- #1
frzncactus
- 10
- 0
Homework Statement
Use a change of variable to show that
[itex]\int_0^{\infty} \frac{dx}{1+x^2} = 2\int_0^1\frac{dx}{1+x^2}[/itex]
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.
Homework Equations
See problem statement.
The Attempt at a Solution
It's obvious to me that the left-hand side (LHS) integral can be split into two:
[itex]\int_0^{\infty} \frac{dx}{1+x^2}[/itex] [1]
= [itex]\int_0^1 \frac{dx}{1+x^2} + \int_1^{\infty}\frac{dx}{1+x^2}[/itex] [2]
What remains is showing that
[itex]\int_1^{\infty}\frac{dx}{1+x^2} = \int_0^1 \frac{dx}{1+x^2}[/itex] [3]
My initial guess for a u-substitution is as follows:
[itex]u = 1/x[/itex]
[itex]du = x^{-2}dx[/itex]
Then the LHS of [3] becomes
[itex]\int_1^{\infty}\frac{dx}{1+x^2}[/itex]
= [itex]\int_1^0\frac{x^2}{1+x^2}du[/itex]
= [itex]\int_1^0 1+\frac{x^2}{1+x^2}-\frac{1+x^2}{1+x^2}du[/itex]
= [itex]\int_1^0 1-\frac{1}{1+x^2}du[/itex]
= [itex]\int_1^0 du -\int_1^0\frac{1}{1+x^2}du[/itex]
...which looked halfway clever, but then the idea ran out of steam because of the incompatible u's and x's.
Last edited: