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We know that the Gaussian integral is

[tex]\int_{-\infty}^{+\infty}e^{-\frac{x^2}{a^2}}dx=a\sqrt{\pi}[/tex]

However, if the gaussian function is discrete in x, what is the result of

[tex]\sum_{n=0}^{+\infty}e^{-\frac{n^2}{a}} = \\?[/tex]

where n is natural number, that is n=0,1,2,3.............

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# Gaussian Summation

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