Math identities, Poisson summation formula 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&gt;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&gt;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} &amp; =  \sum_{\nu=-\infty}^\infty \int_{-\infty}^\infty dn\, e^{-\pi a (n+1/2)^2 + i 2\pi n\nu} \nonumber\\
&amp; =  \sum_{\nu=-\infty}^\infty \frac{1}{\sqrt{a}} e^{-i \nu\pi -\pi \nu^2/a} \nonumber\\
&amp;= \sum_{n=-\infty}^\infty \frac{1}{\sqrt{a}} (-1)^n e^{-\pi n^2/a} \nonumber\\
&amp;= \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} &amp;=&amp;
    \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......\\  
   &amp;&amp;      \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} &amp;=&amp;
    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\\  
     &amp;=&amp;
     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
I think this are wrong formulas. Consider the first one at $a \gg 1$. The left hand side is exponentially small positive. Whereas the sum at the right hand side tends to the limiting value $-1/2$. Thus this formula can't be correct.
I think this are wrong formulas. Consider the first one at $a \gg 1$. The left hand side is exponentially small positive. Whereas the sum at the right hand side tends to the limiting value $-1/2$. Thus this formula can't be correct.