How can the integral of exp(-x^n) be proven using polar coordinates?

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The integral of the function \(\int e^{-(x_1^2+\dots+x_\nu^2)^{k/2}} \mathrm d x\) can be proven using polar coordinates, specifically for the case when \(\nu = 1\). The proof involves transforming the integral into a more manageable form by substituting \(x\) with \(y = x^k\), leading to the result \(\int_{-\infty}^{+\infty} e^{-|x|^k} \mathrm dx = \frac{2}{k} \Gamma \left(\frac{1}{k} \right)\). This ultimately simplifies to \(\frac{\pi^{1/2} \Gamma \left( \frac{1}{k}+1 \right)}{\Gamma \left( \frac{1}{2} +1\right)}\), confirming the validity of the original integral expression.

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nonequilibrium
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Hello,

How do I prove or find a proof for \int e^{-(x_1^2+\dots+x_\nu^2)^{k/2}} \mathrm d x = \frac{\pi^{\nu/2} \Gamma \left( \frac{\nu}{k}+1 \right)}{\Gamma \left( \frac{\nu}{2} +1\right)}?

EDIT: I think I have it, and as I can't delete this thread, I will sketch the solution:
first of all, notice that proving it for the case of \nu = 1 is enough, because we can always go back to that specific case by using polar coordinates (replacing x_1^2 + \dots + x_\nu^2 \mapsto r^2); and then:

\int_{-\infty}^{+\infty} e^{-|x|^k} \mathrm dx = 2 \int_{0}^{+\infty} e^{-x^k} \mathrm dx and if we substitute with y=x^k => \mathrm d y = k y^{(k-1)/k} \mathrm d x, this is equal to 2\int_0^{+\infty} e^{-y} y^{\frac{1}{k}-1} \frac{1}{k} \mathrm d y = \frac{2}{k} \Gamma \left(\frac{1}{k} \right) and using x \Gamma(x) = \Gamma(x+1) we get:
\int_{-\infty}^{+\infty} e^{-|x|^k} \mathrm dx = 2 \Gamma \left( \frac{1}{k} + 1 \right) = \frac{\pi^{1/2} \Gamma \left( \frac{1}{k}+1 \right)}{\Gamma \left( \frac{1}{2} +1\right)}
 
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Thank's a lot for sharing the solution!
 

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