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.
 A: 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.
A: Your second substitution is not quite right. For $s<0$ it should be
\begin{align*}
&\int_0^{+\infty}\frac{1-\cos(t)}{t^2}\mathrm dt\stackrel{\frac t2\mapsto t}=\int_0^{-\infty}\frac{1-\cos(st)}{st^2}\mathrm dt\\
\stackrel{t\mapsto -t}=&\int_0^{\color{red}-(-\infty)}\frac{1-\cos(\color{red}-st)}{s(\color{red}-t)^2}\color{red}-\mathrm dt=\int_0^{+\infty}\frac{1-\cos(st)}{(\color{red}-s)t^2}\mathrm dt
\end{align*}
From hereon you can correctly deduce that
$$\frac2\pi\int_0^{+\infty}\frac{1-\cos(st)}{t^2}\mathrm dt=|s|$$
for $s<0$ aswell.

Another way of showing this would invoke the half-angle formula. To be precise for $s>0$
$$\int_0^{+\infty}\frac{1-\cos(st)}{t^2}\mathrm dt=\int_0^{+\infty}\frac{2\sin^2\left(\frac{st}2\right)}{t^2}\mathrm dt\stackrel{\frac{st}2\mapsto t}=s\int_0^{+\infty}\frac{\sin^2(t)}{t^2}\mathrm dt$$
The latter is also welll-known to equal $\pi/2$. Furthermore for $s<0$ we can deduce that
$$\int_0^{+\infty}\frac{1-\cos(st)}{t^2}\mathrm dt=\int_0^{+\infty}\frac{2\sin^2\left(\frac{st}2\right)}{t^2}\mathrm dt\stackrel{\frac{st}2\mapsto t}=s\int_0^{-\infty}\frac{\sin^2(t)}{t^2}\mathrm dt\stackrel{t\mapsto -t}=-s\int_0^{+\infty}\frac{\sin^2(t)}{t^2}\mathrm dt$$
Where a similiar argumentation is used as above. 

$$\therefore~\frac2\pi\int_0^{+\infty}\frac{1-\cos(st)}{t^2}\mathrm dt~=~|s|$$

