1 6 2 Volume Of A P Dimensional Ball
1.
(1)
\begin{align} \Gamma(1) = \int_0^\infty e^{-x}dx = [-e^{-x}]_0^\infty , \end{align}
which gives $\boxed{\Gamma(1) = 1.}$
- The change of variable $y=\sqrt {x}$ gives:
(2)
\begin{align} \Gamma(1/2) = \int_0^\infty e^{-x}\frac{dx}{\sqrt{x}} = 2\int_0^\infty e^{-y^2}dy = \int_{-\infty}^{+\infty}e^{-y^2}dy \end{align}
which is the Gaussian integral. We can compute it by the following method, using Fubini-Tonelli theorem and a polar change of variables:
(3)
\begin{align} \Gamma(1/2)^2 = \int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}e^{-x^2-y^2}dx\ dy = \int_0^{+\infty}\int_{0}^{2\pi}e^{-r^2}r d\theta dr = 2\pi[-e^{-r^2}/2]_0^{+\infty} = \pi. \end{align}
so $\boxed{\Gamma(1/2) = \sqrt{\pi}.}$
- By partial integration, for $\alpha>0$,
(4)
\begin{align} \Gamma(\alpha +1) = \int_0^{+\infty}x^\alpha e^{-x}dx = [-x^\alpha e^{-x}]_0^{+\infty} + \int_0^{+\infty} \alpha x^{\alpha-1}e^{-x}dx, \end{align}
it gives $\boxed{\Gamma(\alpha+1) = \alpha\Gamma(\alpha).}$
- Then we can deduce by recurrence, that for all $\alpha>0$ and $p\in\mathbb{N}^*$,
(5)
\begin{align} \Gamma(\alpha + p) = \Gamma(\alpha)\Pi_{k=1}^{p}(\alpha+p-k), \end{align}
and as a corollary, if we take $\alpha=1$ and $\alpha=1/2$ in this formula, we get for all $p\in\mathbb{N}$:
(6)
\begin{align} \boxed{\Gamma(p+1) = p!} \ \text{and} \ \Gamma(p+3/2) = \Gamma(1/2)\Pi_{k=0}^{p}(1/2+p-k) \end{align}
so
(7)
\begin{align} \boxed{\Gamma(p+3/2)= \frac{(2p+1)(2p-1)\dots 1}{2^{p+1}}\sqrt{\pi}.} \end{align}
2.
- Here is the integral definition of the volume:
(8)
\begin{align} V_p(r) = \int_{\mathbb{R}^p}\mathbf{1}_{||x||\leq r}\text{Leb}_p(dx), \end{align}
with $\text{Leb}_p$ the Lebesgue measure on $\mathbb{R}^p$. The change of variables $y = r^{-1}x$ gives
(9)
\begin{align} V_p(r) = \int_{\mathbb{R}^p}\mathbf{1}_{||y||\leq 1}\det(r^{-1}\mathbf{I_p})^{-1}\text{Leb}_p(dy) \end{align}
so we have $\boxed{V_p(r)= r^p V_p(1).}$
(10)
\begin{align} V_1(1) = \int_{\mathbb{R}}\mathbf{1}_{|x|\leq 1}dx = \int_{-1}^{1}dx, \end{align}
so $\boxed{V_1(1)=2.}$
- With Fubini-Tonelli theorem, we can write
(11)
\begin{align} V_2(1) = \int_{\mathbb{R}}\int_{\mathbb{R}}\mathbf{1}_{x^2+y^2\leq 1}dx\ dy \end{align}
and with a polar change of variables:
(12)
\begin{align} V_2(1) = \int_0^{+\infty}\int_0^{2\pi}\mathbf{1}_{r^2\leq 1}rd\theta\ dr = \int_0^1 2\pi r \ dr, \end{align}
so $\boxed{V_2(1) = \pi.}$
3. Let $p\geq 3$ be an integer. An application of Fubini theorem gives:
(13)
\begin{align} V_p(1) &= \int_{\mathbb{R}^p}\mathbf{1}_{\sum x_i^2\leq 1}dx_1\ dx_2 \dots dx_p,\\ &= \int_{\mathbb{R}^2}\mathbf{1}_{ x_1^2+x_2^2\leq 1}\int_{\mathbb{R}^{p-2}}\mathbf{1}_{x_3^2+\dots+x_p^2\leq 1 - x_1^2-x_2^2}dx_3 \dots dx_p\ dx_1\ dx_2,\\ &= \int_{x_1^2+x_2^2\leq 1}V_{p-2}(\sqrt{1-x_1^2-x_2^2})dx_1\ dx_2. \end{align}
Now we can compute a polar change of variables:
(14)
\begin{align} V_p(1) &= \int_0^{+\infty}\int_0^{2\pi}\mathbf{1}_{r^2\leq 1}V_{p-2}(\sqrt{1-r^2})rd\theta\ dr,\\ &= 2\pi\int_0^1 r V_{p-2}(\sqrt{1-r^2}) dr,\\ &= 2\pi V_{p-2}(1)\int_0^1 r(1-r^2)^{p/2-1} dr. \end{align}
The term under the integral is the derivative of $r\mapsto -p^{-1}(1-r^2)^{p/2}$.
We finally have $\boxed{V_p(1) = \frac{2\pi}{p}V_{p-2}(1).}$
4. The last result gives by an immediate recurrence, for all $p\in\mathbb{N}^*$,
(15)
\begin{align} V_{2p}(1) =\frac{\pi^{p-1}}{p!}V_2(1)\ \text{and}\ V_{2p-1} = \frac{(2\pi)^{p-1}}{\Pi_{k=1}^{p-1}(2p-1-2k)}V_1(1). \end{align}
So finally $\boxed{V_{2p}(1) = \frac{\pi^p}{p!}}$ and $\boxed{V_{2p+1}(1) = \frac{2^{p+1}\pi^p}{(2p+1)(2p-1)\dots 3}.}$
- Still, the formula is not yet perfect. We notice a redundancy with the expressions of the gamma function from question 1.: for all $p\in\mathbb{N}^*$, we have the following equality
(16)
\begin{align} \boxed{V_p(r) = \frac{\pi^{p/2}r^p}{\Gamma(p/2+1)}.} \end{align}
5. Using Stirling's approximation, we have for $p\rightarrow +\infty$
(17)
\begin{align} \Gamma(p/2+1) &= \left(\frac{p+2}{2}\right)^{\frac{p+1}{2}}e^{-\frac{p+2}{2}}\sqrt{2\pi}(1+O(p^{-1})),\\ &= \left(\frac{p+2}{2e}\right)^{p/2}\frac{\sqrt{(p+2)\pi}}{e}(1+O(p^{-1})). \end{align}
The limited development of the square root in the vicinity of 1 gives $\sqrt{p+2} = \sqrt{p}(1+O(p^{-1}))$, and the one for the logarithm in the vicinity of 1 gives (with a little calculation left for the reader) the asymptotic equality $(p+2)^{p/2} = e p^{p/2}(1+O(p^{-1}))$. So we can compute
(18)
\begin{align} \Gamma(p/2+1) = \left(\frac{p}{2e}\right)^{p/2}\sqrt{p\pi}(1+O(p^{-1})). \end{align}
We know that $\frac{1}{1+O(p^{-1}))} = 1+O(p^{-1})$ so we finally prove (1.1):
(19)
\begin{align} \boxed{V_p(r) = \left(\frac{2\pi e r^2}{p}\right)^{p/2}(p\pi)^{-1/2}(1+O(p^{-1})).} \end{align}
Solution proposed by Marin
