# Integrate $2 \int_0^{\infty} \frac{1- \cos(tX)}{\pi t^2}~dt$

I want to integrate $2 \int_0^{\infty} \frac{1- \cos(tX)}{\pi t^2}~dt$.

From previous exercises I know that ${2\int_0^{\infty} \frac{1-cos(t)}{\pi t^2}~dt=1}.$

The solution says $2 \int_0^{\infty} \frac{1- \cos(tX)}{\pi t^2}~dt = |X|$, but I don't understand where the absolute value comes from. Because I did it like this:

Substitute $y = tX \Rightarrow$ $2 \int_0^{\infty} \frac{1- \cos(tX)}{\pi t^2}~dt = 2 \int_0^{\infty} \frac{1- \cos (y) }{\pi \left(\frac{y}{X} \right)^2}~\frac{1}{X}~dy = 2X \int_0^{\infty} \frac{1-\cos y}{\pi y^2}~dy = X$

So why should it be $|X|$ instead of $X$?

• It's clearly positive. – Lord Shark the Unknown Jun 18 '17 at 9:51
• In the second integral, $\cos x$ should be $\cos t$. – KCd Jun 18 '17 at 10:18
• Wolfram|Alpha? Mathematica? Photomath? These programs will explain you what you need to understand 😊 – Arthur Guiot Jun 18 '17 at 11:18

If $X$ is negative, the limits of your integral change. So you then have to integrate from $y=0$ to $y=-\infty$, so you obtain (by symmetry of the integral), $-X = |X|$. So you obtain (in the case $X$ is negative) $$2 \int_0^{\infty} \frac{1- \cos(tX)}{\pi t^2}~dt = 2 \int_0^{-\infty} \frac{1- \cos (y) }{\pi \left(\frac{y}{X} \right)^2}~\frac{1}{X}~dy = -2X \int_0^{\infty} \frac{1-\cos y}{\pi y^2}~dy = -X,$$ where I used the symmetry of the integral in the last step.