(0)

\(C_{1}\)を正の実軸の上側を無限から0までマイナス方向に進む経路
\(C_{2}\)を原点の周りを小さな円で反時計回りに一周する経路
\(C_{3}\)を正の実軸の下側を0から無限までプラス方向に進む経路
とする。

(1)

ガンマ関数の定義より、\(0<\Re z\)なので、
\begin{align*} \Gamma\left(z\right) & =\int_{0}^{\infty}t^{z-1}e^{-t}dt\\ & =\frac{e^{i\pi z}-e^{-i\pi z}}{2i\sin\left(\pi z\right)}\int_{0}^{\infty}t^{z-1}e^{-t}dt\\ & =-\frac{e^{i\pi\left(z-1\right)}-e^{-i\pi\left(z-1\right)}}{2i\sin\left(\pi z\right)}\int_{0}^{\infty}t^{z-1}e^{-t}dt\\ & =-\frac{1}{2i\sin\left(\pi z\right)}\int_{0}^{\infty}\left(e^{i\pi}t\right)^{z-1}e^{-t}dt-\int_{0}^{\infty}\left(e^{-i\pi}t\right)^{z-1}e^{-t}dt\\ & =\frac{i}{2\sin\left(\pi z\right)}\int_{0}^{\infty}\left(e^{2i\pi}e^{-i\pi}t\right)^{z-1}e^{-t}dt+\int_{\infty}^{0}\left(e^{-i\pi}t\right)^{z-1}e^{-t}dt\\ & =\frac{i}{2\sin\left(\pi z\right)}\int_{0}^{\infty}\left(-e^{2i\pi}t\right)^{z-1}e^{-t}dt+\int_{\infty}^{0}\left(-t\right)^{z-1}e^{-t}dt\\ & =\frac{i}{2\sin\left(\pi z\right)}\left(\int_{0}^{\infty}\left(-e^{2i\pi}t\right)^{z-1}e^{-t}dt+\int_{\infty}^{0}\left(-t\right)^{z-1}e^{-t}dt+\lim_{\epsilon\rightarrow+0}\int_{C\left(0,\epsilon,0,2\pi\right)}\left(-t\right)^{z-1}e^{-t}dt\right)\\ & =\frac{i}{2\sin\left(\pi z\right)}\left(\int_{C_{3}}\left(-t\right)^{z-1}e^{-t}dt+\int_{C_{1}}\left(-t\right)^{z-1}e^{-t}dt+\int_{C_{2}}\left(-\tau\right)^{z-1}e^{-\tau}d\tau\right)\\ & =\frac{i}{2\sin\left(\pi z\right)}\int_{C}\left(-\tau\right)^{z-1}e^{-\tau}d\tau \end{align*}

(2)

ガンマ関数の定義より、\(0<\Re z\)なので、
\begin{align*} \Gamma\left(z\right) & =\frac{i}{2\sin\left(\pi z\right)}\left(\int_{C}\left(-t\right)^{z-1}e^{-t}dt\right)\\ & =\frac{i}{2\sin\left(\pi z\right)}\left(\int_{0}^{\infty}\left(-e^{2i\pi}t\right)^{z-1}e^{-t}dt+\int_{\infty}^{0}\left(-t\right)^{z-1}e^{-t}dt\right)\\ & =\frac{ie^{i\pi\left(z-1\right)}}{2\sin\left(\pi z\right)}\left(\int_{0}^{\infty}\left(e^{2i\pi}t\right)^{z-1}e^{-t}dt+\int_{\infty}^{0}t^{z-1}e^{-t}dt\right)\\ & =\frac{1}{e^{2i\pi z}-1}\left(\int_{0}^{\infty}\left(e^{2i\pi}t\right)^{z-1}e^{-t}dt+\int_{\infty}^{0}t^{z-1}e^{-t}dt+\lim_{\epsilon\rightarrow+0}\int_{C\left(0,\epsilon,0,2\pi\right)}\tau^{z-1}e^{-\tau}d\tau\right)\\ & =\frac{1}{e^{2i\pi z}-1}\int_{C}\tau^{z-1}e^{-\tau}d\tau \end{align*}

(3)

\begin{align*} \frac{1}{\Gamma\left(z\right)} & =\frac{\sin\left(\pi z\right)}{\pi}\Gamma\left(1-z\right)\\ & =\frac{\sin\left(\pi z\right)}{\pi}\frac{i}{2\sin\left(\pi z\right)}\int_{C}\left(-\tau\right)^{-z}e^{-\tau}d\tau\\ & =\frac{i}{2\pi}\int_{C}\left(-\tau\right)^{-z}e^{-\tau}d\tau \end{align*}