Change of variable to show $\int_0^{\infty}\frac{dx}{1+x^2}=2\int_0^1\frac{dx}{1+x^2}$

Question

Which change of variable can be used to show $$\int_0^{\infty}\frac{dx}{1+x^2}=2\int_0^1\frac{dx}{1+x^2}$$ (direct integration isn't allowed, must use change of variable)

Answer 1

The required change of variable is $$x = \tan ( 2\tan^{-1}y),\>\>\>\>\> dx = \sec^2(2\tan^{-1}y)\frac{2dy}{1+y^2} $$ which leads to $$\int_0^{\infty}\frac{dx}{1+x^2} = \int_0^{1}\frac{\sec^2(2\tan^{-1}y)}{1+ \tan^2( 2\tan^{-1}y)} \frac{2dy}{1+y^2} =2\int_0^1\frac{dy}{1+y^2}$$

Answer 2

To show the equivalent result $\int_1^\infty\frac{dx}{1+x^2}=\int_0^1\frac{dx}{1+x^2}$, use $x\mapsto1/x$.

Harder version, requiring no rewriting of the problem in an equivalent form: to show $\int_0^\infty\frac{dx}{1+x^2}=\int_0^1\frac{2dx}{1+x^2}$, use $x\mapsto\exp(-|\ln x|)$.

Answer 3

Alternative

$$\tan(u) ~=~ x ~\implies~$$

$$\text{[informally]} ~\sec^2(u)du ~=~ dx ~~~\text{and}$$

$$~\frac{1}{1 + x^2} ~=~ \cos^2(u) ~\implies $$

$$\int_0^\infty \frac{dx}{1+x^2} = \int_0^{\pi/2} du = 2\int_0^{\pi/4} du = 2\int_0^1 \frac{dx}{1+x^2}.$$

Answer 4

$$I=\int_0^\infty\frac{1}{1+x^2}dx=\int_0^1\frac{1}{1+x^2}dx+\int_1^\infty\frac{1}{1+x^2}dx$$ now: $$I_1=\int_1^\infty\frac{1}{1+x^2}dx$$ $$u=\frac1x,du=-1/x^2dx\Rightarrow dx=-\frac{1}{u^2}du$$ $$I_1=-\int_1^0\frac{1}{1+(1/u)^2}\frac{du}{u^2}=\int_0^1\frac{1}{1+u^2}$$ $$\therefore I=2\int_0^1\frac{1}{1+x^2}dx$$