Montrer que \(\forall x>-1 \, \, \ln (1 + x)\leq x.\)
Soit \(n \in \Nn^{*}.\) Montrer que \(\forall x \in [0, n]\, \, (1-\frac{x}{n})^{n}\leq
e^{-x} \leq (1 + \frac{x}{n})^{-n}.\)
En déduire que
\(\displaystyle{\int _{0}^{\sqrt {n}}\!\left (1-{\frac {{t}^{2}}{n}}\right )^{n}{dt}} \leq
\displaystyle{\int _{0}^{\sqrt {n}}\!{e^{-{t}^{2}}}{dt}} \leq
\displaystyle{\int _{0}^{\sqrt {n}}\!\frac{1}{\left (1+{\frac {{t}^{2}}{n}}\right )^n}{dt}}.\)
Rappel (intégrales de Wallis) : \(I_{n} =
\displaystyle{\int _{0}^{{\frac {\pi }{2}}}\!\left (\cos(\theta)\right )^{n}{d\theta}} \sim
\sqrt {\frac{\pi}{2n}}.\)
Montrer que \(\displaystyle{\int _{0}^{\infty } \frac 1{(1+{u}^{2})^n}{du}}\)
existe et vaut \(I_{2n-2}.\)
Montrer que \(\displaystyle{\int _{0}^{\infty }\!{e^{-{x}^{2}}}{dx}}\) existe et vaut \(\frac{\sqrt\pi}{2}\).