Does there exist an explicit expression for the series (or function) $$f(x)=\sum _{n=1}^\infty e^{-xn^2}\text{ ?}$$
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1Search "theta function". – Sungjin Kim Apr 06 '13 at 17:03
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i mean something that doesn't involve theta function, a function like the limit of the goemetric series – user71282 Apr 06 '13 at 17:11
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3$\theta_{{3}} \left(0,q \right) =1+2,\sum _{n=1}^{\infty }{q}^{{n}^{2 }}$ and that is the simplest known way to write it. – GEdgar Apr 06 '13 at 17:14
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...and what's your objection against the theta function? – J. M. ain't a mathematician Apr 06 '13 at 17:15
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ok, but i'd like to know if there exists a function(x) which is the limit of that series, such as the integral of $exp(-x^2)$ can't be express with elementary function – user71282 Apr 06 '13 at 17:20
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1Yes, the theta function is not elementary. Yes, it's the limit of your series. @GEdgar already gave you the precise expression. Now, what else? – J. M. ain't a mathematician Apr 06 '13 at 17:22
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i'd like to know if exists a simpler expression, and if it is possible to find out – user71282 Apr 06 '13 at 17:25
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1Did you read GEdgar's comment carefully? $\frac12(\vartheta_3(0,\exp(-x))-1)$ is as simple as it gets. – J. M. ain't a mathematician Apr 06 '13 at 17:28
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mmmmmm now i ask an epression without theta function or a proof of the fact that the theta function is the only way for rappresenting it – user71282 Apr 06 '13 at 17:32
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if i would calculate a particular x of this function, how could i do it if the theta function expresses an identity? – user71282 Apr 06 '13 at 17:41
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There is some more information about your sum that can be obtained using Mellin transforms. We have $$\mathfrak{M}(f(x); s) = f^*(s) = \Gamma(s) \zeta(2s).$$
Now invert by calculating the sum of the residues of $f^*(s) x^{-s}$. They are $$ \operatorname{Res}(f^*(s) x^{-s}; s= 1/2) = \frac{\sqrt{\pi}}{2} \frac{1}{\sqrt{x}} \quad \text{and} \quad \operatorname{Res}(f^*(s) x^{-s}; s= 0) = - \frac{1}{2}.$$ The remaining poles of the gamma function are canceled by the zeros of the zeta function, giving $$f(x) \sim \frac{\sqrt{\pi}}{2} \frac{1}{\sqrt{x}} - \frac{1}{2}.$$ This asymptotic expansion holds in a neighborhood of zero.
Marko Riedel
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thanks :) I just want to know why we use an elliptic function to express this function and not a "simpler" function. Which properties has it got? – user71282 Apr 11 '13 at 21:31