I have derived this following formulas using Poisson summation formula. Can anyone kindly refer if this identity is derived or quoted somewhere? I wanted to be sure.
\begin{equation} \sum_{n=0}^{\infty}e^{-\pi\left(n+\frac{1}{2}\right)^2 a }=\frac{1}{2\sqrt{a}}+\sum_{n=1}^{\infty}(-1)^n e^{-\pi n^2/a}, \quad a>0 \end{equation} and \begin{equation} \sum_{n=-\infty}^{\infty}e^{-\pi\left(n+\frac{1}{2}\right)^2 a }=\frac{1}{\sqrt{a}}+2\sum_{n=1}^{\infty}(-1)^n e^{-\pi n^2/a}, \quad a>0 \end{equation}
Here is the proof:
The Poisson summation formula can be stated as \begin{equation} \sum_{n\in\mathbb{Z}} f(n) = \sum_{n\in\mathbb{Z}} \hat{f}(\nu), \end{equation} where $\hat{f}(\nu)$ is the Fourier transform of $f(n)$ and \begin{equation} \hat{f}(\nu) = \int_{-\infty}^\infty dn\, f(n) e^{i 2\pi n\nu}. \end{equation} Then, \begin{align} \sum_{n=-\infty}^{\infty} e^{-\pi a (n+1/2)^2} & = \sum_{\nu=-\infty}^\infty \int_{-\infty}^\infty dn\, e^{-\pi a (n+1/2)^2 + i 2\pi n\nu} \nonumber\\ & = \sum_{\nu=-\infty}^\infty \frac{1}{\sqrt{a}} e^{-i \nu\pi -\pi \nu^2/a} \nonumber\\ &= \sum_{n=-\infty}^\infty \frac{1}{\sqrt{a}} (-1)^n e^{-\pi n^2/a} \nonumber\\ &= \frac{1}{\sqrt{a}} + 2\sum_{n=1}^\infty \frac{1}{\sqrt{a}} (-1)^n e^{-\pi n^2/a} \end{align} Now the left hand side of above eq. can be re written as,
\begin{eqnarray} \sum_{n=-\infty}^{\infty} e^{-\pi a (n+1/2)^2} &=& \underbrace{e^{-\pi a/4} }_\text{$n=0$} + \underbrace{e^{-\pi a(3/2)^2} }_\text{$n=1$} + \underbrace{e^{-\pi a(5/2)^2 } }_\text{$n=2$} + \underbrace{e^{-\pi a(7/2)^2} }_\text{$n=3$} + \underbrace{e^{-\pi a(9/2)^2} }_\text{$n=4$} + \nonumber......\\ && \underbrace{e^{-\pi a/4} }_\text{$n=-1$} + \underbrace{e^{-\pi a(3/2)^2} }_\text{$n=-2$} + \underbrace{e^{-\pi a(5/2)^2 } }_\text{$n=-3$} + \underbrace{e^{-\pi a(7/2)^2} }_\text{$n=-4$} + \underbrace{e^{-\pi a(9/2)^2} }_\text{$n=-5$} + ...... \end{eqnarray} Therefore one can see the $n=0$ term matches with $n=-1$, $n=1$ term matches with $n=-2$, $n=2$ term matches with $n=-3$, $n=3$ term matches with $n=4$ and so on. As a result the eq. above can be written as \begin{eqnarray} \sum_{n=-\infty}^{\infty} e^{-\pi a (n+1/2)^2} &=& 2(\underbrace{e^{-\pi a/4} }_\text{$n=0$} + \underbrace{e^{-\pi a(3/2)^2} }_\text{$n=1$} + \underbrace{e^{-\pi a(5/2)^2 } }_\text{$n=2$} + \underbrace{e^{-\pi a(7/2)^2} }_\text{$n=3$} + \underbrace{e^{-\pi a(9/2)^2} }_\text{$n=4$} + \nonumber......)\nonumber\\ &=& 2\sum_{n=0}^{\infty}e^{-\pi\left(n+\frac{1}{2}\right)^2 a }. \end{eqnarray} Combining equations above we obtain, \begin{eqnarray} \sum_{n=0}^{\infty}e^{-\pi\left(n+\frac{1}{2}\right)^2 a }=\frac{1}{2\sqrt{a}}+\sum_{n=1}^{\infty}(-1)^n e^{-\pi n^2/a}\nonumber \end{eqnarray}
If anyone thinks any error in derivation, kindly let me know. If anyone knows any reference where this formulas are quoted let me know