I have seen an integral \begin{equation}\frac{1}{(2 \pi)^{\frac{n_z}{2}}} \int_{\varepsilon_{v}} e^{-\frac{1}{2}|| \boldsymbol{v} \|^{2}} \,\mathrm{d}v\,,\end{equation} where $\mathcal{E}_{v}:=\left\{v:\|v\|^{2} \leq r^{2}\right\}.$
It results in the integration $$I=\frac{1}{2^{n_{z} / 2} \Gamma\left(n_{z} / 2\right)} \int_{0}^{r^{2}} \chi^{\frac{n_{z}}{2}-1} e^{-\frac{\chi}{2}} d \chi$$
I am wondering why the result of the first integral is not the same as $\int e^{-\frac{x^2}{2}}$, which leads to $\sqrt{2}\Gamma{(\frac{1}{2})}$. And how should I evaluate the integral over $0$ to $r^2$ in the second equation.
Any hints are appreciated. Thanks.
To solve your integral in terms of $\chi$, change variables to $y = \chi / 2$. This removes the factor of $1/2$ in the exponential and then you'll be able to see that it's an incomplete gamma function integral.
The reason that the integral is different from $\int_{-\infty}^\infty e^{-\frac{x^2}{2}}dx$ is that your original integral is over an n-dimensional space. Your integral is spherically symmetric, so to get to your final form you need to first change coordinates to spherical polars. The factor of $\chi^{\frac{n_{z}}{2}-1}$ comes from the volume element in spherical polars, and the $\Gamma\left(n_{z} / 2\right)$ in the answer comes from the integration over the angular coordinates, it's related to the volume of the n-ball.