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I‌ was wondering if the function \begin{equation} f(a,b) = \sum_{n=1}^\infty \frac{e^{-a n^2}}{n^2 + b^2}, \qquad\qquad (a,b>0) \end{equation} has a representation in terms of known elementary/special functions (like theta functions etc)?

Saïd M
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    $\partial f/\partial a-b^2 f$ is a theta function; solving, you get an integral representation. – metamorphy Jan 10 '21 at 21:05
  • This is $S_{1,0} (b,ab^2 )$ in the preprint https://arxiv.org/pdf/2101.01589.pdf. It is stated at the beginning that the second index should be strictly positive but it seems that everything works fine for the case that it is $0$ as well. You can find a representation in therms of the Riemann zeta function and the confluent hypergeometric function, as well as asymptotic expansions in this manuscript. – Gary Feb 04 '21 at 07:52

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It is not too difficult to see that we have the differential equation $$\frac{\partial}{\partial a}f(a,b)-b^2f(a,b)=\frac12(1-\vartheta_3(e^{-a})).$$ Using the integrating factor $M(a)=e^{-ab^2}$, we have $$\frac{\partial}{\partial t}\left(e^{-tb^2}f(t,b)\right)=\frac12e^{-tb^2}(1-\vartheta_3(e^{-t})),$$ then integrating from $t=0$ to $t=a$, we have $$e^{-ab^2}f(a,b)-f(0,b)=\frac12\int_0^a e^{-tb^2}(1-\vartheta_3(e^{-t}))dt.$$ Then from here, we have $$f(0,b)=\frac{\pi}{2b}\coth\pi b-\frac1{2b^2},$$ so that $$f(a,b)=\frac{e^{ab^2}}{2b^2}\left\{\pi b\coth\pi b-1+b^2\int_0^a e^{-tb^2}(1-\vartheta_3(e^{-t}))dt\right\}.$$

clathratus
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