Is $e^{u/2}\sum_{n=-\infty}^{\infty}e^{-\pi n^{2}e^{2u}}$ even?

Question

In this paper: "ON A RESULT OF G.PÓLYA CONCERNING THE RIEMANN $\xi - FUNCTION$ by DENNIS A. HEJHAL"

the author defines

$$ \theta(x)=\sum_{n=-\infty}^{\infty}e^{-\pi n^{2} x} $$ then he says, in the begining of the second page: "Since $e^{u/2} \theta(e^{2u})$" is even, so is ..."

I Can't seen why $e^{u/2} \theta(e^{2u})$ is even. He is saying that the following function is even

$$e^{u/2}\sum_{n=-\infty}^{\infty}e^{-\pi n^{2}e^{2u}}$$

I used mathematica and got the following result

Clearly this is not even. Certainly I'm missing something obvious, or is this a typo? If so what should have been written?

Thanks.

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After the remark of Professor Vetor, I produced computations with more terms, and indeed it seems that it is even. Amazing!!

Answer 1

The clue is the use of $\theta$ for the function. This is a variant of a Jacobi theta function. More precisely, $\;\theta(x)=\theta_3(0,e^{-\pi x}).\;$ By the transformation properties of $\theta_3$ we have $\theta(x)=\theta(1/x)/\sqrt{x},\;$ and combined with $\;e^{-u}=1/e^u,\;\sqrt{e^u}=e^{u/2},\;$ the even property follows.