How can I arrive from $\prod_{i=1}^{D} \int_{-\infty}^{\infty} e^{-x_i^2} dx_i$ to $\pi^{D/2}$?

Question

This is part of the excersise 1.18 of the Bishop's Book, Pattern Recognition and Machine Learning. I attached the excersise and the information. I do not know how he using the result (1.126) in the left side of (1.142) arrives to $\pi^{D/2}$ (That is the solution he provides). I tried several ways but I get nothing.

1.126: $I=(2 \pi \sigma^{2})^{1/2}$

which comes from: $I^{2}= \int_{-\infty}^{\infty} e^{-\frac{1}{2\sigma^{2}}x^{2}-\frac{1}{2\sigma^{2}}y^{2}}dxdy $

And solving transforming form Cartesian coordinates to polar ones, substituing u = $r^{2}$ and performing the integrals over u and $\theta$

I attached the Excersises, 1.18. I can not upload more pictures so here is a drive link to the others: https://drive.google.com/drive/folders/1sc6WFyhb5wEevZOv-6qcTFKuoieCxPGH?usp=sharing

Answer 1

Recall that $\int_{-\infty}^{\infty} e^{-t^2} dt = \sqrt{\pi} = \pi^{1/2}.$ Therefore $$ \prod_{i=1}^{D} \int_{-\infty}^{\infty} e^{-x_i^2} dx_i = \prod_{i=1}^{D} \pi^{1/2} = (\pi^{1/2})^D = \pi^{D/2}. $$

Answer 2

\begin{align*} I &=\int_{-\infty}^{\infty} e^{-x_i^2} dx_i \\ I^2 &= \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x_i^2} e^{-y_i^2} dx_i dy_i \\ &=\int_{0}^{2\pi} \int_{0}^{\infty} e^{-r^2} r dr d\theta \tag{1} \\ &=\int_{0}^{2\pi} -\frac{e^{-r^2}}{2} \bigg \rvert_{r=0}^{\infty} d\theta \\ &=\int_{0}^{2\pi} \frac{1}{2} d\theta \\ I &=\sqrt{\pi} \\ \end{align*}

And so, \begin{align*} \prod_{i=1}^D \int_{-\infty}^{\infty} e^{-x_i^2} dx_i &= \prod_{i=1}^D \sqrt{\pi} \\ &= \boxed{\pi^{\frac{D}{2}}} \end{align*}

Where in $(1)$ the double integral was converted to polar coordinates.