x・exp(-α x^2)の広義積分[0,∞]

\(\int_0^{\infty} x e^{-\alpha x^2} \mathrm{d}x\)

$$\begin{eqnarray} &&\int_0^{\infty} x e^{-\alpha x^2} \mathrm{d}x \\&=&\int_0^{\infty} \sqrt{\frac{t}{\alpha}} e^{-t} \frac{1}{2 \sqrt{\alpha t}}\mathrm{d}t \\&&\;\cdots\;t=\alpha x^2,\;x=0 \rightarrow t=0,\;x=\infty \rightarrow t=\infty, \\&&\;\cdots\;x=\sqrt{\frac{t}{\alpha}}\;(積分範囲からx\geq 0とする(?)) \\&&\;\cdots\;\frac{\mathrm{d}t}{\mathrm{d}x}=2\alpha x,\;\mathrm{d}x=\frac{1}{2\alpha x}\mathrm{d}t=\frac{1}{2\alpha \sqrt{\frac{t}{\alpha}}}\mathrm{d}t=\frac{1}{2 \sqrt{\alpha t}}\mathrm{d}t \\&=&\int_0^{\infty}\left(\frac{t}{\alpha}\right)^{\frac{1}{2}} e^{-t} \frac{1}{2 \sqrt{\alpha t}}\mathrm{d}t \\&=&\frac{1}{2\alpha^{\frac{1}{2}}\sqrt{\alpha}}\int_0^{\infty} \frac{t^{\frac{1}{2}}}{\sqrt{t}}e^{-t}\mathrm{d}t \\&=&\frac{1}{2\alpha}\int_0^{\infty} e^{-t}\mathrm{d}t \\&=&\frac{1}{2\alpha}\int_0^{\infty} 1 \cdot e^{-t}\mathrm{d}t \\&=&\frac{1}{2\alpha}\int_0^{\infty} t^0 \cdot e^{-t}\mathrm{d}t \\&=&\frac{1}{2\alpha}\int_0^{\infty} t^{1-1} e^{-t}\mathrm{d}t \\&=&\frac{1}{2\alpha}\Gamma\left(1\right)\;\cdots\;\Gamma(z)=\int_0^{\infty} t^{z-1} e^{-t}\mathrm{d}t\;(\mathfrak{Re}(z)>0) \\&=&\frac{1}{2\alpha}\;\left(1-1\right)!\;\cdots\;\Gamma(n)=(n-1)!\;(nは自然数) \\&=&\frac{1}{2\alpha}\cdot 1\;\cdots\;0!=1 \\&=&\frac{1}{2\alpha} \end{eqnarray}$$

x^2・exp(-α x^2) の広義積分[0, ∞]

\(\int_0^{\infty} x^2 e^{-\alpha x^2} \mathrm{d}x\)

$$\begin{eqnarray} &&\int_0^{\infty} x^2 e^{-\alpha x^2} \mathrm{d}x \\&=&\int_0^{\infty} \frac{t}{\alpha} e^{-t} \frac{1}{2 \sqrt{\alpha t}}\mathrm{d}t \\&&\;\cdots\;t=\alpha x^2,\;x=0 \rightarrow t=0,\;x=\infty \rightarrow t=\infty, \\&&\;\cdots\;x^2=\frac{t}{\alpha} \\&&\;\cdots\;x=\sqrt{\frac{t}{\alpha}}\;(積分範囲からx\geq 0とする(?)) \\&&\;\cdots\;\frac{\mathrm{d}t}{\mathrm{d}x}=2\alpha x,\;\mathrm{d}x=\frac{1}{2\alpha x}\mathrm{d}t=\frac{1}{2\alpha \sqrt{\frac{t}{\alpha}}}\mathrm{d}t=\frac{1}{2 \sqrt{\alpha t}}\mathrm{d}t \\&=&\int_0^{\infty} \frac{t}{\alpha} e^{-t} \frac{1}{2 \sqrt{\alpha t}}\mathrm{d}t \\&=&\frac{1}{2\alpha\sqrt{\alpha}}\int_0^{\infty} \frac{t}{\sqrt{t}}e^{-t}\mathrm{d}t \\&=&\frac{1}{2\alpha\sqrt{\alpha}}\int_0^{\infty} t t^{-\frac{1}{2}}e^{-t}\mathrm{d}t \\&=&\frac{1}{2\alpha\sqrt{\alpha}}\int_0^{\infty} t^{\frac{1}{2}}e^{-t}\mathrm{d}t \\&=&\frac{1}{2\alpha\sqrt{\alpha}}\int_0^{\infty} t^{\frac{3}{2}-1}e^{-t}\mathrm{d}t \\&=&\frac{1}{2\alpha\sqrt{\alpha}}\Gamma\left(\frac{3}{2}\right)\;\cdots\;\Gamma(z)=\int_0^{\infty} t^{z-1} e^{-t}\mathrm{d}t\;(\mathfrak{Re}(z)>0) \\&=&\frac{1}{2\alpha\sqrt{\alpha}}\Gamma\left(\frac{1}{2}+1\right) \\&=&\frac{1}{2\alpha\sqrt{\alpha}}\frac{(2\cdot 1-1)!!}{2^1}\sqrt{\pi}\;\cdots\;\Gamma\left(\frac{1}{2}+n\right)=\frac{(2n-1)!!}{2^n}\sqrt{\pi}\;(!!は二重階乗) \;(nは自然数) \\&=&\frac{1}{2\alpha\sqrt{\alpha}}\frac{1!!}{2}\sqrt{\pi} \\&=&\frac{1}{2\alpha\sqrt{\alpha}}\frac{1}{2}\sqrt{\pi}\;\cdots\;nの二重階乗は，1 から n まで n と同じ偶奇性を持つものだけを全て掛けた積 \\&=&\frac{1}{2\alpha\sqrt{\alpha}}\frac{1\sqrt{\pi}}{2} \\&=&\frac{1}{4\alpha}\sqrt{\frac{\pi}{\alpha}} \end{eqnarray}$$

x^3・exp(-α x^2)の広義積分[0,∞]

\(\int_0^{\infty} x^3 e^{-\alpha x^2} \mathrm{d}x\)

$$\begin{eqnarray} &&\int_0^{\infty} x^3 e^{-\alpha x^2} \mathrm{d}x \\&=&\int_0^{\infty} \left(\frac{t}{\alpha}\right)^2 e^{-t} \frac{1}{2 \sqrt{\alpha t}}\mathrm{d}t \\&&\;\cdots\;t=\alpha x^2,\;x=0 \rightarrow t=0,\;x=\infty \rightarrow t=\infty, \\&&\;\cdots\;x=\sqrt{\frac{t}{\alpha}}\;(積分範囲からx\geq 0とする(?)) \\&&\;\cdots\;x^3=\left(\sqrt{\frac{t}{\alpha}}\right)^3=\left(\frac{t}{\alpha}\right)^{\frac{3}{2}} \\&&\;\cdots\;\frac{\mathrm{d}t}{\mathrm{d}x}=2\alpha x,\;\mathrm{d}x=\frac{1}{2\alpha x}\mathrm{d}t=\frac{1}{2\alpha \sqrt{\frac{t}{\alpha}}}\mathrm{d}t=\frac{1}{2 \sqrt{\alpha t}}\mathrm{d}t \\&=&\int_0^{\infty}\left(\frac{t}{\alpha}\right)^{\frac{3}{2}} e^{-t} \frac{1}{2 \sqrt{\alpha t}}\mathrm{d}t \\&=&\frac{1}{2\alpha^{\frac{3}{2}}\sqrt{\alpha}}\int_0^{\infty} \frac{t^{\frac{3}{2}}}{\sqrt{t}}e^{-t}\mathrm{d}t \\&=&\frac{1}{2\alpha^2}\int_0^{\infty} t e^{-t}\mathrm{d}t \\&=&\frac{1}{2\alpha^2}\int_0^{\infty} t^{2-1} e^{-t}\mathrm{d}t \\&=&\frac{1}{2\alpha^2}\Gamma\left(2\right)\;\cdots\;\Gamma(z)=\int_0^{\infty} t^{z-1} e^{-t}\mathrm{d}t\;(\mathfrak{Re}(z)>0) \\&=&\frac{1}{2\alpha^2}\;\left(2-1\right)!\;\cdots\;\Gamma(n)=(n-1)!\;(nは自然数) \\&=&\frac{1}{2\alpha^2}\cdot 1 \\&=&\frac{1}{2\alpha^2} \end{eqnarray}$$

x^4・exp(-α x^2) の広義積分[0,∞]

\(\int_0^{\infty} x^4 e^{-\alpha x^2} \mathrm{d}x\)

$$\begin{eqnarray} &&\int_0^{\infty} x^4 e^{-\alpha x^2} \mathrm{d}x \\&=&\int_0^{\infty} \left(\frac{t}{\alpha}\right)^2 e^{-t} \frac{1}{2 \sqrt{\alpha t}}\mathrm{d}t \\&&\;\cdots\;t=\alpha x^2,\;x=0 \rightarrow t=0,\;x=\infty \rightarrow t=\infty, \\&&\;\cdots\;x^2=\frac{t}{\alpha},\;x^4=x^{2^2}=\left(\frac{t}{\alpha}\right)^2 \\&&\;\cdots\;x=\sqrt{\frac{t}{\alpha}}\;(積分範囲からx\geq 0とする(?)) \\&&\;\cdots\;\frac{\mathrm{d}t}{\mathrm{d}x}=2\alpha x,\;\mathrm{d}x=\frac{1}{2\alpha x}\mathrm{d}t=\frac{1}{2\alpha \sqrt{\frac{t}{\alpha}}}\mathrm{d}t=\frac{1}{2 \sqrt{\alpha t}}\mathrm{d}t \\&=&\int_0^{\infty} \left(\frac{t}{\alpha}\right)^2 e^{-t} \frac{1}{2 \sqrt{\alpha t}}\mathrm{d}t \\&=&\frac{1}{2\alpha^2\sqrt{\alpha}}\int_0^{\infty} \frac{t^2}{\sqrt{t}}e^{-t}\mathrm{d}t \\&=&\frac{1}{2\alpha^2\sqrt{\alpha}}\int_0^{\infty} t^2 t^{-\frac{1}{2}}e^{-t}\mathrm{d}t \\&=&\frac{1}{2\alpha^2\sqrt{\alpha}}\int_0^{\infty} t^{\frac{3}{2}}e^{-t}\mathrm{d}t \\&=&\frac{1}{2\alpha^2\sqrt{\alpha}}\int_0^{\infty} t^{\frac{5}{2}-1}e^{-t}\mathrm{d}t \\&=&\frac{1}{2\alpha^2\sqrt{\alpha}}\Gamma\left(\frac{5}{2}\right)\;\cdots\;\Gamma(z)=\int_0^{\infty} t^{z-1} e^{-t}\mathrm{d}t\;(\mathfrak{Re}(z)>0) \\&=&\frac{1}{2\alpha^2\sqrt{\alpha}}\Gamma\left(\frac{1}{2}+2\right) \\&=&\frac{1}{2\alpha^2\sqrt{\alpha}}\frac{(2\cdot 2-1)!!}{2^2}\sqrt{\pi}\;\cdots\;\Gamma\left(\frac{1}{2}+n\right)=\frac{(2n-1)!!}{2^n}\sqrt{\pi}\;(!!は二重階乗) \;(nは自然数) \\&=&\frac{1}{2\alpha^2\sqrt{\alpha}}\frac{3!!}{4}\sqrt{\pi} \\&=&\frac{1}{2\alpha^2\sqrt{\alpha}}\frac{3\cdot 1}{4}\sqrt{\pi}\;\cdots\;nの二重階乗は，1 から n まで n と同じ偶奇性を持つものだけを全て掛けた積 \\&=&\frac{1}{2\alpha^2\sqrt{\alpha}}\frac{3\sqrt{\pi}}{4} \\&=&\frac{3}{8\alpha^2}\sqrt{\frac{\pi}{\alpha}} \end{eqnarray}$$

ガンマ凾数の相反公式の積分表示

ガンマ凾数の相反公式の積分表示

ベータ凾数のガンマ凾数での表示

$$\begin{eqnarray} \left.B\left(p,q\right)\right|_{p=1-q}&=&\href{https://shikitenkai.blogspot.com/2020/05/blog-post_22.html}{\left.\frac{\Gamma\left(p\right)\Gamma\left(q\right)}{\Gamma\left(p+q\right)}\right|_{p=1-q}} \;\ldots\;p,q\in\mathbb{C},\;\Re\left(p\right),\Re\left(q\right)\gt0 \\&=&\frac{\Gamma\left(1-q\right)\Gamma\left(q\right)}{\Gamma\left(1-q+q\right)} \\&=&\frac{\Gamma\left(1-q\right)\Gamma\left(q\right)}{\Gamma\left(1\right)} \\&=&\frac{\Gamma\left(1-q\right)\Gamma\left(q\right)}{1} \\&&\;\ldots\;\Gamma\left(1\right)=\int_0^\infty t^{1-1}e^{-t}\mathrm{d}t=\int_0^\infty t^{0}e^{-t}\mathrm{d}t=\int_0^\infty 1\cdot e^{-t}\mathrm{d}t=\int_0^\infty e^{-t}\mathrm{d}t \\&&\;\ldots\;\href{https://shikitenkai.blogspot.com/2021/07/e-ax0.html}{\int_0^\infty e^{-t}\mathrm{d}t=\frac{1}{1}}=1 \\&=&\Gamma\left(1-q\right)\Gamma\left(q\right) \end{eqnarray}$$

ベータ凾数の定義での表示

$$\begin{eqnarray} \left.B\left(p,q\right)\right|_{p=1-q} &=&\int_0^1 t^{p-1}\left(1-t\right)^{q-1}\mathrm{d}t \\&=&\int_\infty^0 \left(\frac{1}{x+1}\right)^{p-1}\left(1-\frac{1}{x+1}\right)^{q-1}\cdot\frac{-1}{\left(x+1\right)^2}\mathrm{d}x \\&&\;\ldots\;t=\frac{1}{x+1},\;x=\frac{1}{t}-1,\;\frac{\mathrm{d}t}{\mathrm{d}x}=\frac{-1}{\left(x+1\right)^2} \\&&\;\ldots\;t:0\rightarrow1,\;x:\infty\rightarrow0 \\&=&\int_0^\infty \left(\frac{1}{x+1}\right)^{p-1}\left(1-\frac{1}{x+1}\right)^{q-1}\cdot\frac{1}{\left(x+1\right)^2}\mathrm{d}x \\&=&\int_0^\infty \left(\frac{1}{x+1}\right)^{p-1}\left(\frac{x}{x+1}\right)^{q-1}\cdot\frac{1}{\left(x+1\right)^2}\mathrm{d}x \\&=&\int_0^\infty \frac{1^{p-1}\cdot x^{q-1}\cdot 1}{\left(x+1\right)^{p-1}\left(x+1\right)^{q-1}\left(x+1\right)^{2}}\mathrm{d}x \\&=&\int_0^\infty \frac{x^{q-1}}{\left(x+1\right)^{(p-1)+(q-1)+2}}\mathrm{d}x \\&=&\int_0^\infty \frac{x^{q-1}}{\left(x+1\right)^{p+q}}\mathrm{d}x \\&=&\int_0^\infty \frac{x^{q-1}}{\left(x+1\right)^{1-q+q}}\mathrm{d}x \\&=&\int_0^\infty \frac{x^{q-1}}{x+1}\mathrm{d}x \end{eqnarray}$$

ガンマ凾数の相反公式の積分表示

$$\begin{eqnarray} \Gamma\left(1-q\right)\Gamma\left(q\right)&=&\int_0^\infty \frac{x^{q-1}}{x+1}\mathrm{d}x \end{eqnarray}$$

ガンマ凾数の相反公式

ガンマ凾数の相反公式

$$\begin{eqnarray} \Gamma\left(z\right)\Gamma\left(1-z\right) &=&\Gamma\left(z\right)\Gamma\left(-z+1\right) \;\ldots\;\Re\left(z\right)\gt0かつ\Re\left(1-z\right)\gt0なので0\lt\Re\left(z\right)\lt1 \\&=&\Gamma\left(z\right)\left(-z\Gamma\left(-z\right)\right) \;\ldots\;\href{https://shikitenkai.blogspot.com/2020/08/s1ss.html}{\Gamma\left(z+1\right)=z\Gamma\left(z\right)} \\&=&-z\Gamma\left(z\right)\Gamma\left(-z\right) \\&=&-z \left(\lim_{n\rightarrow\infty}\frac{n^zn!}{\prod_{k=0}^n (z+k)}\right) \left(\lim_{n\rightarrow\infty}\frac{n^{-z}n!}{\prod_{k=0}^n (-z+k)}\right) \;\ldots\;\href{https://shikitenkai.blogspot.com/2021/07/blog-post_16.html}{\Gamma\left(z\right)=\lim_{n\rightarrow\infty}\frac{n^zn!}{\prod_{k=0}^n (z+k)}} \\&=&-z \left(\lim_{n\rightarrow\infty}\frac{n^zn! \cdot n^{-z}n!}{\prod_{k=0}^n (z+k)(-z+k)}\right) \\&=&-z\left(\lim_{n\rightarrow\infty}\frac{n!n!}{\prod_{k=0}^n (k^2-z^2)}\right) \;\ldots\;A^BA^{-B}=A^{B-B}=A^{0}=1 \\&=&-z\left(\lim_{n\rightarrow\infty}\frac{\left(\prod_{k=1}^nk\right)\left(\prod_{k=1}^nk\right)}{\prod_{k=0}^n (k^2-z^2)}\right) \;\ldots\;n!=\prod_{k=1}^nk \\&=&-z\left(\lim_{n\rightarrow\infty}\frac{\prod_{k=1}^nk^2}{\prod_{k=0}^n (k^2-z^2)}\right) \;\ldots\;\left(\prod_{k=m}^nA_k\right)\left(\prod_{k=m}^nB_k\right)=\prod_{k=m}^nA_kB_k \\&=&-z\left(\lim_{n\rightarrow\infty}\frac{\prod_{k=1}^nk^2}{(0^2-z^2)\prod_{k=1}^n (k^2-z^2)}\right) \\&=&-z\frac{1}{-z^2}\left(\lim_{n\rightarrow\infty}\prod_{k=1}^n\frac{k^2}{k^2-z^2}\right) \\&=&\frac{1}{z}\prod_{k=1}^\infty\frac{k^2}{k^2-z^2} \\&=&\frac{1}{\frac{1}{\frac{1}{z}\prod_{k=0}^\infty\frac{k^2}{k^2-z^2}}}\;\ldots\;A=\frac{1}{\frac{1}{A}} \\&=&\frac{1}{z\prod_{k=1}^\infty\frac{k^2-z^2}{k^2}} \\&=&\frac{\pi}{\pi}\frac{1}{z\prod_{k=0}^\infty\frac{k^2-z^2}{k^2}} \\&=&\frac{\pi}{\pi z\prod_{k=0}^\infty \left(1-\frac{z^2}{k^2}\right)} \\&=&\frac{\pi}{\sin{\left(\pi z\right)}} \\&&\;\ldots\;\href{https://ja.wikipedia.org/wiki/%E4%B8%89%E8%A7%92%E9%96%A2%E6%95%B0%E3%81%AE%E7%84%A1%E9%99%90%E4%B9%97%E7%A9%8D%E5%B1%95%E9%96%8B}{\sin{\left(\pi z\right)} =\pi z\prod_{k=1}^\infty\left(1-\left(\frac{z}{k}\right)^2\right)\;(三角凾数の無限乗積展開 wikipedia)} \end{eqnarray}$$

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\(\Gamma\left(z\right)\)

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

Γ(s+1)=sΓ(s)

\(\Gamma(s+1)=s\Gamma(s)\)

$$ \begin{eqnarray} \Gamma(s+1)&=&\int_{0}^{\infty}e^{-t}t^{s\color{red}{+1-1}}\mathrm{d}t \;\cdots\;\Gamma(s)=\int_{0}^{\infty}e^{-t}t^{s-1}\mathrm{d}t \\&=&\int_{0}^{\infty}e^{-t}t^{s}\mathrm{d}t \\&=&\int_{0}^{\infty}\left\{-e^{-t}\right\}^\prime t^{s}\mathrm{d}t \;\cdots\;\frac{\mathrm{d}}{\mathrm{d}t}\left(-e^{-t}\right)=\left\{-e^{-t}\right\}^\prime=e^{-t} \\&=&\left[-e^{-t} t^{s}\right]_0^\infty-\int_{0}^{\infty}\left\{-e^{-t}\right\}\left\{ st^{s-1} \right\}\mathrm{d}t \;\cdots\;\href{https://shikitenkai.blogspot.com/2020/02/blog-post_7.html}{\int f^\prime g\;\mathrm{d}x= fg-\int f g^\prime\;\mathrm{d}x} \\&=&\left[\left(\lim_{t\rightarrow \infty} -\frac{t^{s}}{e^{t}}\right)-\left(-\frac{0^{s}}{e^{0}}\right)\right]+s\int_{0}^{\infty}e^{-t}t^{s-1}\mathrm{d}t \;\cdots\;\int cf(x)\mathrm{d}x=c\int f(x)\mathrm{d}x \\&=&\left[0-0\right]+s\int_{0}^{\infty}e^{-t}t^{s-1}\mathrm{d}t \;\cdots\;\href{https://shikitenkai.blogspot.com/2020/08/xnex.html}{\lim_{t\rightarrow \infty} \frac{t^{s}}{e^{t}}=0} \\&=&s\Gamma(s) \;\cdots\;\Gamma(s)=\int_{0}^{\infty}e^{-t}t^{s-1}\mathrm{d}t \end{eqnarray} $$

\(\Gamma(s+2)=(s+1)s\Gamma(s)\)

$$ \begin{eqnarray} \Gamma(s+2)&=&(s+1)\Gamma(s+1) \\&=&(s+1)s\Gamma(s) \end{eqnarray} $$