\(I_a = \int_{x=-a}^a|f(x)|\,d x \le \int_{x=-a}^a \int_{t=-\infty}^{+\infty} \frac{e^{-|x+t|}}{\sqrt{|t|}(1+|t|)}\,d t\,d x = \int_{t=-\infty}^{+\infty} \int_{x=-a}^a \frac{e^{-|x+t|}}{\sqrt{|t|}(1+|t|)}\,d x\,d t\). On a \(\int_{x=-a}^a e^{-|x+t|}\,d x = \begin{cases}2-2e^{-a}\ch t & \text{ si } |t|<a\cr 2e^{-|t|}\sh a & \text{ si } |t|\ge a,\cr\end{cases}\) donc \(I_a \le \int_{t=0}^a \frac{2\,d t}{(1+t)\sqrt t} + \int_{t=a}^{+\infty}\frac{4e^{-t}\sh a}{(1+a)\sqrt a}\,d t = 4\Arctan(\sqrt a) + \frac{4e^{-a}\sh a}{(1+a)\sqrt a} \le \text{cste}\).

exo7 4364