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伽马函数中，Γ(1/2)=√π 怎么来的？
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学生
\begin{align} \Gamma\left(\frac{1}{2}\right)&=2\int_0^\infty e^{-x^2}\mathrm{~d}x\\ &=2\sqrt{\left(\int_0^\infty e^{-x^2}\mathrm{~d}x\right)^2}\\ &=2\sqrt{\int_0^\infty e^{-x^2}\mathrm{~d}x\int_0^\infty e^{-x^2}\mathrm{~d}x}\\ &=2\sqrt{\int_0^\infty e^{-x^2}\mathrm{~d}x\int_0^\infty e^{-y^2}\mathrm{~d}y}\\ &=2\sqrt{\int_0^\infty\int_0^\infty e^{-x^2-y^2}\mathrm{~d}x\mathrm{~d}y}\\ &=2\sqrt{\int_0^{\frac{\pi}{2}}\mathrm{d}\theta\int_0^\infty\rho e^{-\rho^2}\mathrm{~d}\rho}\\ &=2\sqrt{\frac{1}{2}\int_0^{\frac{\pi}{2}}\mathrm{d}\theta\int_0^\infty e^{-\rho^2}\mathrm{~d}\rho^2}\\ &=2\sqrt{\frac{1}{2}\cdot \frac{\pi}{2}\cdot 1}\\ &=\sqrt{\pi}. \end{align}
匿名用户
先给出Gamma函数的Hankel积分
\frac{1}{\Gamma(z)}=\frac{1}{2\pi i}\int_{-\infty}^{(0+)}e^tt^{-z}\text{d}t
计算积分，
\begin{aligned} \frac{1}{\Gamma(1-z)} & = \frac{1}{2\pi i}\int_{-\infty}^{(0+)}e^tt^{z-1}\text{d}t\\ & =\frac{1}{2\pi i} \left [ \int_{\infty}^{0}e^{-t}t^{z-1}e^{-i\pi z}\text{d}t+\int^{\infty}_{0}e^{-t}t^{z-1}e^{i\pi z}\text{d}t\right]\\ & =\frac{(e^{i\pi z}-e^{-i \pi z})}{2\pi i} \int_{0}^{\infty} e^{-t}t^{z-1}\text{d}t\\ & =\frac{\sin(\pi z)}{\pi} \Gamma(z) \end{aligned}
也就是 \Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin(\pi z)}
令 z=\frac{1}{2} 便得到 \Gamma\left(\frac12\right)=\sqrt{\pi}>0