Gaussian Summation - Find Out the Result!

[mfengwang]

Hi,
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....

 

[Mute]

I checked Wolfram alpha; it gives the result as

[tex]\frac{1 + \vartheta_3(0,e^{-1/a})}{2},[/tex]

where [itex]\vartheta_b(x,q)[/itex] is a theta function. Looking up the definition of the theta function on mathworld reveals that the result is pretty much by definition:

[tex]\vartheta_3(z,q) = \sum_{n=-\infty}^\infty q^{n^2} e^{2\pi i z}[/tex]

Rearrangement and plugging in z = 0, q = e^(-1/a) gives the result, although it's not very enlightening. The theta function is a special function that should be available in mathematica and perhaps matlab.

I'm not sure if, setting a = 1 for example, there is a way to compute the resulting value without reference to the theta function.

 

[mfengwang]

Thanks a lot! It's really a great help to me.