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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 $X$?

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  • 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
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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.

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