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I was reading an article in which it was stated that, with a change of variable, one could show that:

$$\int_{0}^{\infty} \frac{dx}{1 + x^2} = 2 \int_0^1 \frac{dx}{1 + x^2}$$

I tried with $t = 1 + \frac{1}{x}$ but that doesn't work out, especially because the lower bound doesn't become $0$.

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    $\begingroup$ Use $u=\frac{1}{x}$. $\endgroup$
    – J.R.
    Apr 17, 2014 at 10:02
  • $\begingroup$ @YourAdHere: Then how do I change the bounds? $\endgroup$
    – rubik
    Apr 17, 2014 at 10:04

3 Answers 3

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With the substitution $u=\frac{1}{x}$ we get

$$\int_1^\infty \frac{dx}{1+x^2} = \int_0^1 \frac{u^{-2}}{1+u^{-2}}du=\int_0^1 \frac{du}{1+u^2}$$

which implies

$$\int_0^\infty \frac{dx}{1+x^2} = \int_0^1 \frac{dx}{1+x^2} + \int_1^\infty \frac{dx}{1+x^2} = 2\int_0^1 \frac{dx}{1+x^2}$$

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  • $\begingroup$ Nice, I got it. In the first passage of the first line you missed some steps; it took me a good 6 minutes to work it out! $\endgroup$
    – rubik
    Apr 17, 2014 at 12:44
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    $\begingroup$ @rubik: We don't want to spoil all the practice :). $\endgroup$
    – J.R.
    Apr 17, 2014 at 12:46
  • $\begingroup$ @downvoter: What is your objection? $\endgroup$
    – J.R.
    Apr 19, 2014 at 10:06
  • $\begingroup$ Just wondering, how do you see downvotes? Clicking on the vote like on StackOverflow does not work. $\endgroup$
    – rubik
    Apr 19, 2014 at 11:56
  • $\begingroup$ @rubik: (1) It does work, but you need a certain amount of reputation. (2) You get -2 reputation for a downvote, that shows up on your summary. $\endgroup$
    – J.R.
    Apr 19, 2014 at 11:57
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You just need to show that $\int_1^\infty \frac{dx}{1+x^2} = \int_0^1\frac{dx}{1+x^2}$; this follows by the substitution $u=1/x$.

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Or just by calculation:

$$\int_0^\infty \frac{dx}{1+x^2}= [\arctan(x)]_{0}^{\infty}=\frac{1}{2}\pi$$ and

$$\int_0^1 \frac{dx}{1+x^2}= [\arctan(x)]_{0}^{1}=\frac{1}{4}\pi$$

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