For all integers $k \geqslant 2$, we denote: $$w_{k} = \frac{1}{2} \int_{k-1}^{k} \frac{(t-k+1)(k-t)}{t^{2}} dt$$ Using yet another integration by parts, show that: $$\left| w_{k} - \frac{1}{12} \int_{k-1}^{k} \frac{\mathrm{~d}t}{t^{2}} \right| \leqslant \frac{1}{6} \int_{k-1}^{k} \frac{dt}{t^{3}}$$

The function $h$ is defined on $\mathbb{R}$ by $$h(u) = u - [u] - 1/2$$ Using integration by parts, justify, for $x > 0$, the convergence of the following integral: $$\int_{0}^{+\infty} \frac{h(u)}{u+x} du$$

Let $f : \mathbb{R} \rightarrow \mathbb{C}$ be a function of class $C^{\infty}$ on $\mathbb{R}$ and 1-periodic. We consider the function $g$ defined on $[-1,1]$ by
$$\forall x \in ]-1,1[\backslash\{0\}, \quad g(x) = \frac{f(x)-f(0)}{\sin(\pi x)} \quad g(0) = \frac{f'(0)}{\pi} \quad g(1) = g(-1) = -g(0)$$
We henceforth admit that $g$ is of class $C^{1}$ on $[-1,1]$.
Using integration by parts, show the existence of a real number $C$ such that
$$\forall n \in \mathbb{N}, \quad \left|\int_{-1/2}^{1/2} g(x) \sin((2n+1)\pi x) \mathrm{d}x\right| \leqslant \frac{C}{2n+1}$$

For $x \in [ 2 , + \infty[$, we set $$I ( x ) = \int _ { 2 } ^ { x } \frac { \mathrm{dt} } { \ln ( t ) }$$ Justify, for $x \in [ 2 , + \infty[$, the relation $$I ( x ) = \frac { x } { \ln ( x ) } - \frac { 2 } { \ln ( 2 ) } + \int _ { 2 } ^ { x } \frac { \mathrm{dt} } { ( \ln ( t ) ) ^ { 2 } }$$ Establish moreover the relation $$\int _ { 2 } ^ { x } \frac { \mathrm{dt} } { ( \ln ( t ) ) ^ { 2 } } \underset { x \rightarrow + \infty } { = } o ( I ( x ) )$$ Deduce finally an equivalent of $I ( x )$ as $x$ tends to $+ \infty$.