# Improper Integral

• Feb 2nd 2013, 09:27 AM
strammer
Improper Integral
Prove that $\int_{-\infty}^{\infty} dx\: e^{-i x^2} = \int_{-\infty}^{\infty} e^{-ix^2} dx$
• Feb 2nd 2013, 06:09 PM
chiro
Re: Improper Integral
Hey strammer.

There is no work to do: they are both the same integral.

Putting a dx on one side doesn't change the nature of the integral.
• Feb 6th 2013, 12:21 AM
strammer
Re: Improper Integral
They apparently are not the same since d(xe^-ix^2) = e^-ix^2(1-2ix^2)dx
• Feb 6th 2013, 01:27 PM
chiro
Re: Improper Integral
The post didn't have a bracket around the differential measure so I missed that.

If the function is differentiable, then d(g(x)) becomes g'(x)dx and you can integrate as per normal.