The polar-coordinate proof for the Gaussian integral ends up with the integral squared equaling $\pi$. From there, I understand that $\sqrt{\pi}$ is the solution to the integral, but why isn't $-\sqrt{\pi}$ also a solution? Is that just not how polar coordinates work?
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1If you consider the function $f(x) = -e^{-x^2}$ then the same method applies and also leads to $(\int_{\mathbb{R}} f(x),{\rm d}x)^2 = \pi$. This time $-\sqrt{\pi}$ is the answer. – Winther May 01 '18 at 23:47
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The integral of a positive function is positive and $x\mapsto e^{-x^2}$ is certainly positive.
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