The Gaussian integral is defined as $$\int_0^\infty e^{-x^2}dx.$$
I know there are many ways to solve it, but what would be the fastest you know?
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2To put it in Sage. – Phicar Oct 05 '20 at 14:02
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2Square the integral & convert to polar coordinates ... This is a duplicate of ? (anyone ?) – Donald Splutterwit Oct 05 '20 at 14:03
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2@Phicar What a fantastic answer ... you have made me laugh out loud $\ddot \smile$ – Donald Splutterwit Oct 05 '20 at 14:05
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what would be the fastest you know?
given that
$$\int_{-\infty}^{+\infty}e^{-x^2}dx=\sqrt{\pi}$$
...your integral is exactly the half: $\frac{\sqrt{\pi}}{2}$
tommik
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Sorry but this is one of the funniest answers I have read. I can not stop laughing while seeing it – tryst with freedom Oct 05 '20 at 14:23
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First of all the gaussian integral is defined as $$\int_{-\infty}^{\infty}e^{-x^2}\mathrm{d}x,$$ but since the function we want to integrate is even, we can easily find the connection between those two. The fastest way I know, is to use Eulers Reflection formula. I would start with the substitiution $x^2 = u$. That means that $\mathrm{d}x = \frac{\mathrm{d}u}{2\sqrt{u}}$. We get: $$\int_{-\infty}^{\infty}e^{-x^2}\mathrm{d}x = 2\int_{0}^{\infty}\cfrac{u^{-\frac{1}{2}}}{2}e^{-u}\mathrm{d}u.$$
If you are familiar with the gamma function you'll see that this integral is equivalent to $\Gamma(\frac{1}{2})$. Now use the Euler Reflection formula $$\Gamma(z)\Gamma(1-z) = \cfrac{\pi}{\sin{\pi z}}\quad{\Longrightarrow}\quad\Gamma\left(\frac{1}{2}\right)\Gamma\left(1-\frac{1}{2}\right) = \cfrac{\pi}{\sin{\frac{\pi}{2}}} $$ Now simplify and take the square root on both sides: $$\int_{-\infty}^{\infty}e^{-x^2}\mathrm{d}x = \Gamma\left(\frac{1}{2}\right) = \sqrt{\pi}$$
Daniel
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