Prove that $\displaystyle \int_{-\infty}^{\infty} dx\: e^{-i x^2} = \int_{-\infty}^{\infty} e^{-ix^2} dx$

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- Feb 2nd 2013, 08:27 AMstrammerImproper Integral
Prove that $\displaystyle \int_{-\infty}^{\infty} dx\: e^{-i x^2} = \int_{-\infty}^{\infty} e^{-ix^2} dx$

- Feb 2nd 2013, 05:09 PMchiroRe: 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 5th 2013, 11:21 PMstrammerRe: Improper Integral
They apparently are not the same since d(xe^-ix^2) = e^-ix^2(1-2ix^2)dx

- Feb 6th 2013, 12:27 PMchiroRe: 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.