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\title{\textbf{Gamma Function and Balls In $\mathbb{R}^n$}}
\author{Wang-Shiuan Pahngerei}
\maketitle


The Gamma function (a generalization of the factor function $n!$) is defined as $\Gamma (x) = \int_0^\infty e^{-t} t^{x-1}dt$ and the Beta funcion is $B(x,y):=\frac{\Gamma (x) \Gamma(y)}{\Gamma (x+y)}$. Then we have these properties:
\begin{itemize}
\item[(i)] $\Gamma (x+1) = x\Gamma (x)$.
\item[(ii)] $\Gamma (n+1) = n!$.
\item[(iii)] $\Gamma (\frac{1}{2}) =\sqrt{\pi}$.
\item[(iv)] $\Gamma (m+\frac{1}{2}) = \frac{(2m)(2m-1)\cdots (m+2)(m+1)}{2^{2m}}\sqrt{\pi}$.
\item[(v)] $B(x,y)=\int_0^1 (1-z)^{x-1} z^{y-1} dz$. 
\end{itemize}

To evidence (iv), 
\begin{align}
\Gamma (m+\frac{1}{2}) &= (m-1+\frac{1}{2})(m-2 + \frac{1}{2})\cdots (0+\frac{1}{2}) \Gamma (\frac{1}{2})\notag\\
&=\frac{2m-1}{2} \cdot \frac{2m-3}{2} \cdots \frac{1}{2} \sqrt{\pi}\notag\\
&=\frac{(2m)(2m-1)(2m-2)\cdots (4)(3)(2)(1)}{2^m \cdot 2^m \cdot (m)(m-1)(m-2)\cdots (2)(1)}\sqrt{\pi}\notag\\
&=\frac{(2m)!}{2^{2m}\cdot m!}=\frac{(2m)(2m-1)\cdots (m+2)(m+1)}{2^{2m}}\sqrt{\pi}.\notag
\end{align}


Supposely we know that the volumn of a ball in $\mathbb{R}^3$ with radius $r$ is $V_3 (r)=\frac{4}{3} \pi r^3$. Find that in $\mathbb{R}^4$ and in $\mathbb{R}^5$, i.e.  $V_4 (r)$ and $V_5 (r)$. 

\begin{align}
V_4 (r) &= \int _{-r} ^r V_3 (\sqrt{r^2 - x^2}) dx= \int_{-r}^r \frac{4}{3}\pi (\sqrt{r^2 - x^2})^3 dx\notag\\
&= \frac{4}{3}\pi\int _{-r}^r (r-x)^\frac{3}{2} (r+x)^\frac{3}{2} dx =_{(x:=2ry\,\, \Rightarrow\,\, dx=2rdy)}\, \frac{4}{3}\pi\int_{-\frac{1}{2}}^\frac{1}{2} (r-2ry)^\frac{3}{2} (r+2ry)^\frac{3}{2} dy\notag\\
&=\frac{4}{3}\pi(2r)^3 (2r)\int_{-\frac{1}{2}} ^\frac{1}{2} (\frac{1}{2} - y )^\frac{3}{2} (\frac{1}{2}+y)^\frac{3}{2} dy=_{(y:=z-\frac{1}{2}\,\, \Rightarrow \,\, dy=dz )} \frac{4}{3}\pi (2r)^4 \int_0^1 (1-z)^\frac{3}{2} z^\frac{3}{2} dz\notag\\
&=\frac{4}{3} \pi  (2r)^4 B(\frac{5}{2}, \frac{5}{2})=\frac{4}{3} \pi (2r)^4 \frac{ (\Gamma (\frac{5}{2}))^2}{  \Gamma (4+1)}=\frac{4}{3} \pi 16 r^4 \frac{ (\frac{3}{2}\cdot \frac{1}{2}\cdot \sqrt{\pi}    )^2      } {4!}\notag\\
&=\frac{1}{2} \pi ^2 r^4\notag
\end{align}


For $V_5 (r)$, 

\begin{align}
V_5 (r) &= \int _{-r} ^r V_4 (\sqrt{r^2 - x^2}) dx= \int_{-r}^r \frac{1}{2}\pi ^2 (\sqrt{r^2 - x^2})^4 dx= \frac{1}{2}\pi ^2 \int _{-r} ^r (r^2 - x ^2 )^2 dx\notag\\
&= \frac{1}{2}\pi ^2 \int _{-r} ^r (r^4 - 2r^2 x^2 +x^4 )dx= \frac{1}{2}\pi ^2 (r^4 x - \frac{2}{3}r^2 x^3 + \frac{1}{5} x^5) _{-r}^r=\frac{8}{15} \pi ^2 r^5 \notag
\end{align}

If we try $V_6 (r)$, we'll find $V_6 (r) = \frac{1}{6} \pi ^3 r^6$. It seems that $V_n (r)$ is of the form $V_n (r) = a_n \pi^{k_n} r^n = C_n r^n$, where $\{k_n\}_{n=2}^\infty = \langle 1,1,2,2,3,3,4,4,\cdots \rangle$. 


To reach a recurrance relation between $V_n (r)$ and $V_{n+1} (r)$, Assume $V_n (r)=C_n r^n$, we perform the following:
\begin{align}
V_{n+1} (r) &= \int_{-r}^r C_n (\sqrt{r^2 - x^2 })^n dx\notag\\
&=C_n \int_{-r}^r (r-x)^\frac{n}{2} (r+x)^\frac{n}{2} dx\notag\\
&=(2r)^n C_n (2r) \int_{\frac{1}{2}} ^\frac{1}{2} (\frac{1}{2} -y)^\frac{n}{2} (\frac{1}{2}+y)^\frac{n}{2} dy \notag\\
&=(2r)^{n+1} C_n \int_0^1 (1-z)^\frac{n}{2} z^\frac{n}{2} dz\notag\\
&=(2r)^{n+1} C_n B(\frac{n}{2} +1, \frac{n}{2} +1 )\notag\\
&=(2r)^{n+1} C_n \frac{   (\Gamma (\frac{n}{2} +1))^2      }{\Gamma (n+2)} = \frac{(2r)^{n+1} (\Gamma (\frac{n}{2} +1))^2}{(n+1)!}C_n\notag\\
&= \frac{2^{n+1}C_n}{(n+1)!} (\Gamma (\frac{n}{2} +1))^2 r^{n+1}\notag
\end{align}

This means:
\begin{align}
C_{n+1} = \frac{2^{n+1}}{(n+1)!} (\Gamma (\frac{n}{2}+1))^2 C_n\notag
\end{align}



Let us try more cases. In fact, we have: 
\begin{align}
C_2 &= \pi= \frac{1}{1!}\pi\notag\\
C_3 &= \frac{4}{3}\pi = \frac{2^4}{3\cdot 4}\pi\notag\\
C_4 &=\frac{1}{2}\pi ^2 = \frac{1}{2!}\pi^2\notag\\
C_5 &= \frac{2^3}{3\cdot 5}\pi^2 = \frac{2^6}{4\cdot 5\cdot 6}\pi^2\notag\\
C_6 &=\frac{1}{3!}\pi^3\notag\\
C_7 &=  \frac{2^5}{7\cdot 6\cdot 5}\pi^3 = \frac{2^8}{5\cdot 6\cdot 7\cdot 8}\pi^3\notag\\
C_8 &= \frac{1}{4!}\pi^4\notag\\
C_9 &= \frac{2^9}{5\cdot 6\cdot 7\cdot 8\cdot 9}\pi^4 = \frac{2^{10}}{6\cdot 7 \cdot 8 \cdot 9 \cdot 10}\pi^4.\notag
\end{align}
Observing more, we might find that 
\begin{align}
C_{2j} &= \frac{\pi^j}{j!} = \frac{\pi^\frac{2j}{2}}{\Gamma (\frac{2j}{2}+1)}.\notag\\
C_{2j+1} &= \frac{2^{2(j+1)}}{(j+1+1)(j+1+2)\cdots (j+1+j+1)\sqrt{\pi}}\pi^{j+\frac{1}{2}} \notag\\
&= \frac{\pi^\frac{2j+1}{2}}{\Gamma (j+1+\frac{1}{2})} = \frac{\pi^\frac{2j+1}{2}}{\Gamma (\frac{2j+1}{2}+1)}.\notag
\end{align}


Hence, for general case, 

\begin{align}
V_n (r) = \frac{\pi ^{\frac{n}{2}}}{\Gamma (\frac{n}{2} +1)} r^n\notag
\end{align}

To show this, we know that when $n=2$, the formula holds, so now if we also know that $V_k(r) = \frac{\pi ^\frac{k}{2}}{\Gamma (\frac{k}{2} +1)} r^k$, we hope to evalute $V_{k+1}$. 
\begin{align}
V_{k+1} (r) &= \int_{-r}^r V_k (\sqrt{r^2 - x^2 }) dx\notag\\
&=\int_{-r}^r \frac{\pi ^\frac{k}{2}}{\Gamma (\frac{k}{2}+1)} (r^2-x^2)^\frac{k}{2} dx\notag\\
&=\frac{\pi ^\frac{k}{2}}{\Gamma (\frac{k}{2} +1 ) } \int_{-r}^r (r-x)^\frac{k}{2} (r+x)^\frac{k}{2} dx\notag\\
&=_{(  x:=2ry \,\,\Rightarrow \,\, dx=2rdy )} \frac{\pi ^\frac{k}{2}}{\Gamma (\frac{k}{2} +1 )} \int_{-\frac{1}{2}}^\frac{1}{2} (r-2ry)^\frac{k}{2} (r+2ry)^\frac{k}{2} 2rdy\notag\\
&=_{(Similarly)} \frac{\pi ^\frac{k}{2}}{\Gamma (\frac{k}{2} +1 )} (2r)^{k+1} B(\frac{k}{2} +1, \frac{k}{2} +1) = \frac{\pi ^\frac{k}{2}}{\Gamma (\frac{k}{2} +1 )} (2r)^{k+1} \frac{   (\Gamma (\frac{k}{2} +1 ))^2 }{\Gamma (k+2)}\notag\\
&= \frac{\pi ^\frac{k}{2} (2r)^{k+1}  \Gamma (\frac{k}{2} +1 )        }{(k+1)! }  \notag
\end{align}





If $k=2p+1$, namely, odd, then 
\begin{align}
(\ast)\,\,&=\frac{\pi^\frac{k}{2} r^{k+1} 2^{2p+2} \Gamma (p+1+\frac{1}{2})} {(2p+2)!}\notag\\
&=\pi^\frac{k}{2} r^{k+1}\frac{2^{2p+2} (p+\frac{1}{2})\cdots (0+\frac{1}{2})\Gamma (\frac{1}{2})}{(2p+2)!}\notag\\
&=\pi^\frac{k}{2} r^{k+1} \frac{2^{2p+2} \frac{2p+1}{2}\cdot \frac{2p-1}{2}\cdots\frac{1}{2}\Gamma (\frac{1}{2})}{(2p+2)!}\notag\\
&=\pi ^\frac{k}{2} r^{k+1}\frac{2^{2p+2} (\frac{1}{2})^{p+1} (2p+1)\cdots (1)\Gamma (\frac{1}{2}) (2p)(2p-2)\cdots (4)(2)}{(2p+2)!\cdot (2p)(2p-2)\cdots (4)(2)}\notag\\
&=\frac{\pi^\frac{k}{2} r^{k+1} 2^{2p+2} (\frac{1}{2})^p (2p+1)! \sqrt{\pi}}{(2p+2)! 2^p p!} = \frac{\pi ^\frac{k+1}{2} r^{k+1}}{(p+1)!} = \frac{\pi ^\frac{k+1}{2} r^{k+1}}{\Gamma (\frac{k+1}{2} +1)}.\notag
\end{align}

Else if $k=2p$, even, then
\begin{align}
(\ast)\,\, &= \frac{\pi ^\frac{k}{2} r^{k+1} 2^{2p+1} \Gamma (p+1)}{(2p+1)!}\notag\\
&=\frac{\pi ^\frac{k}{2} r^{k+1} 2^{2p+1} p!}{(2p+1)!}= \frac{\,\,\pi ^\frac{k}{2} r^{k+1} 2^{p+1}\,\,}{\frac{ (2p+1)(2p)(2p-1)\cdots (4)(3)(2)(1)}{(2p)(2p-2)\cdots (4)(2)}}\notag\\
&=\frac{\pi ^\frac{k+1}{2} r^{k+1} }{ (p+\frac{1}{2}) (p-1+\frac{1}{2})\cdots (\frac{1}{2})\sqrt{\pi}}=\frac{\pi ^\frac{k+1}{2} r^{k+1}}{\Gamma (p+1+\frac{1}{2})}= \frac{\pi ^\frac{k+1}{2} r^{k+1}}{\Gamma (\frac{k+1}{2} +1)}.\notag
\end{align}


Mathematical induction then yields that $V_n (r)=\frac{\pi ^\frac{n}{2}}{\Gamma (\frac{n}{2} +1)}$. 









\end{document}