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

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 $0$.

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

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

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

• 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! Apr 17, 2014 at 12:44
• @rubik: We don't want to spoil all the practice :).
– J.R.
Apr 17, 2014 at 12:46
• @downvoter: What is your objection?
– J.R.
Apr 19, 2014 at 10:06
• Just wondering, how do you see downvotes? Clicking on the vote like on StackOverflow does not work. Apr 19, 2014 at 11:56
• @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.
– J.R.
Apr 19, 2014 at 11:57

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 $u=1/x$.

Or just by calculation:

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

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