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The calculus text book says

$\int_{-\infty}^{\infty} \left\vert e^{-ax} \right\vert^{2} dx = \int_{0}^{\infty} e^{-2ax}dx$ but does not explain how this happened, and I am not able to figure it out. Could someone please show step by step how this transformation happened?

screen shot from the book (into to applied math, by Strang, page 314)

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ps. to answer comment below that the book has type in lower limit, and that it should be 0 and not negative infinity. The book definition uses negative infinity and not zero. So the lower limit is not a type. Screen shot:

enter image description here

The above is just before the example. So the book is using this example to illustrate the Plancherel's formula.

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The part inside the integral follows from $e^{a}e^{b} = e^{a+b}$. I'm not sure about the limits of integration changing, that looks like a typo.

Edit: yes that is a typo, otherwise the energy would be infinite.

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  • $\begingroup$ thanks. But I do not understand where is the typo. Are you saying the final answer $\frac{1}{2 a}$ is wrong also? if not, how would you integrate this then? $\endgroup$ – Steve H Nov 12 '14 at 4:57
  • $\begingroup$ The lower limit of integration as $-\infty$ is a typo, it should be 0. $\endgroup$ – Suzu Hirose Nov 12 '14 at 4:58
  • $\begingroup$ The lower limit of integration as −∞ is a typo, it should be 0 But this is not how the book defines it. It actually has the definition from negative infinity to infinity. I can post screen shot of the definition. I am sure the definition is correct. $\endgroup$ – Steve H Nov 12 '14 at 5:02
  • $\begingroup$ @SteveH - it cannot be, that gives infinite energy. $\endgroup$ – Suzu Hirose Nov 12 '14 at 5:03
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    $\begingroup$ @SteveH The author must be assuming that you turn the pulse on at t=0. If you have a pulse of exponential height, of course the energy is infinite if it goes all the way to negative infinity. $\endgroup$ – Kevin Driscoll Nov 12 '14 at 5:03

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