好久没写博客了，就从知乎上瞎翻点东西水一篇吧。据我所知，知乎上经常会推送一些积分题目（csl貌似也是这样），并且解答我基本上都看不懂。这次好不容易有一个看懂的了，很高兴。但这都是些什么神仙操作啊，为什么人能想得出如此精妙的推导啊！

这篇文章试图求$\int_{0}^{1}\log(\Gamma(x))\textrm{d}x$。默认大家知道伽马函数是啥，以及分部积分可以马上得出的$\Gamma(a+1)=a\Gamma(a)$。

\[\int_{0}^{1}\log(\Gamma(x)) \textrm{d}x=\dfrac{1}{2}\int_{0}^{1}\log(\Gamma(x)\Gamma(1-x))\textrm{d}x\]

引理：余元公式

这比正文还长$\dots$记$a=x,b=1-x$

\[\begin{aligned} \Gamma(a)\Gamma(b)&=\int_{0}^{\infty}e^{-x}x^{a-1}\left(\int_{0}^{\infty}e^{-y}y^{b-1}\textrm{d}y\right)\textrm{d}x\\ &=\int_{0}^{\infty}\int_{0}^{\infty}e^{-p^2}p^{2a-2}2p\textrm{d}pe^{-q^2}q^{2b-2}2q\textrm{d}q,\qquad(x=p^2,y=q^2)\\ &=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-p^2}\vert p\vert^{2a-1}e^{-q^2}\vert q\vert^{2b-1}\textrm{d}p\textrm{d}q\\ &=\int_{0}^{\infty}\int_{0}^{2\pi}e^{-r^2}\vert r\cos\theta\vert^{2a-1}\vert r\sin\theta\vert^{2b-1}r\textrm{d}r\textrm{d}\theta,\qquad(p=r\cos\theta,q=r\sin \theta)\\ &=4\left(\int_{0}^{\infty}r^{2(a+b)-1}e^{-r^2}\textrm{d}r\right)\int_{0}^{\pi/2}(\cos\theta)^{2a-1}(\sin\theta)^{2b-1}\textrm{d}\theta\\ &=2\left(\int_{0}^{\infty}t^{a+b-1}e^{-t}\textrm{d}t\right)\int_{1}^{0}\frac{t^{a}}{\cos\theta}\frac{(1-t)^{b}}{\sin\theta}\frac{\textrm{d}t}{-2\cos(\theta)\sin(\theta)},\qquad(t=\cos^2\theta)\\ &=\Gamma(a+b)\left(\int_{0}^{1}t^{a-1}(1-t)^{b-1}\textrm{d}t\right)\\ \end{aligned}\] \[\Gamma(a+b)=\Gamma(1)=0!=1\] \[\begin{aligned} \Gamma(x)\Gamma(1-x)&=\int_{0}^{1}t^{x-1}(1-t)^{-x}\textrm{d}t\\ &=\int_{0}^{1}\frac{\left(\frac{t}{1-t}\right)^{x}}{t} \textrm{d}t\\ &=\int_{0}^{\infty}\frac{s^x}{(1+s)^2}\frac{1+s}{s} \textrm{d}s,\qquad(s=\frac{t}{1-t})\\ &=\int_{0}^{\infty}\frac{s^{x-1}}{1+s}\textrm{d}s\\ &=\int_{0}^{1}\frac{s^{x-1}}{1+s}\textrm{d}s+\int_{1}^{\infty}\frac{s^{x-1}}{1+s}\textrm{d}s\\ &=\int_{0}^{1}\frac{s^{x-1}}{1+s}\textrm{d}s+\int_{1}^{0}\frac{s^{-x+1}}{1+1/s}\frac{-1}{s^2}\textrm{d}s.\qquad(s=\frac{1}{s})\\ &=\int_{0}^{1}\frac{s^{x-1}}{1+s}\textrm{d}s+\int_{0}^{1}\frac{s^{-x}}{1+s}\textrm{d}s\\ &=\int_{0}^{1}s^{x-1}\sum_{k=0}^{\infty}(-s)^k\textrm{d}s+\int_{0}^{1}s^{-x}\sum_{k=0}^{\infty}(-s)^k\textrm{d}s\\ &=\sum_{k=0}^{\infty}(-1)^{k}\int_{0}^{1}s^{x+k-1}\textrm{d}s+\sum_{k=0}^{\infty}(-1)^{k}\int_{0}^{1}s^{k-x}\textrm{d}s\\ &=\sum_{k=0}^{\infty}(-1)^{k}\frac{s^{x+k}}{x+k}\vert_{0}^{1}+\sum_{k=0}^{\infty}(-1)^{k}\frac{s^{k-x+1}}{k-x+1}\vert_{0}^{1}\\ &=\sum_{k=0}^{\infty}\frac{(-1)^{k}}{x+k}+\sum_{k=0}^{\infty}\frac{(-1)^{k}}{k-x+1}\\ &=\frac{1}{x}+\sum_{k=1}^{\infty}(-1)^{k}(\frac{1}{x+k}+\frac{1}{x-k})\\ &=\frac{1}{x}+2\sum_{k=1}^{\infty}(\frac{1}{x+2k}+\frac{1}{x-2k})-\sum_{k=1}^{\infty}(\frac{1}{x+k}+\frac{1}{x-k})\\ &=\left(\frac{1}{x/2}+\sum_{k=1}^{\infty}(\frac{1}{x/2+k}+\frac{1}{x/2-k})\right)-\left(\frac{1}{x}+\sum_{k=1}^{\infty}(\frac{1}{x+k}+\frac{1}{x-k})\right)\\ \end{aligned}\]

众所周知的结论，在 Proofs from THE BOOK 一书中有个优美的证明

\[\frac{1}{x}+\sum_{k=1}^{\infty}(\frac{1}{x+k}+\frac{1}{x-k})=\pi\cot(\pi x)\]

所以

\[\begin{aligned} \Gamma(x)\Gamma(1-x)&=\pi(\cot(\pi x/2)-\cot(\pi x))\\ &=\pi(\frac{1+\cos \pi x}{\sin \pi x}-\frac{\cos \pi x}{\sin \pi x})\\ &=\frac{\pi}{\sin{\pi x}} \end{aligned}\]

继续

\[\begin{aligned} \int_{0}^{1}\log(\Gamma(x))\textrm{d}x&=\frac12\int_{0}^{1}\log(\frac{\pi}{\sin\pi x})\textrm{d}x\\ &=\frac12\left(\log(\pi)-\int_{0}^{1}\log(\sin \pi x)\textrm{d}x\right)\\ &=\frac12\left(\log(\pi)-\frac{1}{\pi}\int_{0}^{\pi}\log(\sin(x))\textrm{d}x\right),\qquad(x=\pi x) \end{aligned}\] \[\begin{aligned} \int_{0}^{\pi}\log\sin x\textrm{d}x&=\int_{0}^{\pi/2}\log\sin x\textrm{d}x+\int_{\pi/2}^{\pi}\log\sin x\textrm{d}x\\ &=\int_{0}^{\pi/2}\log\sin x \textrm{d}x+\int_{0}^{\pi/2}\log\cos x \textrm{d}x\\ &=\int_{0}^{\pi/2}\log(\sin x\cos x) \textrm{d}x\\ &=\int_{0}^{\pi/2}(\log\sin 2x-\log 2)\textrm{d}x\\ &=\frac{1}{2}\int_{0}^{\pi}\log\sin x\textrm{d}x-\frac{\pi\log{2}}{2},\qquad(x=2x) \end{aligned}\]

$I=\dfrac{I}{2}-\dfrac{\pi\log{2}}{2}$，即$I=-\pi\log{2}$。

\[\int_{0}^{1}\log(\Gamma(x))\textrm{d}x=\frac12\left(\log(\pi)+\log(2)\right)=\frac{\log(2\pi)}{2}\]

拓展

求$\int_{a}^{a+1}\log(\Gamma(x))\textrm{d}x$。记$F(a)=\int_{a}^{a+1}\log(\Gamma(x))\textrm{d}x$，则$F’(a)=\log(\Gamma(a+1))-\log(\Gamma(a))=\log(a)$。所以积分得$F(a)=a(\log(a)-1)+C$，由初始条件$F(0)=\dfrac{\log(2\pi)}{2}$可以求个极限确定$C$，得：

\[\int_{a}^{a+1}\log(\Gamma(x))\textrm{d}x=a(\log(a)-1)+\frac{\log(2\pi)}{2}\]