I was trying to derive the Volume of an $n$-ball. With pure intuition, I could get to the following recurrence relation: $$ V_{n + 1}\left(r\right)=\int_{-r}^{r} V_{n}\left(\,\sqrt{\,{r^{2} - x^{2}}}\,\right){\rm d}x $$ The function $V_{n}\left(r\right)$ gives the volume of the ball in the $n$th dimension given a radius $r$.
- If you cut a $3$d sphere into infinitely thin slices and add up the combined area of all those slices then it gives you the volume of the sphere.
- Here I just generalized that idea to the $n$th dimension.
- We can actually calculate the volume of any $n$-ball because we know the area of a circle which is $\pi r^{2}$ and with that, we can figure the volume of a sphere, hypersphere, and so on.
Here is where I got stuck I have no idea how to find an explicit formula:
- Is there any heuristic approach you can do to solve recurrent relations?
- The actual explicit formula when I searched it up involves the Gamma Function: How did Euler arrive at that formula? Is it by solving the recurrent relation?
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{{\displaystyle #1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\sr}[2]{\,\,\,\stackrel{{#1}}{{#2}}\,\,\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ The following evaluation is a "direct" one. Namely, it starts from the volume $\ds{V_{n}}$ definition. Hereafter, $\ds{\bracks{\cdots}}$ is an Iverson Bracket. \begin{align} \left.\color{#44f}{\large V_{n}\pars{r}}\right\vert_{r\ >\ 0} & \equiv \color{#44f}{% \int_{-\infty}^{\infty}\cdots\int_{-\infty}^{\infty}\bracks{x_{1}^{2} + \cdots + x_{n}^{2} < r^{2}}\dd x_{1}\ldots\dd x_{n}} \\[1cm] & = \int_{-\infty}^{\infty}\cdots\int_{-\infty}^{\infty} \\[2mm] & \underbrace{\int_{0^{+}\ -\ \infty\ic}^{0^{+}\ +\ \infty\ic}\,\, {\exp\pars{\pars{r^{2} - x_{1}^{2} - \cdots - x_{n}^{2}}s} \over s}{\dd s \over 2\pi\ic}} _{\ds{\equiv \bracks{x_{1}^{2} + \cdots + x_{n}^{2} < r^{2}}}} \\[2mm] & \dd x_{1}\ldots\dd x_{n} \\[1cm] & = \int_{0^{+}\ -\ \infty\ic}^{0^{+}\ +\ \infty\ic}\,\, {\exp\pars{r^{2}\,s} \over s} \pars{\int_{-\infty}^{\infty}\expo{-sx^{2}}\dd x}^{n}{\dd s \over 2\pi\ic} \\[5mm] & = \int_{0^{+}\ -\ \infty\ic}^{0^{+}\ +\ \infty\ic}\,\, {\exp\pars{r^{2}\,s} \over s} \pars{\root{\pi} \over s^{1/2}}^{n}{\dd s \over 2\pi\ic} \\[5mm] & = \pi^{n/2}\int_{0^{+}\ -\ \infty\ic}^{0^{+}\ +\ \infty\ic}\,\, {\exp\pars{r^{2}\,s} \over s^{1 + n/2}}{\dd s \over 2\pi\ic} \\[5mm] & \sr{r^{2}\,s\ \mapsto\ s}{=} \pi^{n/2}\,r^{n}\int_{0^{+}\ -\ \infty\ic}^{0^{+}\ +\ \infty\ic}\,\, {\expo{s} \over s^{1 + n/2}}\,{\dd s \over 2\pi\ic} \end{align} I'll deform the integration path such that the integral runs along a Hankel Contour $\ds{\cal H}$. The factor $\ds{\left.\expo{s}\right\vert_{\Re\pars{s}\ <\ 0}}\,\,\,$ enforces the vanishing out of the integration along two quarter circle in $\ds{\braces{s \mid \Re\pars{s} < 0}}$. Namely, $$ \int_{0^{+}\ -\ \infty\ic}^{0^{+}\ +\ \infty\ic}\,\, {\expo{s} \over s^{1 + n/2}}\,{\dd s \over 2\pi\ic} = \oint_{\cal H}\,\, {\expo{s} \over s^{1 + n/2}}\,{\dd s \over 2\pi\ic} = {1 \over \Gamma\pars{1 + n/2}} $$ See the Gamma Function Integral representation "along $\ds{\cal H}$"
Therefore, $$ \color{#44f}{\large V_{n}\pars{r}} = \bbx{\color{#44f}{{\pi^{n/2} \over \Gamma\pars{1 + n/2}}\,r^{n}}} $$