# why $\int_{-\infty}^{\infty} \left\vert e^{-ax} \right\vert^{2} dx = \int_{0}^{\infty} e^{-2ax}dx$

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)

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:

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

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.
• 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? – Steve H Nov 12 '14 at 4:57
• The lower limit of integration as $-\infty$ is a typo, it should be 0. – Suzu Hirose Nov 12 '14 at 4:58
• 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. – Steve H Nov 12 '14 at 5:02