Zeta Function - Part 10 - Jacobi Theta Function

What people said

Recommends Max Echendu · 2016-02 · Mathematics Stack Exchange · on “Why is this not a complex variable?”

I'm watching a video [this video] about the Jacobi Theta Function on YouTube. At around 6:14 in the video, he has shown that... $$\vartheta (x) = \sum_{n\in \mathbb Z} e^{-\pi {n^2} x} = \sum_{k\in \mathbb Z} {e^{-\pi {k^2} {1\over x}}} \int_{-\infty}^{\infty} {e^{-\pi x \big( y+ i {k\over x}\big)^2}} \space dy$$ I understand the steps he took to get to this point. But here is where I got confused. He then looked at $y + i {k\over x}$ and he Let... $y + i {k\over x} = z$ But then he said... $dy = dz$ Because $i {k\over x}$ is not a complex variable. I don't understand why it isn't. Isn't a complex number defined as $a+bi$ where $a,b\in \mathbb R$ ? Am I missing something?

Cites Peter Driscoll · 2017-08 · Mathematics Stack Exchange · on “Riemann zeta-function functional equation proof”

I spelled this argument out in some detail, so that I could understand it. The argument is based on [this video] , but elaborated to my tastes. Define $\theta(x) = 2 \psi(x) + 1 $ . Then, $$\theta(x)=(\sum_{n=-\infty}^1 e^{-n^2\pi x}) + 1 + (\sum_{n=1}^\infty e^{-n^2\pi x}) =\sum_{n=\infty}^\infty e^{-n^2\pi x}$$ The Poisson summation formula [LINK] . is, $$ \sum_{n-\infty}^\infty{f(n)} = \sum_{k=-\infty}^\infty \int_{-\infty}^\infty f(y) e^{-2\pi iky} x dy$$ Substituting $ f(n) = e^{-n^2\pi x}$ and $ f(y) = e^{-y^2\pi x}$ gives, $$ \theta(x) = \sum_{n-\infty}^\infty{e^{-n^2\pi x}} = \sum_{k=-\infty}^\infty \int_{-\infty}^\infty e^{-y^2\pi x} e^{-2\pi iky} dy = \sum_{k=-\infty}^\infty \int_{-\inf

Cites Michal Dvořák · 2017-12 · Mathematics Stack Exchange · on “Improper definite integration with complex bounds”

I am looking to prove the functional equation for theta function. Source: [this video] Time about 8.00. We are to integrate: $$\int_{-\infty+\frac{ik}{x}}^{+\infty+\frac{ik}{x}}e^{-\pi xz^2}dz$$ The argument is to change the bounds to $-\infty$ and $+\infty$ using something called Estimation Lemma or "ML-Inequality". I have absolutely no knowledge of complex analysis and integrals with complex variables. I've read in the comments, that one could say that $\infty+k = \infty $ for any finite number $k$, equivallently for $-\infty$ but as there is $i$, this argument seems invalid to me. Would anyone give me some deeper explanation of what we have actually done?