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Ramanujan theta function defined as-$$f(a,b)=\sum_{n=0}^\infty a^{\frac{n(n+1)}{2}}b^{\frac{n(n-1)}{2}}$$ And it's integral representation:$$f(a,b)=1+\int_0^\infty \frac{2ae^{-t^{2}/2}}{\sqrt{2\pi}}\left[\frac{1-a\sqrt{ab}\operatorname{cosh}(\sqrt{\operatorname{log}(ab)}t)}{a^3b-2a\sqrt{ab}\operatorname{cosh}(\sqrt{\operatorname{log}(ab)}t)+1}\right]dt+\int_0^\infty \frac{2be^{-t^{2}/2}}{\sqrt{2\pi}}\left[\frac{1-b\sqrt{ab}\operatorname{cosh}(\sqrt{\operatorname{log}(ab)}t)}{ab^3-2b\sqrt{ab}\operatorname{cosh}(\sqrt{\operatorname{log}(ab)}t)+1}\right]dt$$how is it derived?

I checked this paper but still did not understand.

pjmathematician
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1 Answers1

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For $B>A >0$ $$\sqrt{\pi}\sum_{n=0}^N (AB)^{n^2}(A/B)^n=\sum_{n=0}^N (A/B)^ne^{(n\sqrt{\log AB})^2}\int_{-\infty}^\infty e^{-t^2}dt$$ Using Cauchy integral theorem we can move the line of integration obtaining $$=\sum_{n=0}^N (A/B)^ne^{(n\sqrt{\log AB})^2}\int_{-\infty}^\infty e^{-(t+n\sqrt{\log AB})^2}dt =\sum_{n=0}^N (A/B)^n\int_{-\infty}^\infty e^{-t^2-2nt\sqrt{\log AB}}dt $$ $$=\int_{-\infty}^\infty e^{-t^2}\sum_{n=0}^N (A/B)^ne^{-2nt\sqrt{\log AB}}dt=\int_{-\infty}^\infty e^{-t^2}\frac{1- (A/B)^{N+1}e^{-2(N+1)t\sqrt{\log AB}}}{1- (A/B)e^{-2t\sqrt{\log AB}}}dt$$

For $A,B$ fixed, since $\sqrt{\log AB}$ is purely imaginary then $|e^{-t^2} \frac{e^{-2(N+1)t\sqrt{\log AB}}}{1- (A/B)e^{-2t\sqrt{\log AB}}}|$ is bounded in $L^1$ norm so that $$\int_{-\infty}^\infty e^{-t^2} \frac{(A/B)^{N+1}e^{-2(N+1)t\sqrt{\log AB}}}{1- (A/B)e^{-2t\sqrt{\log AB}}}dt\to 0$$ thus

$$\sqrt{\pi}\sum_{n=0}^\infty (AB)^{n^2}(A/B)^n=\int_{-\infty}^\infty e^{-t^2}\frac{1}{1- (A/B)e^{-2t\sqrt{\log AB}}}dt$$ Note sure if the last result can be extended by analytic continuation.

reuns
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