h ( x ) \mathrm { d } x .$$
You may use integration by parts.

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 $]0;1[$. Deduce that:
$$\forall \theta \in ] 0 ; \pi \left[ , \quad g ( \theta ) \sin ( \theta x ) = \int _ { \cot ( \theta ) } ^ { + \infty } \frac { ( u \sin ( \theta ) - \cos ( \theta ) ) ^ { x } } { 1 + u ^ { 2 } } \mathrm {~d} u , \right.$$
where $\cot ( \theta ) = \frac { \cos ( \theta ) } { \sin ( \theta ) }$.

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: $$\int_0^1 h(t)\, dt = 2\int_0^{\frac{1}{2}} \tilde{h}(t)\, dt.$$

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 $[ k - 1 , k [$.
Prove that, for all $k$ in $\mathbb { N } ^ { * }$,
$$\int _ { k - 1 } ^ { k } \exp \left( \frac { 1 } { x } \int _ { 0 } ^ { x } \ln ( f ( t ) ) \mathrm { d } t \right) \mathrm { d } x \geqslant \exp \left( \frac { 1 } { k } \sum _ { i = 1 } ^ { k } \ln \left( a _ { i } \right) \right)$$
You may use the previous question.

Deduce that:
$$\int _ { 0 } ^ { + \infty } \frac { 1 - ( \cos ( t ) ) ^ { 2 p + 1 } } { t ^ { 2 } } \mathrm {~d} t = \frac { \pi } { 2 } \frac { ( 2 p + 1 ) ! } { 2 ^ { 2 p } \cdot ( p ! ) ^ { 2 } }$$

Show that for all $s \in \mathbf { R }$
$$\int _ { 0 } ^ { + \infty } \frac { 1 - \cos ( s t ) } { t ^ { 2 } } \mathrm {~d} t = \frac { \pi } { 2 } | s |$$

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.

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 $$I _ { p , q } ( t ) := \int _ { 0 } ^ { 1 } \frac { x ^ { ( t + 1 ) \alpha _ { p , q } } } { 1 + x ^ { \alpha _ { p , q } } } d x$$ For all $x \in [ 0,1 ]$, calculate $\sum _ { k = 0 } ^ { n } \left( - x ^ { \alpha _ { p , q } } \right) ^ { k }$ then deduce that $$\phi _ { p , q } ( n ) = \frac { 1 } { q } \left( \int _ { 0 } ^ { 1 } \frac { 1 } { 1 + x ^ { \alpha _ { p , q } } } d x + ( - 1 ) ^ { n } I _ { p , q } ( n ) \right)$$

Let $I _ { n } = \int _ { 0 } ^ { 1 } ( 1 - x ^ { 3 } ) ^ { n } d x$, where $n \in \mathbb { N }$. Then $\frac { 3 n + 1 } { 3 n } \cdot \frac { I _ { n + 1 } } { I _ { n } }$ is equal to:
(1) 1
(2) $\frac { n } { n + 1 }$
(3) $\frac { n + 1 } { n }$
(4) $\frac { 3 n + 1 } { 3 n - 2 }$

Let $I _ { n } = \int _ { 1 } ^ { e } x ^ { 19 } ( \log | x | ) ^ { n } d x$, where $n \in N$. If (20) $I _ { 10 } = \alpha I _ { 9 } + \beta I _ { 8 }$, for natural numbers $\alpha$ and $\beta$, then $\alpha - \beta$ equal to $\_\_\_\_$ .

Let $I _ { n } ( x ) = \int _ { 0 } ^ { x } \frac { 1 } { \left( t ^ { 2 } + 5 \right) ^ { n } } d t , n = 1,2,3 , \ldots$. Then
(1) $50 I _ { 6 } - 9 I _ { 5 } = x I _ { 5 } ^ { \prime }$
(2) $50 I _ { 6 } - 11 I _ { 5 } = x I _ { 5 } ^ { \prime }$
(3) $50 I _ { 6 } - 9 I _ { 5 } = I _ { 5 } ^ { \prime }$
(4) $50 I _ { 6 } - 11 I _ { 5 } = I _ { 5 } ^ { \prime }$

Given the function in $x$
$$f _ { n } ( x ) = \sin ^ { n } x \quad ( n = 1,2,3 , \cdots ) ,$$
answer the following questions.
(1) Consider the cases in which the equality
$$\lim _ { x \rightarrow 0 } \frac { a - x ^ { 2 } - \left( b - x ^ { 2 } \right) ^ { 2 } } { f _ { n } ( x ) } = c$$
holds for three real numbers $a , b$ and $c$.
(i) We have $a = b$.
(ii) When $n = 2$, if $c = 6$, then $b = \frac { \mathbf { P } } { \mathbf { Q } }$.
(iii) When $n = 4$, then $b = \frac { \mathbf { R } } { \mathbf { S } }$ and $c = - \mathbf { T }$.
(2) For this $f _ { n } ( x )$, consider the definite integral
$$I _ { n } = \int _ { 0 } ^ { \frac { \pi } { 2 } } f _ { n } ( x ) \sin 2 x \, d x \quad ( n = 1,2,3 , \cdots )$$
When the integral is calculated, we have
$$I _ { n } = \frac { \mathbf { U } } { n + \mathbf { V } } .$$
Hence we obtain
$$\lim _ { n \rightarrow \infty } \left( I _ { n - 1 } + I _ { n } + I _ { n + 1 } + \cdots + I _ { 2 n - 2 } \right) = \int _ { 0 } ^ { \mathbf { W } } \frac { \mathbf { X } } { \mathbf { Y } + x } \, dx = \log \mathbf { Z }$$

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) $2 n + 2$

For every integer $m$ greater than 1
$$\int \tan ^ { m } x d x = \frac { 1 } { m - 1 } \tan ^ { m - 1 } x - \int \tan ^ { m - 2 } x d x$$
the equality is satisfied. Accordingly, what is the value of the integral $\int _ { 0 } ^ { \frac { \pi } { 4 } } \tan ^ { 4 } \mathrm { xdx }$?
A) $\frac { 2 \pi + 3 } { 4 }$
B) $\frac { 4 \pi - 3 } { 8 }$
C) $\frac { 3 \pi - 8 } { 12 }$
D) $\pi + 2$
E) $2 \pi + 1$