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 $X$?

  • 2
    $\begingroup$ It's clearly positive. $\endgroup$ – Lord Shark the Unknown Jun 18 '17 at 9:51
  • $\begingroup$ In the second integral, $\cos x$ should be $\cos t$. $\endgroup$ – KCd Jun 18 '17 at 10:18
  • $\begingroup$ Wolfram|Alpha? Mathematica? Photomath? These programs will explain you what you need to understand 😊 $\endgroup$ – 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.