(0)

\(0<\Re z\)とする。
\begin{align*} \Gamma(z+1) & =\int_{0}^{\infty}t^{z+1-1}e^{-t}dt\\ & =-\left[t^{z}e^{-t}\right]_{t=0}^{t=\infty}+z\int_{0}^{\infty}t^{z-1}e^{-t}dt\\ & =z\Gamma(z) \end{align*}

(0)-2

ガンマ関数のハンケル積分表示を使う。
\begin{align*} \Gamma\left(z\right) & =\frac{i}{2\sin\left(\pi z\right)}\int_{C}\left(-\tau\right)^{z-1}e^{-\tau}d\tau\\ & =\frac{i}{2\sin\left(\pi z\right)}\lim_{R\rightarrow\infty}\int_{C\left(0,R,0,2\pi\right)}\left(-\tau\right)^{z-1}e^{-\tau}d\tau\\ & =\frac{i}{2\sin\left(\pi z\right)}\lim_{R\rightarrow\infty}\int_{0}^{2\pi}\left(-Re^{i\theta}\right)^{z-1}e^{-Re^{i\theta}}iRe^{i\theta}d\theta\\ & =\frac{1}{2\sin\left(\pi z\right)}\lim_{R\rightarrow\infty}\int_{0}^{2\pi}\left(-Re^{i\theta}\right)^{z}e^{-Re^{i\theta}}d\theta\\ & =\frac{1}{2\sin\left(\pi z\right)}\lim_{R\rightarrow\infty}\left(\left[\frac{\left(-Re^{i\theta}\right)^{z}}{iz}e^{-Re^{i\theta}}\right]_{0}^{2\pi}-\int_{0}^{2\pi}\frac{\left(-Re^{i\theta}\right)^{z}}{iz}e^{-Re^{i\theta}}\left(-iRe^{i\theta}\right)d\theta\right)\\ & =\frac{1}{2\sin\left(\pi z\right)}\lim_{R\rightarrow\infty}\left(\frac{e^{-R}}{iz}\left(\left(-Re^{2\pi i}\right)^{z}-\left(-R\right)^{z}\right)-\frac{1}{z}\int_{0}^{2\pi}\left(-Re^{i\theta}\right)^{z+1}e^{-Re^{i\theta}}d\theta\right)\\ & =\frac{1}{2\sin\left(\pi z\right)}\lim_{R\rightarrow\infty}\left(\frac{e^{-R}R^{z}}{iz}\left(\left(-e^{2\pi i}\right)^{z}-\left(-1\right)^{z}\right)-\frac{1}{z}\int_{0}^{2\pi}\left(-Re^{i\theta}\right)^{z+1}e^{-Re^{i\theta}}d\theta\right)\\ & =\frac{-1}{2z\sin\left(\pi z\right)}\lim_{R\rightarrow\infty}\int_{0}^{2\pi}\left(-Re^{i\theta}\right)^{z+1}e^{-Re^{i\theta}}d\theta\\ & =\frac{1}{2z\sin\left(\pi\left(z+1\right)\right)}\lim_{R\rightarrow\infty}\int_{0}^{2\pi}\left(-Re^{i\theta}\right)^{z+1}e^{-Re^{i\theta}}d\theta\\ & =\frac{\Gamma\left(z+1\right)}{z} \end{align*} これより、
\[ \Gamma(z+1)=z\Gamma(z) \] この場合、\(z\ne0\)で成り立つ。