Throughout the problem, we denote for every integer $n \geqslant 1$, $H _ { n } = \sum _ { k = 1 } ^ { n } \frac { 1 } { k }$ and $S _ { r } = \sum _ { n = 1 } ^ { + \infty } \frac { H _ { n } } { ( n + 1 ) ^ { r } }$ when the series converges.
Deduce from the previous questions that for every integer $r \geqslant 2$, $$S _ { r } = \sum _ { n = 1 } ^ { + \infty } \frac { H _ { n } } { ( n + 1 ) ^ { r } } = \frac { ( - 1 ) ^ { r } } { ( r - 1 ) ! } \int _ { 0 } ^ { 1 } ( \ln t ) ^ { r - 1 } \frac { \ln ( 1 - t ) } { 1 - t } \mathrm {~d} t$$

Throughout the problem, we denote for every integer $n \geqslant 1$, $H _ { n } = \sum _ { k = 1 } ^ { n } \frac { 1 } { k }$ and $S _ { r } = \sum _ { n = 1 } ^ { + \infty } \frac { H _ { n } } { ( n + 1 ) ^ { r } }$ when the series converges.
Establish that we then have $S _ { r } = \frac { ( - 1 ) ^ { r } } { 2 ( r - 2 ) ! } \int _ { 0 } ^ { 1 } \frac { ( \ln t ) ^ { r - 2 } ( \ln ( 1 - t ) ) ^ { 2 } } { t } \mathrm {~d} t$.

Throughout the problem, we denote for every integer $n \geqslant 1$, $H _ { n } = \sum _ { k = 1 } ^ { n } \frac { 1 } { k }$, $S _ { r } = \sum _ { n = 1 } ^ { + \infty } \frac { H _ { n } } { ( n + 1 ) ^ { r } }$ when the series converges, and $\zeta ( x ) = \sum _ { n = 1 } ^ { + \infty } \frac { 1 } { n ^ { x } }$ for $x > 1$.
Deduce that $S _ { 2 } = \frac { 1 } { 2 } \int _ { 0 } ^ { 1 } \frac { ( \ln t ) ^ { 2 } } { 1 - t } \mathrm {~d} t$ then find the value of $S _ { 2 }$ as a function of $\zeta ( 3 )$.

We denote $S _ { r } = \sum _ { n = 1 } ^ { + \infty } \frac { H _ { n } } { ( n + 1 ) ^ { r } }$ for $r \geqslant 2$, and $B ( x ) = \int _ { 0 } ^ { 1 } ( \ln ( 1 - t ) ) ^ { 2 } t ^ { x - 1 } \mathrm {~d} t$. We have shown that $S _ { r } = \frac { ( - 1 ) ^ { r } } { 2 ( r - 2 ) ! } \int _ { 0 } ^ { 1 } \frac { ( \ln t ) ^ { r - 2 } ( \ln ( 1 - t ) ) ^ { 2 } } { t } \mathrm {~d} t$.
Deduce that for every integer $r \geqslant 2 , S _ { r } = \frac { ( - 1 ) ^ { r } } { 2 ( r - 2 ) ! } \lim _ { x \rightarrow 0 ^ { + } } B ^ { ( r - 2 ) } ( x )$.

For every natural integer $n$ and every real $x$, set $I_n(x) = \int_0^{\pi/2} \cos(2xt)(\cos t)^n \, \mathrm{d}t$, and for $x \in J = [0,1/2[$, $$S_n(x) = \sum_{p=1}^{+\infty} \left(\sum_{k=n+1}^{+\infty} \frac{2^{2p+1} x^{2p-1}}{(2k-1)^{2p}}\right).$$ Show $$\forall n \in \mathbb{N}^{\star}, \forall x \in J, \quad \pi \tan(\pi x) + S_n(x) = -\frac{2I_{4n}^{\prime}(2x)}{I_{4n}(2x)} + \frac{I_{2n}^{\prime}(x)}{I_{2n}(x)} + \sum_{p=1}^{+\infty} 2\left(2^{2p} - 1\right) \zeta(2p) x^{2p-1}.$$

Prove, without using what precedes, that
$$\int _ { 0 } ^ { + \infty } \frac { x e ^ { - x } } { 1 - e ^ { - x } } \mathrm {~d} x = \frac { \pi ^ { 2 } } { 6 }$$

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 $$\phi _ { p , q } ( n ) = \frac { 1 } { q } \left( \int _ { 0 } ^ { 1 } \frac { 1 } { 1 + x ^ { \alpha _ { p , q } } } d x + ( - 1 ) ^ { n } I _ { p , q } ( n ) \right)$$