Interesting sum-integral equality Is there an elementary proof of
$$\lim_{n \to \infty} \int_0^\infty e^{-\alpha x^2} \frac{\sin((2n + 1)x)}{\sin x} dx = \pi\left(\frac{1}{2} + \sum_{k = 1}^\infty e^{-\alpha k^2 \pi^2}\right),$$
where $\alpha > 0$? I remember seeing a proof somewhere, but I cannot remember where or how it was done.
Thank you
 A: Consider 
$$I_n=\int_0^{\infty} e^{-\alpha x^2} \frac{\sin((2n+1)x)}{\sin x}\,dx \Rightarrow I_{n-1}=\int_0^{\infty} e^{-\alpha x^2} \frac{\sin((2n-1)x)}{\sin x}\,dx$$
$$\Rightarrow I_n-I_{n-1}=2\int_0^{\infty} e^{-\alpha x^2}\cos(2n x)\,dx=\sqrt{\frac{\pi}{\alpha}}e^{-n^2/\alpha}\,\,\,\,\,\,\,(*)$$
Hence,
$$I_n-I_0=\sqrt{\frac{\pi}{\alpha}}\sum_{k=1}^n e^{-k^2/\alpha}$$
Since, $I_0=\frac{1}{2}\sqrt{\frac{\pi}{\alpha}}$, 
$$I_n=\sqrt{\frac{\pi}{\alpha}}\left(\frac{1}{2}+\sum_{k=1}^n e^{-k^2/\alpha}\right)$$
$$\Rightarrow \lim_{n\rightarrow \infty} I_n=\sqrt{\frac{\pi}{\alpha}}\left(\frac{1}{2}+\sum_{k=1}^{\infty}e^{-k^2/\alpha}\right)=\frac{1}{2}\sqrt{\frac{\pi}{\alpha}}\sum_{k=-\infty}^{\infty} e^{-k^2/\alpha}$$
From the Poisson summation formula:
$$ \begin{aligned}
\sum_{k=-\infty}^{\infty} e^{-k^2/\alpha} &=\sum_{k=-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^2/\alpha}e^{-2\pi ikx}\,dx \\ &=\sum_{k=-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^2/\alpha}\cos(2\pi kx)\,dx \\
&=\sum_{k=-\infty}^{\infty} \sqrt{\pi \alpha}\,e^{-\alpha\pi^2k^2}\,\,\,\,\,\,\,(**) \\ &=\sqrt{\alpha\pi}\left(1+2\sum_{k=1}^{\infty} e^{-\alpha k^2\pi^2}\right)\\
\end{aligned}$$
Hence,
$$\boxed{\displaystyle  \lim_{n\rightarrow \infty} I_n=\pi \left(\dfrac{1}{2}+\sum_{k=1}^{\infty} e^{-\alpha k^2\pi^2} \right)}$$

The integrals in $(*)$ and $(**)$ can be evaluated using the techniques mentioned here:
Gaussian-like integral : $\int_0^{\infty} e^{-x^2} \cos( a x) \ \mathrm{d}x$
