In the last step of the proof of Theorem 10.1 in the book Multiplicative number theory I: Classical theory by Hugh L. Montgomery, Robert C. Vaughan I couldn't understand what exactly "turn the path through an angle" means. I reparametrize $x$ to $xe^{-\frac12 \arg z}$ and I got $$\int_{-\infty}^{\infty} e^{- \pi x^2 e^{-\arg z}z} e^{-\frac12 \arg z} dx$$ which is not equal to the result that the authors claim, i.e. $$z^{-1/2} \int_{-\infty}^{\infty} e^{- \pi x^2} dx = |z|^{-1/2} e^{-\frac12 \arg z} \int_{-\infty}^{\infty} e^{- \pi x^2} dx.$$
My question is how to show that $$\int_{-\infty}^{\infty} e^{- \pi x^2 z} dx = z^{-1/2} \int_{-\infty}^{\infty} e^{- \pi x^2} dx$$ is true?
It means (I think) to consider the following path:
That is, the rotated path is $\gamma(t) = \frac{t}{\sqrt{z}} = \frac{t}{ \sqrt{|z|}} e^{-i\frac{ arg z}{2}}$, so we also scale by $\frac{1}{\sqrt{|z|}}$.
The vertical parts will tend to $0$ so we get
$$ \int_{-\infty}^\infty e^{-\pi x^2z} dx = \int_{-\infty}^\infty e^{-\pi \left(\frac{t}{\sqrt{z}}\right)^2z} \frac{1}{\sqrt{z}} dt = z^{-\frac{1}{2}} \int_{-\infty}^\infty e^{-\pi t^2} dt. $$