Legendre 加倍公式的一个证明
Legendre 加倍公式的一个证明
\[\sqrt{\pi}\Gamma (2s)=2^{2s-1}\Gamma (s)\Gamma \left(s+\frac{1}{2}\right),s>0，\]
其中$\Gamma$是Gamma函数,
\[\Gamma (s)=\int_0^{+\infty}x^{s-1}e^{-x}\mathrm{d}x, s>0.\]
证.记
\[I(s)=\int_0^1\frac{\mathrm{d}x}{(1+x^{\frac{1}{s}})^{2s}}.\]
令$x=\tan^{2s} t$, 则$\mathrm{d}x=2s\tan^{2s-1}t\sec^2t\mathrm{d}t=\sin^{2s-1}t\cos^{-2s-1}t\mathrm{d}t$, $(1+x^{\frac{1}{s}})^{2s}=\sec^{4s}t$, 从而
\[\begin{array}{rl}I(s)&=\int_0^1\frac{\mathrm{d}x}{(1+x^{\frac{1}{s}})^{2s}}\\&=2s\int_0^{\frac{\pi}{4}} (\sin t\cos t)^{2s-1}\mathrm{d}t\\&=s2^{1-2s} \int_0^{\frac{\pi}{2}}\sin^{2s-1}u\mathrm{d}u\\&=2^{-2s}sB(\frac{1}{2},s)\\&=2^{-2s}s\frac{\Gamma (\frac{1}{2})\Gamma (s)}{\Gamma(\frac{1}{2}+s)}\\&= 2^{-2s}\sqrt{\pi}s\frac{\Gamma (s)}{\Gamma(\frac{1}{2}+s)}.\end{array}\]
另一方面
\[I(s)=\int_0^1\frac{\mathrm{d}x}{(1+x^{\frac{1}{s}})^{2s}}=\int_1^{+\infty}\frac{\mathrm{d}x}{(1+x^{\frac{1}{s}})^{2s}},\]
从而
\[I(s)=\frac{1}{2}\int_0^{+\infty}\frac{\mathrm{d}x}{(1+x^{\frac{1}{s}})^{2s}}=s\int_0^{\frac{\pi}{2}}(\sin t\cos t)^{2s-1}\mathrm{d}t=\frac{sB(s,s)}{2}=\frac{s\Gamma^2(s)}{2\Gamma (2s)}.\]
因此
\[2^{-2s}\sqrt{\pi}s\frac{\Gamma (s)}{\Gamma(\frac{1}{2}+s)}=\frac{s\Gamma^2(s)}{2\Gamma (2s)}.\] 从而
\[\sqrt{\pi}\Gamma (2s)=2^{2s-1}\Gamma (s)\Gamma \left(s+\frac{1}{2}\right), s>0.\]
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