TITLE: Showing that $\int_{0}^{\infty} \frac{dx}{1 + x^2} = 2 \int_0^1 \frac{dx}{1 + x^2}$
QUESTION [7 upvotes]: 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 $0$.

REPLY [5 votes]: Or just by calculation:
$$\int_0^\infty \frac{dx}{1+x^2}= [\arctan(x)]_{0}^{\infty}=\frac{1}{2}\pi$$ and 
$$\int_0^1 \frac{dx}{1+x^2}= [\arctan(x)]_{0}^{1}=\frac{1}{4}\pi$$