Volume of a hypersphere

In this appendix we will compute the volume of an $\ell$$\ell$-dimensional hypersphere of radius $R$$R$, which we call $V_{\ell }(R)$$V_{\ell }(R)$. This is defined as:

$$V_{\ell }(R)=\int \Theta (R-|{\vec {p}}|)d{\vec {p}}$$

$$V_{\ell }(R)=\int \Theta (R-|{\vec {p}}|)d{\vec {p}}$$

In order to calculate it we will use a "trick". First, we change variable defining ${\vec {z}}$${\vec {z}}$ so that $p_{i}=Rz_{i}$$p_{i}=Rz_{i}$ and thus:

$$V_{\ell }(R)=R^{\ell }\int \Theta (1-|{\vec {z}}|)d{\vec {z}}=R^{\ell }V_{\ell }(1)$$

$$V_{\ell }(R)=R^{\ell }\int \Theta (1-|{\vec {z}}|)d{\vec {z}}=R^{\ell }V_{\ell }(1)$$

We then have:

$${\frac {d}{dR}}V_{\ell }(R)=\ell R^{\ell -1}V_{\ell }(1)=\int \delta (|{\vec {p}}|-R)d{\vec {p}}$$

$${\frac {d}{dR}}V_{\ell }(R)=\ell R^{\ell -1}V_{\ell }(1)=\int \delta (|{\vec {p}}|-R)d{\vec {p}}$$

Multiplying both sides by ${\textstyle e^{-R^{2}}}$${\textstyle e^{-R^{2}}}$ and integrating over $R$$R$ from $0$$0$ to infinity:

$$V_{\ell }(1)\int _{0}^{\infty }e^{-R^{2}}\ell R^{\ell -1}dR=\int _{0}^{\infty }dR\int e^{-R^{2}}\delta (|{\vec {p}}|-R)d{\vec {p}}$$

$$V_{\ell }(1)\int _{0}^{\infty }e^{-R^{2}}\ell R^{\ell -1}dR=\int _{0}^{\infty }dR\int e^{-R^{2}}\delta (|{\vec {p}}|-R)d{\vec {p}}$$

The right hand side is equal to:

$$\int _{0}^{\infty }dR\int e^{-R^{2}}\delta (|{\vec {p}}|-R)d{\vec {p}}=\int d{\vec {p}}\int _{0}^{\infty }e^{-R^{2}}\delta (|{\vec {p}}|-R)=\int d{\vec {p}}e^{-|{\vec {p}}|^{2}}$$

$$\int _{0}^{\infty }dR\int e^{-R^{2}}\delta (|{\vec {p}}|-R)d{\vec {p}}=\int d{\vec {p}}\int _{0}^{\infty }e^{-R^{2}}\delta (|{\vec {p}}|-R)=\int d{\vec {p}}e^{-|{\vec {p}}|^{2}}$$

Therefore:

$$\ell V_{\ell }(1)\int _{0}^{\infty }R^{\ell -1}e^{-R^{2}}dR=\int d{\vec {p}}\prod _{i=1}^{\ell }e^{-p_{i}^{2}}=\prod _{i=1}^{\ell }\underbrace {\int dp_{i}e^{-p_{i}^{2}}} _{\sqrt {\pi }}=\pi ^{\ell /2}$$

$$\ell V_{\ell }(1)\int _{0}^{\infty }R^{\ell -1}e^{-R^{2}}dR=\int d{\vec {p}}\prod _{i=1}^{\ell }e^{-p_{i}^{2}}=\prod _{i=1}^{\ell }\underbrace {\int dp_{i}e^{-p_{i}^{2}}} _{\sqrt {\pi }}=\pi ^{\ell /2}$$

If we now define ${\textstyle t=R^{2}}$${\textstyle t=R^{2}}$, in the integral in the left hand side of the equation we recognise the definition of Euler's $\Gamma$$\Gamma$ function:

$$\ell \int _{0}^{\infty }R^{\ell -1}e^{-R^{2}}dR={\frac {\ell }{2}}\int _{0}^{\infty }e^{-t}t^{{\frac {\ell }{2}}-1}dt={\frac {\ell }{2}}\Gamma \left({\frac {\ell }{2}}\right)=\Gamma \left({\frac {\ell }{2}}+1\right)$$

$$\ell \int _{0}^{\infty }R^{\ell -1}e^{-R^{2}}dR={\frac {\ell }{2}}\int _{0}^{\infty }e^{-t}t^{{\frac {\ell }{2}}-1}dt={\frac {\ell }{2}}\Gamma \left({\frac {\ell }{2}}\right)=\Gamma \left({\frac {\ell }{2}}+1\right)$$

This way the volume of an $\ell$$\ell$-dimensional sphere of unit radius is ${\textstyle V_{\ell }(1)=\pi ^{\ell /2}/\Gamma \left({\frac {\ell }{2}}+1\right)}$${\textstyle V_{\ell }(1)=\pi ^{\ell /2}/\Gamma \left({\frac {\ell }{2}}+1\right)}$, and thanks to the fact that $V_{\ell }(R)=R^{\ell }V_{\ell }(1)$$V_{\ell }(R)=R^{\ell }V_{\ell }(1)$, in the end we have:

$$V_{\ell }(R)={\frac {R^{\ell }\pi ^{\ell /2}}{\Gamma \left({\frac {\ell }{2}}+1\right)}}$$

$$V_{\ell }(R)={\frac {R^{\ell }\pi ^{\ell /2}}{\Gamma \left({\frac {\ell }{2}}+1\right)}}$$

Let us verify that this formula returns the values we expect for $\ell =2$$\ell =2$ and $\ell =3$$\ell =3$. We first note that:

$$\Gamma \left({\frac {1}{2}}\right)=\int _{0}^{\infty }e^{-t}t^{-1/2}dt={\sqrt {\pi }}$$

$$\Gamma \left({\frac {1}{2}}\right)=\int _{0}^{\infty }e^{-t}t^{-1/2}dt={\sqrt {\pi }}$$

$$\Gamma \left({\frac {5}{2}}\right)={\frac {3}{2}}\Gamma \left({\frac {3}{2}}\right)={\frac {3}{2}}\cdot {\frac {1}{2}}\Gamma \left({\frac {1}{2}}\right)={\frac {3}{4}}{\sqrt {\pi }}$$

$$\Gamma \left({\frac {5}{2}}\right)={\frac {3}{2}}\Gamma \left({\frac {3}{2}}\right)={\frac {3}{2}}\cdot {\frac {1}{2}}\Gamma \left({\frac {1}{2}}\right)={\frac {3}{4}}{\sqrt {\pi }}$$

This way:

$$V_{2}(R)={\frac {R^{2}\pi ^{2/2}}{\Gamma (2)}}={\frac {\pi R^{2}}{1!}}=\pi R^{2}\quad \quad V_{3}(R)={\frac {R^{3}\pi ^{3/2}}{{\frac {3}{4}}{\sqrt {\pi }}}}={\frac {4}{3}}\pi R^{3}$$

$$V_{2}(R)={\frac {R^{2}\pi ^{2/2}}{\Gamma (2)}}={\frac {\pi R^{2}}{1!}}=\pi R^{2}\quad \quad V_{3}(R)={\frac {R^{3}\pi ^{3/2}}{{\frac {3}{4}}{\sqrt {\pi }}}}={\frac {4}{3}}\pi R^{3}$$

which is exactly what we expected.