Summation of exponential terms

[A_s_a_d]

I found the following identity in a paper:
$\sum_{l=1}^{\infty}exp(-\pi\alpha l^2)=(\frac{1}{2\sqrt{\alpha}}-\frac{1}{2})+\frac{1}{\sqrt{\alpha}}\sum_{l=1}^{\infty}exp(\frac{-\pi l^2}{\alpha})$
Someone please let me give some hints on how to prove this.

 

[jfizzix]

It looks like it's the same sum on both sides, but scaled.

I would see what happens to the exponential sum when you change \alpha to \alpha\beta. Can you make the result identical with the right scaling factor in front (twice as wide, but half as tall)?

 

[A_s_a_d]

Yes, it's the same sum but scaled. One thing I have found that if you approximate the summation as integral, it can be proved easily as both are usual Gaussian Integral. But I was worrying about the factor that involves to transform the summation into integral. Any thoughts?

 

[jfizzix]

So, you can write it alternatively as:
\frac{1}{2} +\sum_{l=1}^{\infty}exp(-\pi\alpha l^2)=\frac{1}{\sqrt{\alpha}}\big(\frac{1}{2}+\sum_{l=1}^{\infty}exp(\frac{-\pi l^2}{\alpha})\big)
or better yet
\sum_{l=-\infty}^{\infty}exp(-\pi\alpha l^2)=\frac{1}{\sqrt{\alpha}}\sum_{l=-\infty}^{\infty}exp(\frac{-\pi l^2}{\alpha})
this sum is exactly the integral of a discrete gaussian. With the normalization constant in front, they should have the same area..

At least, that is how I think it should work out.

Does \alpha have to be a positive integer?