10-Euler 积分

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Euler 积分在一些定积分的计算中十分有用. 除了 Beta (贝塔) 函数和 Gamma (伽马) 函数, 让我们对所谓双 Gamma 函数和多重对数函数加以一定关注.

双 Gamma 函数

定义双 Gamma 函数 $\psi$ 如下:

$$\psi (x):=(\ln \Gamma (x){)}^{\prime },\forall x>0.\text{\hspace{1em}(10.1)}$$

该函数也称为 $\psi$ 函数 ${}^{(1)}$ .

关于双 Gamma 函数, 有如下基本性质:

1. 利用 $\Gamma (x+1)=x\Gamma (x)$ 得到

$$\psi (x+1)=\psi (x)+\frac{1}{x},x>0.\text{\hspace{1em}(10.2)}$$

2. 设 $x,y>0$ , 则

$$\psi (x)-\psi (y)=\sum _{k=0}^{\infty }\left(\frac{1}{y+k}-\frac{1}{x+k}\right).\text{\hspace{1em}(10.3)}$$

证明 利用 (i) 可得

$$\psi (x+1)=\frac{1}{x}+\psi (x),\forall x>0.\text{\hspace{1em}(10.4)}$$

从而 $\forall n⩾1,$

$$\begin{array}{rl}\psi (x)-\psi (y)& =\psi (x+1)-\psi (y+1)-\left(\frac{1}{x}-\frac{1}{y}\right)\\ & =\dots \\ & =\psi (x+n)-\psi (y+n)-\sum _{k=0}^{n-1}\left(\frac{1}{x+k}-\frac{1}{y+k}\right).\end{array}\text{\hspace{1em}(10.5)}$$

另一方面, 由于 $\psi$ 单调增加, 所以当 $y⩽x⩽y+1$ 时,

$$\begin{array}{rl}0⩽\psi (x+n)-\psi (y+n)& =\frac{1}{x+n-1}+\psi (x+n-1)-\psi (y+n)\\ & ⩽\frac{1}{x+n-1},\forall n⩾1.\end{array}$$

由此得到

$$\underset{n\to +\infty }{\lim }[\psi (x+n)-\psi (y+n)]=0.\text{\hspace{1em}(10.6)}$$

结合 (10.4) 式知 (10.6) 式对所有 $x,y>0$ 成立. 于是, 在 (10.5) 式中令 $n\to +\infty$ 即得 (10.3) 式.

3. 对 (10.3) 式关于 $x$ 求导即得

$${\psi }^{\prime }(x)=\sum _{n=0}^{\infty }\frac{1}{(n+x{)}^2},x>0.\text{\hspace{1em}(10.7)}$$

4. 对 (10.7) 式求导得

$${\psi }^{\prime \prime }(x)=-\sum _{n=0}^{\infty }\frac{2}{(n+x{)}^3},x>0.\text{\hspace{1em}(10.8)}$$

22. 对余元公式两边取对数后求导可得

$$\psi (x)-\psi (1-x)=-\pi \cot \pi x,\forall x\in (0,1).\text{\hspace{1em}(10.9)}$$

6. 利用

$$\Gamma (x)\Gamma \left(x+\frac{1}{k}\right)\dots \Gamma \left(x+\frac{k-1}{k}\right)=(2\pi {)}^{\frac{k-1}{2}}{k}^{\frac{1}{2}-kx}\Gamma (kx)\text{\hspace{1em}(10.10)}$$

直接得到: 当 $k⩾2$ 时, 成立

$$\sum _{j=0}^{k-1}\psi \left(x+\frac{j}{k}\right)=k(\psi (kx)-\ln k),\forall x>0.\text{\hspace{1em}(10.11)}$$

7. 可以由上述性质和 ${\Gamma }^{\prime }(1)=-\gamma$（其中 $\gamma :={\lim }_{n\to +\infty }\left({\sum }_{k=1}^n\frac{1}{k}-\ln n\right)$ 为 Euler 常数）得到 $\psi$ 在 $\frac{1}{2},\frac{1}{3},\frac{2}{3},\frac{1}{4},\frac{3}{4},\frac{1}{6},\frac{5}{6}$ 等点的值. 例如 $\psi \left(\frac{1}{2}\right)=-2\ln 2-\gamma$ .

多重对数函数

对于 $p>1$ ，可引入多重对数函数

$${\mathrm{Li}}_p(x):=\sum _{n=1}^{\infty }\frac{x^n}{n^p},|x|⩽1.$$

当 p = 2 时, 通过两边求导容易证明以下的 Euler 公式:

$${\mathrm{Li}}_2(x)+{\mathrm{Li}}_2(1-x)=\frac{{\pi }^2}{6}-\ln x\ln (1-x),x\in (0,1).\text{\hspace{1em}(10.12)}$$

这是多重对数函数的余元公式.

证明

$${\mathrm{Li}}_2(x)+{\mathrm{Li}}_2(1-x)={\mathrm{Li}}_2(0)+{\mathrm{Li}}_2(1)+{\int }_0^x[{\mathrm{Li}}_2(t)+{\mathrm{Li}}_2(1-t){]}^{\prime }\mathrm{d}t$$ $$=\frac{{\pi }^2}{6}+{\int }_0^x\sum _{n=1}^{+\infty }\left[\frac{t^{n-1}}{n}-\frac{(1-t{)}^{n-1}}{n}\right]\mathrm{d}x$$ $$=\frac{{\pi }^2}{6}+{\int }_0^x\left(-\frac{\ln (1-t)}{t}+\frac{\ln t}{1-t}\right)\mathrm{d}t$$ $$=\frac{{\pi }^2}{6}-[\ln t\ln (1-t)]{|}_0^x$$ $$=\frac{{\pi }^2}{6}-\ln x\ln (1-x),x\in (0,1).$$

例 10.1

例 10.1

设 $p>-1,q>-1$ ，则

$$\begin{array}{rlr}{\int }_0^{\frac{\pi }{2}}{\sin }^{p}x{\cos }^{q}x\mathrm{d}x& ={\int }_0^1{t}^p(1-t^2{)}^{\frac{q-1}{2}}\mathrm{d}t& \\ & =\frac{1}{2}{\int }_0^1{s}^{\frac{p-1}{2}}(1-s{)}^{\frac{q-1}{2}}\mathrm{d}s& =\frac{1}{2}\mathrm{B}\left(\frac{p+1}{2},\frac{q+1}{2}\right)\\ & =\frac{1}{2}\frac{\Gamma \left(\frac{p+1}{2}\right)\Gamma \left(\frac{q+1}{2}\right)}{\Gamma \left(\frac{p+q}{2}+1\right)}.& \end{array}$$

例 10.2

例 10.2

设 $p,q,r>0,qr>p+1$ ，则

$${\int }_0^{+\infty }\frac{x^p}{(1+x^q{)}^r}\mathrm{d}x={\int }_1^0{\left(\frac{1}{t}-1\right)}^{\frac{p}{q}}{t}^r\left[-\frac{1}{q}{\left(\frac{1}{t}-1\right)}^{\frac{1}{q}-1}\frac{1}{t^2}\right]\mathrm{d}t$$ $$=\frac{1}{q}{\int }_0^1{\left(\frac{1}{t}-1\right)}^{\frac{p+1}{q}-1}{t}^{r-2}\mathrm{d}t$$ $$\begin{array}{rl}& =\frac{1}{q}{\int }_0^1(1-t{)}^{\frac{p+1}{q}-1}{t}^{r-1-\frac{p+1}{q}}dt\\ & =\frac{1}{q}\mathrm{B}\left(\frac{p+1}{q},r-\frac{p+1}{q}\right)\\ & =\frac{\Gamma \left(\frac{p+1}{q}\right)\Gamma \left(r-\frac{p+1}{q}\right)}{q\Gamma (r)}.\end{array}$$

例 10.3

例 10.3

设 $a,b\in \mathbb{R},p>0,q>0$ ，则

$$\begin{array}{rl}{\int }_a^b(b-x{)}^{p-1}(x-a{)}^{q-1}\mathrm{d}x& =(b-a{)}^{p+q-1}{\int }_0^1(1-t{)}^{p-1}{t}^{q-1}\mathrm{d}t(\text{作变量代换}x\\ & =a+t(b-a))\\ & =(b-a{)}^{p+q-1}\mathrm{B}(p,q).\end{array}$$

特别地，

$${\int }_{-1}^1(1-x{)}^{p-1}(1+x{)}^{q-1}\mathrm{d}x=2^{p+q-1}\mathrm{B}(p,q).$$

例 10.4

例 10.4

证明:

$${\Gamma }^{\prime }(1)=-\gamma$$

证明 法 I 我们有

$${\Gamma }^{\prime }(1)=\psi (1)=\psi (n+1)-\sum _{k=1}^n\frac{1}{k},\forall n⩾1.$$

另一方面, 因为 $\ln \Gamma$ 是凸函数, 可得

$$\psi (x)⩽\ln \Gamma (x+1)-\ln \Gamma (x)⩽\psi (x+1),\forall x>0.$$

特别地，

$$\ln n⩽\psi (n+1)⩽\ln (n+1),\forall n⩾1.$$

因此，

$$\ln n-\sum _{k=1}^n\frac{1}{k}⩽{\Gamma }^{\prime }(1)⩽\ln (n+1)-\sum _{k=1}^n\frac{1}{k},\forall n⩾1.$$

在上式中令 $n\to +\infty$ 即得 ${\Gamma }^{\prime }(1)=-\gamma$ .

法II

$$\gamma =\underset{n\to +\infty }{\lim }\left(\frac{1}{1}+\frac{1}{2}+\cdots +\frac{1}{n}-\ln n\right)$$ $$\begin{array}{rlr}& =\underset{n\to +\infty }{\lim }\left(\sum _{k=1}^n{\int }_0^1{x}^{k-1}\mathrm{d}x-{\int }_1^n\frac{1}{x}\mathrm{d}x\right)& \\ & =\underset{n\to +\infty }{\lim }\left({\int }_0^1\frac{1-x^n}{1-x}\mathrm{d}x-{\int }_1^n\frac{1}{x}\mathrm{d}x\right)& \\ & =\underset{n\to +\infty }{\lim }\left[{\int }_0^n\frac{1-{\left(1-\frac{t}{n}\right)}^n}{t}\mathrm{d}t-{\int }_1^n\frac{1}{x}\mathrm{d}x\right]& \\ & =\underset{n\to +\infty }{\lim }\left[{\int }_0^1\frac{1-{\left(1-\frac{x}{n}\right)}^n}{x}\mathrm{d}x-{\int }_1^n\frac{{\left(1-\frac{x}{n}\right)}^n}{x}\mathrm{d}x\right]& \\ & ={\int }_0^1\frac{1-{\mathrm{e}}^{-x}}{x}\mathrm{d}x-{\int }_1^{+\infty }\frac{{\mathrm{e}}^{-x}}{x}\mathrm{d}x& \\ & =-{\int }_0^{+\infty }{\mathrm{e}}^{-x}\ln x\mathrm{d}x& =-{\Gamma }^{\prime }(1).\end{array}$$

法III

$$\gamma =\sum _{n=1}^{+\infty }\left[\frac{1}{n}-\ln \left(1+\frac{1}{n}\right)\right]=\sum _{n=1}^{+\infty }{\int }_0^1\frac{x}{n(n+x)}\mathrm{d}x$$ $$=\sum _{n=1}^{+\infty }{\int }_0^1\mathrm{d}x{\int }_0^1\frac{x{t}^{n+x-1}}{n}\mathrm{d}t=-{\int }_0^1\mathrm{d}x{\int }_0^{1}x{t}^{x-1}\ln (1-t)\mathrm{d}t$$ $$=-{\int }_0^{1}x\frac{\partial \mathrm{B}(x,y)}{\partial y}{|}_{y=1}\mathrm{d}x=-{\int }_0^{1}x\frac{\partial }{\partial y}\frac{\Gamma (x)\Gamma (y)}{\Gamma (x+y)}{|}_{y=1}\mathrm{d}x$$ $$=-{\int }_0^1\left[{\Gamma }^{\prime }(1)-\frac{{\Gamma }^{\prime }(x+1)}{\Gamma (x+1)}\right]\mathrm{d}x=-{\Gamma }^{\prime }(1)+\ln \Gamma (x+1){|}_{x=0}^1$$ $$=-{\Gamma }^{\prime }(1).$$

例 10.5

例 10.5

设 $\alpha >0$ 计算

$$\sum _{n=1}^{\infty }\frac{1}{n(n+\alpha )}$$

解 法 I 利用双 Gamma 函数, 由 (10.3) 式立即得到

$$\sum _{n=1}^{\infty }\frac{1}{n(n+\alpha )}=\frac{1}{\alpha }\sum _{n=1}^{\infty }\left(\frac{1}{n}-\frac{1}{n+\alpha }\right)=\frac{1}{\alpha }[\psi (1+\alpha )-\psi (1)].$$

法II

$$\sum _{n=1}^{\infty }\frac{1}{n(n+\alpha )}=\sum _{n=1}^{\infty }{\int }_0^1\frac{t^{n+\alpha -1}}{n}\mathrm{d}t=-{\int }_0^1{t}^{\alpha -1}\ln (1-t)\mathrm{d}t$$ $$\begin{gathered} \begin{array}{rl}& =-\left[\frac{\partial }{\partial y}{\int }_0^1{t}^{\alpha -1}(1-t{)}^{y-1}\mathrm{d}t\right]{|}_{y=1}\\ & =-\frac{\partial \mathrm{B}(\alpha ,y)}{\partial y}{|}_{y=1}\\ & =-\frac{\partial }{\partial y}\frac{\Gamma (\alpha )\Gamma (y)}{\Gamma (\alpha +y)}{|}_{y \\ =1}\\ & =-\frac{{\Gamma }^{\prime }(1)}{\alpha }+\frac{{\Gamma }^{\prime }(\alpha +1)}{\alpha \Gamma (\alpha +1)}.\end{array} \end{gathered}$$

例 10.6

例 10.6

设 $0<\alpha <1$ , 计算

$$\sum _{n=1}^{\infty }\frac{1}{n(n-\alpha )}$$

解 利用双 Gamma 函数, 有

$$\sum _{n=1}^{\infty }\frac{1}{n(n-\alpha )}=\frac{1}{\alpha }\sum _{n=1}^{\infty }\left(\frac{1}{n-\alpha }-\frac{1}{n}\right)=\frac{1}{\alpha }[\psi (1)-\psi (1-\alpha )].$$

例 10.7

例 10.7

设 $0<\alpha <1$ ，计算

$$\sum _{n=1}^{\infty }\frac{1}{n^2-{\alpha }^2}$$

解

$$\sum _{n=1}^{\infty }\frac{1}{n^2-{\alpha }^2}=\frac{1}{2\alpha }\sum _{n=1}^{\infty }\left(\frac{1}{n-\alpha }-\frac{1}{n+\alpha }\right)$$ $$\begin{array}{rl}& =\frac{1}{2\alpha }[\psi (1+\alpha )-\psi (1-\alpha )]\\ & =\frac{1}{2\alpha }\left[\frac{1}{\alpha }+\psi (\alpha )-\psi (1-\alpha )\right]\\ & =\frac{1}{2{\alpha }^2}+\frac{1}{2\alpha }\frac{\mathrm{d}}{\mathrm{d}\alpha }\ln [\Gamma (\alpha )\Gamma (1-\alpha )]\\ & =\frac{1}{2{\alpha }^2}-\frac{\pi \cot (\alpha \pi )}{2\alpha }.\end{array}$$

注 10.1 ^zhu-10-1 另一方面,

$$\begin{gathered} \begin{array}{rl}\sum _{n=1}^{\infty }\frac{1}{n^2-{\alpha }^2}& \\ & =\sum _{n \\ =1}^{\infty }\frac{1}{n^2}\sum _{k \\ =1}^{\infty }\frac{{\alpha }^{2(k-1)}}{n^{2(k-1)}}\\ & =\sum _{k=1}^{\infty }\sum _{n \\ =1}^{\infty }\frac{{\alpha }^{2(k-1)}}{n^{2k}}\\ & =\sum _{k \\ =1}^{\infty }\zeta (2k){\alpha }^{2(k-1)}.\end{array} \end{gathered}$$

因此

$$\alpha \pi \cot (\alpha \pi )=1-2\sum _{k=1}^{\infty }\zeta (2k){\alpha }^{2k},|\alpha |<1.\text{\hspace{1em}(10.13)}$$

例 10.8

例 10.8

设 $0<\alpha ⩽1$ , 计算

$$\sum _{n=1}^{\infty }\frac{1}{n^2+{\alpha }^2}$$

解 可以由例 10.7 的结果作如下猜想:

$$\begin{array}{rl}\sum _{n=1}^{\infty }\frac{1}{n^2+{\alpha }^2}& \\ & =\frac{1}{2(\mathrm{i}\alpha {)}^2}-\frac{\pi \cos (\mathrm{i}\alpha \pi )}{2\mathrm{i}\alpha \sin (\mathrm{i}\alpha \pi )}\\ & =-\frac{1}{2{\alpha }^2}+\frac{\pi }{2\alpha \tanh (\alpha \pi )}.\end{array}$$

由解析函数理论即可直接得到上述结果, 在实数范围内则可作如下解答:

对于 $0<\alpha <1$ ，有

$$\begin{gathered} \begin{array}{rlr}\sum _{n=1}^{\infty }\frac{1}{n^2+{\alpha }^2}& & \\ & =\sum _{n \\ =1}^{\infty }\frac{1}{n^2}\sum _{k \\ =1}^{\infty }\frac{(-1{)}^{k-1}{\alpha }^{2(k-1)}}{n^{2(k-1)}}& \\ & =\sum _{k=1}^{\infty }\sum _{n \\ =1}^{\infty }\frac{(-1{)}^{k-1}{\alpha }^{2(k-1)}}{n^{2k}}& \\ & =\sum _{k \\ =1}^{\infty }(-1{)}^{k-1}\zeta (2k){\alpha }^{2(k-1)}& \\ & =\frac{\pi }{2\alpha }\left[\coth (\alpha \pi )-\frac{1}{\alpha \pi }\right]& =-\frac{1}{2{\alpha }^2}+\frac{\pi }{2\alpha }\coth (\alpha \pi ).\end{array} \end{gathered}$$

利用连续性, 上式对于 $\alpha =1$ 也成立. 利用 (实) 解析性, 可以说明上式对所有 $\alpha >0$ 成立 (详细讨论请见 [10]).

例 10.9

例 10.9

对于 $\alpha >-1$ ，有

\begin{aligned} \int_ {0} ^ {1} \frac {x ^ {\alpha} - 1}{1 - x} \mathrm{d} x &= \int_ {0} ^ {1} \sum_ {n \\[6pt] &= 0} ^ {\infty} (x ^ {\alpha} - 1) x ^ {n} \mathrm{d} x \\[6pt] &= \sum_ {n = 0} ^ {\infty} \left(\frac {1}{n + 1 + \alpha} - \frac {1}{n + 1}\right) \\[6pt] &= \psi (1) - \psi (1 + \alpha). \end{aligned}

例 10.10

例 10.10

对于 $\alpha >0$ , 有

$$\begin{array}{rlr}{\int }_0^1\frac{x^{\alpha -1}\ln x}{1-x}\mathrm{d}x& =\frac{\mathrm{d}}{\mathrm{d}\alpha }\left({\int }_0^1\frac{x^{\alpha -1}-1}{1-x}\mathrm{d}x\right)& \\ & =\frac{\mathrm{d}}{\mathrm{d}\alpha }[\psi (1)-\psi (\alpha )]& =-{\psi }^{\prime }(\alpha ).\end{array}$$

例 10.11

例 10.11

计算

$$I={\int }_0^{\frac{\pi }{2}}x\ln (\sin x)\ln (\cos x)\mathrm{d}x.$$

解 易见

$$I={\int }_0^{\frac{\pi }{2}}x\ln (\sin x)\ln (\cos x)\mathrm{d}x=\frac{\pi }{4}{\int }_0^{\frac{\pi }{2}}\ln (\sin x)\ln (\cos x)\mathrm{d}x.$$

法I

$$I=\frac{\pi }{4}{\int }_0^1\frac{\ln t\ln \sqrt{1-t^2}}{\sqrt{1-t^2}}\mathrm{d}t=\frac{\pi }{32}{\int }_0^1\frac{\ln s\ln (1-s)}{\sqrt{s}\sqrt{1-s}}\mathrm{d}s$$ $$\begin{gathered} \begin{array}{rl}& ={\frac{\pi }{32}\left[\frac{{\partial }^2}{\partial \alpha \partial \beta }{\int }_0^1{s}^{-\frac{1}{2}+\alpha }(1-s{)}^{-\frac{1}{2}+\beta }ds\right]|}_{\alpha =0,\beta \\ =0}\\ & ={\frac{\pi }{32}\left[\frac{{\partial }^2}{\partial \alpha \partial \beta }\frac{\Gamma \left(\frac{1}{2}+\alpha \right)\Gamma \left(\frac{1}{2}+\beta \right)}{\Gamma (1+\alpha +\beta )}\right]|}_{\alpha =0,\beta \\ =0}\\ & =\frac{\pi }{32}{\Gamma }^2\left(\frac{1}{2}\right)\left\{{\left[\psi \left(\frac{1}{2}\right)-\psi (1)\right]}^2-{\psi }^{\prime }(1)\right\}\\ & =\frac{{\pi }^2}{32}\left[(-2\ln 2{)}^2-\frac{{\pi }^2}{6}\right]\\ & =\frac{{\pi }^2}{8}{\ln }^{2}2-\frac{{\pi }^4}{192}.\end{array} \end{gathered}$$

法II 利用

$$\ln \left(2\sin \frac{x}{2}\right)=-\sum _{n=1}^{\infty }\frac{\cos nx}{n},\forall x\in (0,2\pi ).$$

我们有

$$I=\frac{\pi }{8}{\int }_0^{\frac{\pi }{2}}\left\{{\left[\ln (\sin x)+\ln (\cos x)\right]}^2-{\ln }^2(\sin x)-{\ln }^2(\cos x)\right\}dx$$ $$=\frac{\pi }{8}{\int }_0^{\frac{\pi }{2}}\left[{\ln }^2\frac{\sin 2x}{2}-2{\ln }^2(\sin x)\right]\mathrm{d}x$$ $$=\frac{\pi }{8}{\int }_0^{\frac{\pi }{2}}\left\{{\left[2\ln 2+\sum _{n=1}^{\infty }\frac{\cos (4nx)}{n}\right]}^2-2{\left[\ln 2+\sum _{n=1}^{\infty }\frac{\cos (2nx)}{n}\right]}^2\right\}dx$$ $$=\frac{\pi }{8}\left(\pi {\ln }^{2}2-\frac{\pi }{4}\sum _{n=1}^{\infty }\frac{1}{n^2}\right)$$ $$=\frac{{\pi }^2}{8}{\ln }^{2}2-\frac{{\pi }^4}{192}.$$

例 10.12

例 10.12

计算

$${\int }_0^1\frac{\ln \frac{1}{x}}{2-x}\mathrm{d}x.$$

解

$${\int }_0^1\frac{\ln \frac{1}{x}}{2-x}\mathrm{d}x=-{\int }_0^1\frac{1}{2}\sum _{n=1}^{\infty }{\left(\frac{x}{2}\right)}^{n-1}\ln x\mathrm{d}x$$ $$=\sum _{n=1}^{\infty }\frac{1}{2^n{n}^2}={\mathrm{Li}}_2\left(\frac{1}{2}\right)=\frac{{\pi }^2}{12}-\frac{1}{2}{\ln }^{2}2.$$

例 10.13

例 10.13

计算

$${\int }_0^{+\infty }\frac{\ln x}{x^2-1}\mathrm{d}x.$$

法 I

$$\begin{array}{rl}{\int }_0^{+\infty }\frac{\ln x}{x^2-1}\mathrm{d}x& ={\int }_1^{+\infty }\frac{\ln x}{x^2-1}\mathrm{d}x+{\int }_0^1\frac{\ln x}{x^2-1}\mathrm{d}x\\ & =2{\int }_0^1\frac{\ln x}{x^2-1}\mathrm{d}x\\ & =-{\int }_0^1\frac{\ln x}{2\sqrt{x}(1-x)}\mathrm{d}x\\ & =-\frac{1}{2}\underset{\alpha \to \frac{1}{2}}{\lim }\underset{\beta \to 0^{+}}{\lim }\frac{\partial }{\partial \alpha }{\int }_0^1{x}^{\alpha -1}(1-x{)}^{\beta -1}\mathrm{d}x\\ & =-\frac{1}{2}\underset{\alpha \to \frac{1}{2}}{\lim }\underset{\beta \to 0^{+}}{\lim }\frac{\partial }{\partial \alpha }\frac{\Gamma (\alpha )\Gamma (\beta )}{\Gamma (\alpha +\beta )}\\ & =-\frac{1}{2}\underset{\alpha \to \frac{1}{2}}{\lim }\underset{\beta \to 0^{+}}{\lim }\frac{\Gamma (\alpha )\Gamma (\beta )}{\Gamma (\alpha +\beta )}[\psi (\alpha )-\psi (\beta +\alpha )]\\ & =-\frac{1}{2}\underset{\alpha \to \frac{1}{2}}{\lim }\underset{\beta \to 0^{+}}{\lim }\frac{1}{\beta }[\psi (\alpha )-\psi (\beta +\alpha )]\\ & =\frac{1}{2}{\psi }^{\prime }\left(\frac{1}{2}\right).\end{array}$$

另一方面，对

$$\ln \left[\Gamma \left(\frac{1}{2}-s\right)\Gamma \left(\frac{1}{2}+s\right)\right]=\ln \frac{\pi }{\sin \left(\frac{1}{2}-s\right)\pi },s\in \left(-\frac{1}{2},\frac{1}{2}\right)$$

求导两次得到 ${\psi }^{\prime }\left(\frac{1}{2}\right)=\frac{{\pi }^2}{2}$ . 所以

$${\int }_0^{+\infty }\frac{\ln x}{x^2-1}\mathrm{d}x=\frac{{\pi }^2}{4}.$$

法II

$${\int }_0^{+\infty }\frac{\ln x}{x^2-1}\mathrm{d}x=2{\int }_0^1\frac{\ln x}{x^2-1}\mathrm{d}x$$ $$=-2{\int }_0^1\sum _{n=0}^{\infty }{x}^{2n}\ln x\mathrm{d}x=-2\sum _{n=0}^{\infty }{\int }_0^1{x}^{2n}\ln x\mathrm{d}x$$ $$=2\sum _{n=0}^{\infty }\frac{1}{(2n+1{)}^2}=\frac{{\pi }^2}{4}.$$