The following sum may it is easy for computation
$$-\int_{0}^{\infty} \bigg( \exp(-\dfrac{\pi}{4}x²)) \bigg) \;dx+\bigg(\sum_{n=0}^{\infty }\bigg( \exp(-\dfrac{\pi}{4}n²)\bigg)=\dfrac12$$
The sum LHS can be computed by theta function and the integral by error function which is equal 1 , Wolfram alpha assumed that difference close to the half integer , But i want to know if the above identity can be considered be true analytically ?
The Poisson Summation Formula says $$ \sum_{n\in\mathbb{Z}}e^{-\frac\pi4n^2}=2\sum_{n\in\mathbb{Z}}e^{-4\pi n^2}\implies \sum_{n=0}^\infty e^{-\frac\pi4n^2}=\frac32+2\sum_{n=1}^\infty e^{-4\pi n^2} $$ Furthermore, we have the Gaussian Integral $$ \int_{-\infty}^\infty e^{-\pi x^2}\,\mathrm{d}x=1\implies \int_0^\infty e^{-\frac\pi4x^2}\,\mathrm{d}x=1 $$ Thus, $$ -\int_0^\infty e^{-\frac\pi4x^2}\,\mathrm{d}x+\sum_{n=0}^\infty e^{-\frac\pi4n^2} =\frac12+2\sum_{n=1}^\infty e^{-4\pi n^2} $$ Thus, the expression in the question exceeds $\frac12$ by approximately $2e^{-4\pi}\doteq0.000006975$.
Wolfy says
$1/2 (1 + ϑ_3(0, e^{-π/4})) - 1≈0.500006974684712417991279357455722773386084811819343959670243423623882370819559454961925300924629951 $.
So, no.
