[Calculus] For two functions $f ( x )$ and $g ( x )$ that have second derivatives on the set of all real numbers, consider the definite integral $$\int _ { 0 } ^ { 1 } \left\{ f ^ { \prime } ( x ) g ( 1 - x ) - g ^ { \prime } ( x ) f ( 1 - x ) \right\} d x$$ Let the value of this integral be $k$. Which of the following statements in are correct? [4 points]
ㄱ. $\int _ { 0 } ^ { 1 } \left\{ f ( x ) g ^ { \prime } ( 1 - x ) - g ( x ) f ^ { \prime } ( 1 - x ) \right\} d x = - k$ ㄴ. If $f ( 0 ) = f ( 1 )$ and $g ( 0 ) = g ( 1 )$, then $k = 0$. ㄷ. If $f ( x ) = \ln \left( 1 + x ^ { 4 } \right)$ and $g ( x ) = \sin \pi x$, then $k = 0$.
(1) ㄴ
(2) ㄷ
(3) ㄱ, ㄴ
(4) ㄱ, ㄷ
(5) ㄱ, ㄴ, ㄷ

For all integers $k \geqslant 2$, we set: $$u_{k} = \ln k - \int_{k-1}^{k} \ln t \, dt$$ Using two integrations by parts, show that: $$u_{k} = \frac{1}{2}(\ln k - \ln(k-1)) - \frac{1}{2} \int_{k-1}^{k} \frac{(t-k+1)(k-t)}{t^{2}} dt$$

We fix $x > 0$ and for all natural integers $n$, we define $F_{n}(x)$ by: $$F_{n}(x) = \ln\left(\frac{n! \, n^{x+1}}{(x+1)(x+2) \ldots (x+n+1)}\right)$$ Show that for all natural integers $i$: $$\int_{x+i}^{x+i+1} \ln t \, dt = \ln(x+i) - \int_{i}^{i+1} \frac{u-i-1}{u+x} du$$

If $T$ is a distribution on $\mathcal{D}$, we define the derivative distribution $T'$ by $$\forall \varphi \in \mathcal{D}, \quad T'(\varphi) = -T(\varphi')$$
Let $f$ be a continuous function from $\mathbb{R}$ to $\mathbb{R}$. If $f$ is of class $\mathcal{C}^1$, show that $(T_f)' = T_{f'}$. Adapt this result to the case where $f$ is piecewise of class $\mathcal{C}^1$.

Let $r$ be a non-zero natural integer and $f$ a function expandable as a power series on $] - 1,1 [$. We assume that for every $x$ in $] - 1,1 \left[ , f ( x ) = \sum _ { n = 0 } ^ { + \infty } a _ { n } x ^ { n } \right.$ and that $\sum _ { n \geqslant 0 } \frac { a _ { n } } { ( n + 1 ) ^ { r } }$ converges absolutely.
Show that $\int _ { 0 } ^ { 1 } ( \ln t ) ^ { r - 1 } f ( t ) \mathrm { d } t = ( - 1 ) ^ { r - 1 } ( r - 1 ) ! \sum _ { n = 0 } ^ { + \infty } \frac { a _ { n } } { ( n + 1 ) ^ { r } }$.

Justify that, if $\varphi$ is a $\mathcal{C}^1$ application from $[0, \pi]$ to $\mathbf{R}$, then $$\lim_{x \to +\infty} \int_0^{\pi} \varphi(t) \sin(xt) \mathrm{d}t = 0$$ and conclude that $$\sigma(1) = \frac{\pi^2}{6}$$

Deduce that:
$$\int _ { 0 } ^ { + \infty } ( \cos ( t ) ) ^ { 2 p } \frac { \sin ( t ) } { t } \mathrm {~d} t = \int _ { 0 } ^ { \frac { \pi } { 2 } } ( \cos ( t ) ) ^ { 2 p } \mathrm {~d} t$$
In the case $p = 0$, this integral is commonly called the ``Dirichlet Integral''.

We define on $[ 0,1 ]$ the function $P _ { n }$ by:
$$\forall x \in [ 0,1 ] , \quad P _ { n } ( x ) = \frac { 1 } { n ! } \frac { \mathrm { d } ^ { n } \left( x ^ { n } ( 1 - x ) ^ { n } \right) } { \mathrm { d } x ^ { n } } .$$
Let $n \in \mathbb { N } ^ { * }$. Show that for all $y \in ] 0,1 [$,
$$\int _ { 0 } ^ { 1 } \frac { P _ { n } ( x ) } { 1 - x y } \mathrm {~d} x = ( - y ) ^ { n } \int _ { 0 } ^ { 1 } \frac { x ^ { n } ( 1 - x ) ^ { n } } { ( 1 - x y ) ^ { n + 1 } } \mathrm {~d} x$$

We define on $[ 0,1 ]$ the function $P _ { n }$ by:
$$\forall x \in [ 0,1 ] , \quad P _ { n } ( x ) = \frac { 1 } { n ! } \frac { \mathrm { d } ^ { n } \left( x ^ { n } ( 1 - x ) ^ { n } \right) } { \mathrm { d } x ^ { n } } .$$
We set $I _ { n } = \int _ { 0 } ^ { 1 } \int _ { 0 } ^ { 1 } \frac { ( 1 - y ) ^ { n } P _ { n } ( x ) } { 1 - x y } \mathrm {~d} x \mathrm {~d} y$.
Deduce that
$$I _ { n } = ( - 1 ) ^ { n } \int _ { 0 } ^ { 1 } \int _ { 0 } ^ { 1 } \frac { x ^ { n } ( 1 - x ) ^ { n } y ^ { n } ( 1 - y ) ^ { n } } { ( 1 - x y ) ^ { n + 1 } } \mathrm {~d} x \mathrm {~d} y$$