It isn't an ordinary one!

Consider the function: $\varphi \left( x \right) =\sum _{ n=-\infty }^{ \infty }{ { x }^{ { n }^{ 2 } } }$

We need to find: $\displaystyle \varphi(e^{-\pi})=\frac{\pi^{\frac{1}{4}}}{\Gamma \left(\frac{3}{4} \right)}$

We take the following integral: $\displaystyle I(a)= \int_0^{\frac{\pi}{2}} \frac{1}{\sqrt{1-a^2\sin^2\theta}}d\theta = \frac{\pi}{2}\varphi^2(q)=\frac{\pi_2F_1 \left(\frac{1}{2},\frac{1}{2};1;a^2 \right)}{2}$

Here $\displaystyle q=e^{-\pi \frac{I'}{I}}$ .

I'm working on the proof of this Identity

On inserting appropriate values, we get: $\displaystyle \varphi^2(e^{-\pi})=_2F_1 \left(\frac{1}{2},\frac{1}{2};1;\frac{1}{2} \right)$

We use the identity: $\displaystyle _2F_1 \left( a,b;\frac{1}{2}(a+b+1);\frac{1}{2}\right)= \frac{\sqrt{\pi} \Gamma \left[ \frac{1}{2}(a+b+1) \right]}{\Gamma \left( \frac{a+1}{2}\right)\Gamma \left( \frac{b+1}{2}\right)}$

I'm working on the proof of this Identity

Hence, $\boxed{\displaystyle \varphi(e^{-\pi})=\frac{\pi^{\frac{1}{4}}}{\Gamma \left(\frac{3}{4} \right)}}$

Adapted from the original proof of $\varphi \left( { -e }^{ -\pi } \right)$ by Ramanujan .

Please provide appropriate proofs of some identities used here.