How does one show $$ \int_0^{\infty} e^{-t^2/2} \left[ \frac{e^{2\pi} - \cos\left(\sqrt{2\pi} t \right)}{e^{4\pi} - 2 e^{2\pi} \cos\left(\sqrt{2\pi} t\right) + 1} \right] dt = \frac{\pi^{1/4}}{4e^\pi\Gamma\left(\frac{3}{4}\right)}-\frac{1}{e^\pi}\sqrt{\frac{\pi}{8}} \,? $$ This comes directly from here. The integral is related to one of Ramanujan's mock theta functions by $$ \varphi(e^{-\pi}) =\sum_{n=-\infty}^\infty e^{-\pi n^2}. $$ How do I show that the original integral is related to this series? The only possibly way I can reasonably see this being done is via contour integration. Is there a somewhat systematic way to reduce the integral into the sum?
As for the explicit result: what references can I look into to read more about these functions? I have yet to find a solid article or book containing detailed evaluations of these functions and would be delighted to find some.
Thank you!
For $|r|<1$ we have $$\frac{1-r\cos\phi}{1-2r\cos\phi+r^2}=\frac12\left(\frac{1}{1-re^{i\phi}}+\frac{1}{1-re^{-i\phi}}\right)=\sum_{n=0}^\infty r^n\cos n\phi$$ (we'll apply it with $r=e^{-2\pi}$ and $\phi=t\sqrt{2\pi}$) and, for $a>0$ (in our case $a=1/2$), $$\int_0^\infty e^{-at^2}\cos bt\,dt=\sum_{n=0}^\infty\frac{(-b^2)^n}{(2n)!}\int_0^\infty t^{2n}e^{-at^2}\,dt=\ldots=\sqrt\frac{\pi}{4a}e^{-\frac{b^2}{4a}}.$$ So, the given integral is equal to $$e^{-2\pi}\sum_{n=0}^\infty e^{-2n\pi}\int_0^\infty e^{-t^2/2}\cos(nt\sqrt{2\pi})\,dt=e^{-\pi}\sqrt{\pi/2}\sum_{n=1}^\infty e^{-\pi n^2}.$$ The sum is $(S-1)/2$, where $S$ is the sum you seem to know: $$S=\sum_{n=-\infty}^\infty e^{-\pi n^2}=\frac{\pi^{1/4}}{\Gamma(3/4)}.$$