Throughout this problem, $I = ]-1, +\infty[$, and $f(x) = \int_0^{\pi/2} (\sin(t))^x \mathrm{~d}t$. We call $\tilde{f}$ the application from $\mathbf{R}_+$ to $\mathbf{R}$, defined by: $$\forall x \in \mathbf{R}^+, \tilde{f}(x) = \ln(f(2x))$$ Suppose here that $x \in \mathbf{R}_+^*$, $(n,p) \in (\mathbf{N}^*)^2$ and $x \leq p$. Verify that $$\tilde{f}(n) - \tilde{f}(n-1) \leq \frac{\tilde{f}(n+x) - \tilde{f}(n)}{x} \leq \frac{\tilde{f}(n+p) - \tilde{f}(n)}{p}$$ and that $(\tilde{f}(n+x) - \tilde{f}(n))$ has a limit as $n$ tends to $+\infty$.
Throughout this problem, $I = ]-1, +\infty[$, and $f(x) = \int_0^{\pi/2} (\sin(t))^x \mathrm{~d}t$. We call $\tilde{f}$ the application from $\mathbf{R}_+$ to $\mathbf{R}$, defined by:
$$\forall x \in \mathbf{R}^+, \tilde{f}(x) = \ln(f(2x))$$
Suppose here that $x \in \mathbf{R}_+^*$, $(n,p) \in (\mathbf{N}^*)^2$ and $x \leq p$. Verify that
$$\tilde{f}(n) - \tilde{f}(n-1) \leq \frac{\tilde{f}(n+x) - \tilde{f}(n)}{x} \leq \frac{\tilde{f}(n+p) - \tilde{f}(n)}{p}$$
and that $(\tilde{f}(n+x) - \tilde{f}(n))$ has a limit as $n$ tends to $+\infty$.