ガンマ凾数の極限表示

ガンマ凾数の極限表示

ガンマ凾数の定義と極限表示

$$\begin{eqnarray} \Gamma\left(z\right) &=&\int_0^\infty t^{z-1}e^{-t}\mathrm{d}t\;\;\;\left(\mathrm{Re}\left(z\right)\gt0\right)\;\ldots\;定義 \\&=&\lim_{n\rightarrow\infty}\frac{n^z n!}{\prod_{k=0}^n\left(z+k\right)}\;\ldots\;極限表示 \end{eqnarray}$$

\(G_n\)の導入

$$\begin{eqnarray} G_n\left(z\right)&=&\int_0^n t^{z-1}\left(1-\frac{t}{n}\right)^{n}\mathrm{d}t \end{eqnarray}$$

\(G_n\)の総乗表示

$$\begin{eqnarray} G_n\left(z\right)&=&\int_0^n t^{z-1}\left(1-\frac{t}{n}\right)^{n}\mathrm{d}t \\&=&\int_0^1 \left(nu\right)^{z-1}\left(1-\frac{nu}{n}\right)^{n}n\mathrm{d}u \\&&\;\ldots\;t=nu,\;u=\frac{t}{n},\;\frac{\mathrm{d}t}{\mathrm{d}u}=n,\;\mathrm{d}t=n\mathrm{d}u \\&&\;\ldots\;t:0\rightarrow n,\;u:0\rightarrow1 \\&=&n^z\int_0^1 u^{z-1}\left(1-u\right)^{n}\mathrm{d}u \\&=&n^z\cdot g_n\left(z\right)\;\ldots\;g_n\left(z\right)=\int_0^1 u^{z-1}\left(1-u\right)^{n}\mathrm{d}u \\g_0\left(z\right)&=&\int_0^1 u^{z-1}\left(1-u\right)^{0}\mathrm{d}u \\&=&\int_0^1 u^{z-1}\mathrm{d}u \\&=&\left[\frac{u^z}{z}\right]_{u=0}^1 \\&=&\frac{1}{z} \\g_n\left(z\right)&=&\int_0^1 u^{z-1}\left(1-u\right)^{n}\mathrm{d}u \\&=&\int_0^1 \left(\frac{u^{z}}{z}\right)^\prime\left(1-u\right)^{n}\mathrm{d}u \;\ldots\;\frac{\mathrm{d}}{\mathrm{d}u}\frac{u^{z}}{z}=\frac{1}{z}zu^{z-1}=u^{z-1} \\&=&\left[\frac{u^z}{z}\left(1-u\right)^{n}\right]_{u=0}^1 -\int_0^1 \frac{u^{z}}{z}\left(\left(1-u\right)^{n}\right)^\prime\mathrm{d}u \\&=&\left[\frac{1^z}{z}\left(1-1\right)^{n}-\frac{0^z}{z}\left(1-0\right)^{n}\right] -\int_0^1 \frac{u^{z}}{z}\left(n\left(1-u\right)^{n-1}(-1)\right)\mathrm{d}u \\&=&0+\frac{n}{z}\int_0^1 u^{z}\left(1-u\right)^{n-1}\mathrm{d}u \\&=&\frac{n}{z}g_{n-1}\left(z+1\right) \\&=&\frac{n}{z}\frac{n-1}{z+1}g_{n-2}\left(z+2\right) \\&=&\frac{n}{z}\frac{n-1}{z+1}\frac{n-2}{z+2}g_{n-3}\left(z+3\right) \\&=&\frac{n}{z}\frac{n-1}{z+1}\frac{n-2}{z+2}\ldots\frac{n-(n-1)}{z+(n-1)}g_{n-n}\left(z+n\right) \\&=&\frac{n}{z}\frac{n-1}{z+1}\frac{n-2}{z+2}\ldots\frac{1}{z+(n-1)}g_{0}\left(z+n\right) \\&=&\frac{n}{z}\frac{n-1}{z+1}\frac{n-2}{z+2}\ldots\frac{1}{z+(n-1)}\frac{1}{z+n} \\&=&\frac{n!}{\prod_{k=0}^n (z+k)} \\G_n\left(z\right)&=&n^z\cdot g_n\left(z\right) \\&=&n^z\frac{n!}{\prod_{k=0}^n (z+k)} \\&=&\frac{n^zn!}{\prod_{k=0}^n (z+k)} \end{eqnarray}$$

\(G_n\)の極限

$$\begin{eqnarray} \lim_{n\rightarrow\infty}G_n\left(z\right) &=&\lim_{n\rightarrow\infty}\int_0^n t^{z-1}\left(1-\frac{t}{n}\right)^{n}\mathrm{d}t \\&=&\int_0^\infty t^{z-1} e^{-t}\mathrm{d}t \\&&\;\ldots\;\lim_{n\rightarrow\infty}\left(1-\frac{t}{n}\right)^{n} =\lim_{n\rightarrow\infty}\left(1+\frac{-t}{n}\right)^{n} =e^{-t} \\&=&\Gamma\left(z\right) \end{eqnarray}$$

よって\(G_n\)の総乗表示での極限も\(\Gamma\left(z\right)\)

$$\begin{eqnarray} \lim_{n\rightarrow\infty}G_n\left(z\right) &=&\lim_{n\rightarrow\infty}\frac{n^zn!}{\prod_{k=0}^n (z+k)} \\&=&\Gamma\left(z\right) \end{eqnarray}$$