# Change of Variables

1. Feb 19, 2015

### frzncactus

1. The problem statement, all variables and given/known data
Use a change of variable to show that
$\int_0^{\infty} \frac{dx}{1+x^2} = 2\int_0^1\frac{dx}{1+x^2}$

Please note: the point of this exercise is to change the bounds of the integral to be finite to allow numerical estimation, as opposed to directly solving the integral.

2. Relevant equations
See problem statement.

3. The attempt at a solution
It's obvious to me that the left-hand side (LHS) integral can be split into two:
$\int_0^{\infty} \frac{dx}{1+x^2}$ [1]
= $\int_0^1 \frac{dx}{1+x^2} + \int_1^{\infty}\frac{dx}{1+x^2}$ [2]

What remains is showing that
$\int_1^{\infty}\frac{dx}{1+x^2} = \int_0^1 \frac{dx}{1+x^2}$ [3]

My initial guess for a u-substitution is as follows:
$u = 1/x$
$du = x^{-2}dx$

Then the LHS of [3] becomes
$\int_1^{\infty}\frac{dx}{1+x^2}$
= $\int_1^0\frac{x^2}{1+x^2}du$
= $\int_1^0 1+\frac{x^2}{1+x^2}-\frac{1+x^2}{1+x^2}du$
= $\int_1^0 1-\frac{1}{1+x^2}du$
= $\int_1^0 du -\int_1^0\frac{1}{1+x^2}du$

...which looked halfway clever, but then the idea ran out of steam because of the incompatible u's and x's.

Last edited: Feb 19, 2015
2. Feb 19, 2015

### frzncactus

Aha! The last step was counterproductive. Instead, swapping x's for 1/u's and doing some algebra at the second-to-last step finishes the problem. Sorry about answering my own question, this problem has embarrassingly been bothering me for weeks. I'll leave this up here for the curious.

(Aside)
For anyone interested, it was from the book Street-Fighting Mathematics: The Art of Educated Guessing and Opportunistic Problem Solving by Sanjoy Mahajan.