Volume of the $n$-ball

The following is a simple and beautiful proof, shown to me to my great delight while I was in high school, that the {n}-ball of radius {r} has {n}-volume

\displaystyle  \frac{\pi^{n/2} r^n}{\Gamma(\frac{n}{2}+1)}

Although I expect most readers will know it, I believe that everybody should see it. I don’t know the history of it, and would be interested in learning.

Suppose that the {n}-ball {B^n} of radius {1} has {n}-volume {V_n(1)}. Then, by considering linear transformations, the {n}-ball {rB^n} of radius {r} has {n}-volume {V_n(r) = V_n(1) r^n}. Moreover, differentiating this with respect to {r} should produce the surface {(n-1)}-volume of the {(n-1)}-sphere {S^{n-1} = \partial B^n}: thus we expect {\text{vol}^{(n-1)}(S^{n-1}) = V_n(1) n r^{n-1}}.

Now consider the integral

\displaystyle I = \int_{\mathbf{R}^n} e^{-\pi|x|^2}\,dx.

We will compute {I} in two different ways. On the one hand,

\displaystyle I = \int_{-\infty}^{+\infty}\cdots\int_{-\infty}^{+\infty} e^{-\pi x_1^2 -\cdots - \pi x_n^2}\,dx_1\cdots dx_n = \left(\int_{-\infty}^{+\infty} e^{-\pi x^2}\,dx\right)^n = 1.

On the other hand, the form of {I} suggests introducing a radial coordinate {r=|x|}. Computing this way,

\displaystyle I = \int_0^\infty e^{-\pi r^2} (V_n(1) n r^{n-1})\,dr = \frac{n V_n(1)}{\pi^{n/2}} \int_0^\infty e^{-t^2} t^{n-1}\, dt\qquad\quad

\displaystyle \qquad\quad= \frac{n V_n(1)}{2 \pi^{n/2}} \int_0^\infty e^{-s} s^{n/2-1}\,ds = \frac{n V_n(1)}{2 \pi^{n/2}} \Gamma(n/2) = \frac{V_n(1)\Gamma(n/2+1)}{\pi^{n/2}}.

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