Prove that $\displaystyle \int_{-\infty}^{\infty} dx\: e^{-i x^2} = \int_{-\infty}^{\infty} e^{-ix^2} dx$
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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.
They apparently are not the same since d(xe^-ix^2) = e^-ix^2(1-2ix^2)dx
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.
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