Show that $\forall s \in \mathbb R, \frac{2}{\pi}\int\limits_{0}^{+\infty} \frac{1 - \cos (st)}{t^2} dt = |s|$ Knowing that $$\int_{0}^{+\infty} \frac{1 - \cos (t)}{t^2} dt = \frac{\pi}{2}$$
Show that $$\forall s \in \mathbb R, \frac{2}{\pi}\int_{0}^{+\infty} \frac{1 - \cos (st)}{t^2} dt = |s|$$
Let $s \in \mathbb R$.
If $s=0$, the formula is right.
Else, 
let $u = \frac{t}{s}$, then, if $s > 0$,  $$\int_{0}^{+\infty} \frac{1 - \cos (t)}{t^2} dt = \int_{0}^{+\infty} \frac{1 - \cos (su)}{su^2} du$$ 
so $$\frac{2}{\pi}\int_{0}^{+\infty} \frac{1 - \cos (su)}{u^2} du = |s|.$$
I'm having trouble when $s<0$, because, with $u = \frac{t}{s}$, $$\int_{0}^{+\infty} \frac{1 - \cos (t)}{t^2} dt = \int_{0}^{-\infty} \frac{1 - \cos (su)}{su^2} du$$
Because of $u \mapsto u^2$ and $\cos$ are even, I can say
$$\frac{2}{\pi}\int_{0}^{-\infty} \frac{1 - \cos (su)}{u^2} du = \frac{2}{\pi}\int_{0}^{+\infty} \frac{1 - \cos (su)}{u^2} du = s$$ 
I should get $|s|$.
Any help would be appreciated.

So you showed it for $s>0$. The case $s=0$ is trivial.
For $s<0$, use the fact that $\cos$ is an even function.
