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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.

mrtaurho
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anni
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2 Answers2

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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.

Scientifica
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  • The fact that $\cos$ is even lets met put $+\infty$ instead of $-\infty$ in the integral but I can't figure out how to get $|s|$. – anni Feb 10 '19 at 15:54
  • If $s<0$, then $-s>0$ and you use the formula you already proved for the positive case ;) – Scientifica Feb 10 '19 at 15:57
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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.

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

mrtaurho
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