Problem in deriving the Volume of an $n$-ball 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?

 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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The Tradidional Derivation is given by
\begin{align}
& \pi^{n/2} =
\pars{\int_{-\infty}^{\infty}\expo{-x^{2}}\,\dd x}^{n} =
\int_{-\infty}^{\infty}\cdots\int_{-\infty}^{\infty}
\exp\pars{-x_{1}^{2} - \cdots - x_{n}^{2}}\dd x_{1}\ldots\dd x_{n}
\\[5mm] = & \
\int_{0}^{\infty}C_{n}\expo{-R^{2}}\,R^{n - 1}\,\,\dd R =
C_{n}\bracks{{1 \over 2}\,\Gamma\pars{n \over 2}} \implies
\bbx{C_{n} = {2\pi^{n/2} \over \Gamma\pars{n/2}}}
\\[5mm] & \mbox{Therefore,}
\\ &\color{#44f}{\large V_{n}} = \int_{0}^{r}C_{n}\, R^{n - 1}\,\,\dd R = 
{2\pi^{n/2} \over \Gamma\pars{n/2}}\,{r^{n} \over n} =
\bbx{\color{#44f}{{\pi^{n/2} \over \Gamma\pars{1 + n/2}}\,r^{n}}} \\ & 
\end{align}
$\ds{\underline{A\ few\ examples:}}$

A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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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,
$$
\color{#44f}{\large V_{n}\pars{r}} =
\bbx{\color{#44f}{{\pi^{n/2} \over \Gamma\pars{1 + n/2}}\,r^{n}}}
$$
