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On this answer: https://math.stackexchange.com/a/282675/65097, we see that

$$\int_{-\infty}^{\infty} \: \frac{t^2}{t^4+1} dt = \int_0^{\infty} \frac{\sqrt{x}}{x^2+1} dx$$

from the change of variables $x = t^2$.

This is a dumb question but, how did the bounds change from $0$ to $\infty$ becoming $-\infty$ to $\infty$.

Community
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a a
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1 Answers1

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Before you change variables, you have to halve the interval:

$$\int_{-\infty}^{\infty} dt \: \frac{t^2}{t^4+1} = 2 \int_{0}^{\infty} dt \: \frac{t^2}{t^4+1}$$

Now change variables: $t = \sqrt{x}$, $dt = \frac{dx}{2 \sqrt{x}}$

The result follows.

Ron Gordon
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  • Thanks! I thought the act of changing of variables (in your answer) caused the bounds to change. – a a Mar 11 '13 at 13:31
  • @aa: if this answer is useful, you can upvote it by clicking on the arrow pointing up next to it. If this answers your question best, you can accept it by clicking on the check mark next to it. – robjohn Mar 11 '13 at 13:39