\(\Gamma函数\)

  在数学中，\(\Gamma\)函数，是阶乘函数在实数与复数域上的扩展。

  \(\Gamma\)函数可通过第二类\(Euler\)积分定义:

\[\Gamma(z)=\int_0^{\infty}t^{z-1}e^{-t}\mathrm{d}t\]

简单性质

  1. \(\displaystyle\Gamma(z+1)=\int_0^{\infty}t^{z}e^{-t}\mathrm{d}t=t^z(-e^{-t})|_0^\infty+z\int_0^\infty t^{z-1}e^{-t}\mathrm{d}t=z\Gamma(z)\ (z>0)\)

  从而\(\Gamma(z)=(z-1)!\)

  1. \(\Gamma(z)=2\displaystyle\int_0^{+\infty}t^{2z-1}e^{-t^2}\mathrm{d}t\)

  2. \(\Gamma(z)\)在\((0,+\infty)\)上为严格下凸函数, 它及其任意阶导数都连续, 且 \[\Gamma^{(n)}(z)=\int_0^{+\infty}t^{z-1}(\ln t)^ne^{-t}\mathrm{d}t\]

命题[Bohr-Mollerup定理] 如果\(f:(0,\infty)\to(0,\infty)\)，且满足： 1. \(f(z)>0,f(1)=1\) 2. \(f(z+1)=zf(z)\) 3. \(\ln{f(z)}\)是凸函数

则\(f(z)\equiv \Gamma(z),z\in(0,\infty)\)

欧拉无穷乘积公式

  我们考察被积函数\(f(t)=t^{z-1}e^{-t}\)，设函数序列\(f_n(t):[0,+\infty)\rightarrow\Bbb{C}\),且\(\displaystyle f_n(t)=\begin{cases}t^{z-1}(1-\frac{t}{n})^n\quad ,t\in[0,n]\\0\quad,\text{其他}\end{cases}\)，显然\(\displaystyle\lim_{n\to\infty}f_n(t)=f(t)\)，且\(|Re[f_n(t)]|\leq t^{Re(z)-1}e^{-t}, |Im[f_n(t)]|\leq t^{Im(z)-1}e^{-t}\)，由控制收敛定理得

\[\lim_{n\to\infty}\int_0^{\infty}f_n(t)\mathrm{d}t=\int_0^{\infty}f(t)\mathrm{d}t\]

又因为\(\displaystyle f_n(t)\equiv0\forall t\notin[0,n],\therefore \int_0^{\infty}f_n(t)\mathrm{d}t=\int_0^{n}f_n(t)\mathrm{d}t\),，所以

\[\lim_{n\to\infty}\int_0^{n}f_n(t)\mathrm{d}t=\int_0^{\infty}f(t)\mathrm{d}t\]

定义\(\displaystyle P_n(z)=\int_0^{n}t^{z-1}\left(1-\frac{t}{n}\right)^n\mathrm{d}t\)，我们由以上结论可得

\[\lim_{n\to\infty}P_n(z)=\Gamma(z)\]

  当然我们也可以按如下方式证明：

\[\Gamma(z)-P_n(z)=\int_0^n[e^{-t}-(1-\frac{t}{n})]t^{z-1}\mathrm{d}t+\int_n^{\infty}e^{-t}t^{z-1}\mathrm{d}t\]

易知\(RHS\)第二项极限\(n\to\infty\)为0，对第一项，我们由以下引理

\[0\leq e^{-t}-\left(1-\frac{t}{n}\right)^n\leq \frac{t^2}{n}e^{-t}\]

得

\[\left|\int_0^n[e^{-t}-(1-\frac{t}{n})]t^{z-1}\mathrm{d}t\right|\leq\int_0^n\frac{1}{n}e^{-t}t^{x+1}\mathrm{d}t<\frac{1}{n}\int_0^{\infty}e^{-t}t^{x+1}\mathrm{d}t\rightarrow0\]

其中\(x=Re(z)\)，这样就证明了\(\displaystyle\lim_{n\to\infty}P_n(z)=\Gamma(z)\)

  下面证明引理，\(0<y<1\)时有\(1+y\leq e^y\leq(1-y)^{-1}\)，令\(t=\frac{t}{n}\)得

\[\left(1+\frac{t}{n}\right)^{-n}\geq e^{-t}\geq\left(1-\frac{t}{n}\right)^n\]

故

\[0\leq e^{-t}-\left(1-\frac{t}{n}\right)^n=e^{-t}\left[1-e^t\left(1-\frac{t}{n}\right)^n\right]\leq e^{-t}\left[1-\left(1-\frac{t^2}{n^2}\right)^n\right]\]

又由伯努利不等式\(0\leq\alpha\leq1\)有\((1-n\alpha)^n\geq1-n\alpha\)得

\[1-\left(1-\frac{t^2}{n^2}\right)^n\leq\frac{t^2}{n}\]

即证

在\(P_n(z)\)中令\(t=nx\)得

\[\int_0^{n}t^{z-1}\left(1-\frac{t}{n}\right)^n\mathrm{d}t=n^z\int_0^1x^{z-1}(1-x)^n\mathrm{d}x=n^zI_{n,z}\]

分部积分得

\[\begin{array}{l}I_{n,z}&=&\displaystyle\int_0^1(1-x)^n\mathrm{d}\left(\frac{x^z}{z}\right)\\ &=&\displaystyle\frac{x^z(1-x)^n}{z}|+\frac{n}{z}\int_0^1x^z(1-x)^{n-1}\mathrm{d}x\\ &=&\displaystyle\frac{n}{z}I_{n-1,z+1}\end{array}\]

而

\[I_{0,z+n}=\int_0^1x^{z+n-1}\mathrm{d}x=\frac{1}{z+n}\]

从而

\[I_{n,z}=\frac{n!}{z(z+1)\cdots(z+n)}\]

得到\(\Gamma\)函数另一个定义: \(Gauss\)无穷乘积分解

\[\Gamma(z)=\lim_{n\to\infty}\frac{n!n^z}{z(z+1)\cdots(z+n)}\]

由于\(\displaystyle\lim_{n\to\infty}\frac{n}{z+n}=1\)，又可以写成

\[\Gamma(z)=\lim_{n\to\infty}\frac{(n-1)!n^z}{z(z+1)\cdots(z+n-1)}\]

这就是著名的\(Euler-Gauss\)公式

又因为\(\displaystyle n^z=\prod_{k=1}^{n-1}\left(1+\frac{1}{k}\right)\), 极限可以改写成

\[\Gamma(z)=\lim_{n\to\infty}\frac{1}{z}\prod_{k=1}^n\left(1+\frac{z}{k}\right)^{-1}\left(1+\frac{1}{k}\right)^z\]

我们得到\(\Gamma\)函数的欧拉乘积形式

\[\Gamma(z)=\frac{1}{z}\prod_{k=1}^\infty\frac{\left(1+\frac{1}{k}\right)^z}{1+\frac{z}{k}}\]

\(Weierstrass\)无穷乘积

  由于\(\Gamma\)函数恒不为\(0\)，有

\[\begin{array}{l} \displaystyle\frac{1}{\Gamma(z)}&=&\displaystyle\lim_{n\to\infty}\frac{z(z+1)\cdots(z+n)}{n!n^z}\\ &=&\displaystyle\lim_{n\to\infty}zn^{-z}\prod_{k=1}^n\left(1+\frac{z}{k}\right)\\ &=&\displaystyle\lim_{n\to\infty}ze^{-z\ln n}\prod_{k=1}^n\left(1+\frac{z}{k}\right)\\ &=&\displaystyle\lim_{n\to\infty}ze^{-z(\ln n-H_n)}e^{-zH_n}\prod_{k=1}^n\left(1+\frac{z}{k}\right)\\ &=&\displaystyle\lim_{n\to\infty}ze^{z(H_n-\ln n)}\prod_{k=1}^n\left(1+\frac{z}{k}\right)e^{-\frac{z}{k}}\\ &=&\displaystyle ze^{z\lim\limits_{n\to\infty}(H_n-\ln n)}\prod_{k=1}^n\left(1+\frac{z}{k}\right)e^{-\frac{z}{k}}\\ &=&\displaystyle ze^{\gamma z}\prod_{k=1}^n\left(1+\frac{z}{k}\right)e^{-\frac{z}{k}} \end{array}\]

上式即为\(\Gamma\)函数的\(Weierstrass\)定义，其中\(\displaystyle H_n=\sum_{k=1}^n\frac{1}{k}\)，欧拉常数\(\displaystyle\gamma=\lim_{n\to\infty}(H_n-\ln n)=0.577215\cdots\)

与三角函数的关系

  由欧拉无穷乘积公式

\[\Gamma(z)\Gamma(-z)=-\frac{1}{z^2}\prod_{k=1}^\infty\left(1-\frac{z^2}{n^2}\right)^{-1}\]

由正弦函数无穷乘积公式

\[\frac{\sin z}{z}=\prod_{k=1}^\infty\left(1-\frac{z^2}{\pi^2k^2}\right)\]

得

\[\Gamma(z)\Gamma(-z)=-\frac{\pi}{z\sin{\pi z}}\]

由\(\Gamma(1-z)=-z\Gamma(-z)\)得

\[\Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin{\pi z}}\]

为余元公式，写成对称形式得

\[\Gamma(1+z)\Gamma(1-z)=z!*(-z)!=\frac{\pi z}{\sin{\pi z}}\]

在正弦函数无穷乘积公式中令\(\displaystyle z=\frac{\pi}{2}\)得\(Wallis\)乘积

\[\frac{\pi}{2}=\prod_{k=1}^\infty\frac{(2k)^2}{(2k-1)(2k+1)}=\lim_{n\to\infty}\left(\prod_{k=1}^n\frac{2k}{2k-1}\right)^2\frac{1}{2n+1}\]

欧拉乘积公式

  我们应用余元公式证明

\[\Gamma(z)\Gamma\left(z+\frac{1}{m}\right)\Gamma\left(z+\frac{2}{m}\right)\cdots\Gamma\left(z+\frac{m-1}{m}\right)=(2\pi)^{\frac{m-1}{2}}m^{\frac{1}{2}-mz}\Gamma(mz)\]

令

\[C =\frac{m^{mz}}{m\Gamma(mz)}\prod_{k=0}^{m-1}\Gamma\left(z+\frac{k}{m}\right)\]

由\(Euler-Gauss\)公式

\[\begin{array}{l} C&=&\displaystyle m^{mz-1}\lim_{n\to\infty}\frac{\prod\limits_{k=0}^{m-1}\frac{(n-1)!}{(z+\frac{k}{m})(z+\frac{k}{m}+1)\cdots(z+\frac{k}{m}+n-1)}n^{z+\frac{k}{m}}}{\frac{(mn-1)!}{mz(mz+1)\cdots(mz+mn-1)}(mn)^{mz}}\\ &=&\displaystyle m^{mz-1}\lim_{n\to\infty}\frac{[(n-1)!]^mn^{mz+\frac{1}{2}(m-1)}m^{(n+1)m}}{(mn-1)!(mn)^{mz}}\\ &=&\displaystyle\lim_{n\to\infty}\frac{[(n-1)!]^mn^{\frac{1}{2}(m-1)}m^{nm-1}}{(mn-1)!} \end{array}\]

从而\(C\)与\(z\)无关，令\(z=\frac{1}{n}\)得

\[C=\prod_{k=0}^{m-1}\Gamma\left(\frac{k+1}{m}\right)=\prod_{k=1}^{m-1}\Gamma\left(\frac{k}{m}\right)=\prod_{k=1}^{m-1}\Gamma\left(1-\frac{k}{m}\right)\]

由余元公式得

\[C^2=\prod_{k=1}^{m-1}[\Gamma\left(\frac{k}{m}\right)\Gamma\left(1-\frac{k}{m}\right)]=\pi^{m-1}\prod_{k=1}^{m-1}\left(\sin \frac{k\pi}{m}\right)^{-1}\]

由于\(z^m-1=0\)的根为\(z=e^{\frac{2k\pi i}{m}},k=0,1,\cdots,m-1\)

\[\frac{z^m-1}{z-1}=\sum_{k=0}^{m-1}z^k=\prod_{k=1}^{m-1}\left(z-e^{\frac{2k\pi i}{m}}\right)\]

令\(z=1\)得

\[\begin{array}{l} m&=&\displaystyle\prod_{k=1}^{m-1}\left(1-e^{\frac{2k\pi i}{m}}\right)\\ &=&\displaystyle\prod_{k=1}^{m-1}e^{\frac{k\pi i}{m}}\left(-2i\sin{\frac{k\pi}{m}}\right)\\ &=&\displaystyle e^{\frac{m-1}{2}\pi i}2^{m-1}(-i)^{m-1}\prod_{k=1}^{m-1}\sin{\frac{k\pi}{m}}\\ &=&\displaystyle2^{m-1}\prod_{k=1}^{m-1}\sin{\frac{k\pi}{m}} \end{array}\]

从而

\[C^2=\frac{(2\pi)^{m-1}}{m}\]

取平方根代入即得

\[\prod_{k=0}^{m-1}\Gamma\left(z+\frac{k}{m}\right)=(2\pi)^{\frac{m-1}{2}}m^{\frac{1}{2}-mz}\Gamma(mz)\]

令\(m=2\)得\(Legendre\)加倍公式

\[\Gamma(2z)=\frac{2^{2z-1}}{\sqrt{\pi}}\Gamma(z)\Gamma\left(z+\frac{1}{2}\right)\]

\(\Psi\)函数

  定义\(Digamma\)函数

\[\Psi(z)=\frac{\mathrm{d}\ln{\Gamma(z)}}{\mathrm{d}z}=\frac{\Gamma'(z)}{\Gamma(z)}\]

它有一个漂亮的性质

\[\Psi(z+1)=\Psi(z)+\frac{1}{z}\]

由\(Weierstrass\)对\(\Gamma\)函数的定义\(\displaystyle\Gamma(z)=\frac{e^{-\gamma z}}{z}\prod_{k=1}^\infty\left(1+\frac{z}{k}\right)^{-1}e^{\frac{z}{k}}\)得

\[\ln\Gamma(z)=-\gamma z-\ln z+\sum_{k=1}^\infty[\frac{z}{k}-\ln{(k+z)}+\ln k]\]

两边求导得

\[\frac{\Gamma'(z)}{\Gamma(z)}=-\gamma-\frac{1}{z}+\sum_{k=1}^\infty\left(\frac{1}{k}-\frac{1}{k+z}\right)\]

从而

\[\Psi(z)=-\gamma+\sum_{k=0}^\infty\left(\frac{1}{k+1}-\frac{1}{k+z}\right)\]

令\(z=1\)结合\(\Gamma(1)=1\)得

\[\Psi(1)=\Gamma'(1)=-\gamma\]

注意到

\[\frac{1}{k+1}-\frac{1}{k+z}=\int_0^1(x^k-x^{k+z-1})\mathrm{d}x=\int_0^1x^k(1-x^{z-1})\mathrm{d}x\]

那么

\[\Psi(z)=-\gamma+\int_0^1(1-x^{z-1})\sum_{k=0}^\infty x^k\mathrm{d}x=-\gamma+\int_0^1\frac{1-x^{z-1}}{1-x}\mathrm{d}x\]

\(\rm{B}函数\)

  \({\rm{B}}\)函数可通过第一类\(Euler\)积分定义：

\[{\rm{B}}\left(p,q\right)=\int_0^1x^{p-1}(1-x)^{q-1}\mathrm{d}x,\quad p,q>0\]

简单性质

  1. \({\rm{B}}(p,q)={\rm{B}}(q,p)\)
  2. \({\rm{B}}(p,q)=\displaystyle\frac{p-1}{p+q-1}{\rm{B}}(p-1,q)\)
  3. \({\rm{B}}(p,q)=2\displaystyle\int_0^\frac{\pi}{2}\cos^{2p-1}\theta\sin^{2q-1}\theta\mathrm{d}\theta\)
  4. \({\rm{B}}(p,q)=\displaystyle\int_0^{+\infty}\frac{x^{q-1}}{(1+x)^{p+q}}\mathrm{d}x\)

与\(\Gamma\)函数的关系

  设\(p>0,q>0\)则有\({\rm{B}}(p,q)=\displaystyle\frac{\Gamma(p)\Gamma(q)}{\Gamma(p+q)}\)

  证明:  由性质知

\[\Gamma(p)=2\int_0^{+\infty}x^{2p-1}e^{-x^2}\mathrm{d}x,\quad\Gamma(q)=2\int_0^{+\infty}x^{2q-1}e^{-x^2}\mathrm{d}x\]

令\(D=\left\{(x,y):0\leq x<+\infty,0\leq y<+\infty\right\}\), 则有

\[\Gamma(p)\Gamma(q)=4\iint_Dx^{2p-1}y^{2q-1}e^{-(x^2+y^2)}\mathrm{d}x\mathrm{d}y\]

利用极坐标变换, 记\(D_1=\displaystyle\left\{(r,\theta):0<r<+\infty,0\leq\theta\leq\frac{\pi}{2}\right\}\), 则有

\[\begin{array}{l} \Gamma(p)\Gamma(q)&=&4\displaystyle\iint_{D1}r^{2(p+q)-1}e^{-r^2}\cos^{2p-1}\theta\sin^{2q-1}\theta\mathrm{d}r\mathrm{d}\theta\\ &=&\displaystyle\left(2\int_0^\frac{\pi}{2}\cos^{2p-1}\theta\sin^{2q-1}\theta\mathrm{d}\theta\right)\left(2\int_0^{+\infty}r^{2(p+q)-1}e^{-r^2}\mathrm{d}r\right)\\ &=&\displaystyle{\rm{B}}(p,q)\Gamma(p+q) \end{array}\]

证毕.

   (余元公式) 设\(0<p<1\), 则有

\[{\rm{B} }(p,1-p)=\Gamma(p)\Gamma(1-p)=\frac{\pi}{\sin p\pi}.\]

  证明:  由于\({\rm{B}}(p,1-p)=\displaystyle\int_0^\infty\frac{x^{p-1}}{1+x}\mathrm{d}x\), 利用变量替换\(x=\frac{1}{t}\)有:

\[\int_1^\infty\frac{x^{p-1}}{1+x}\mathrm{d}x=\int_0^1\frac{x^{-p}}{1+x}\mathrm{d}x,\]

将\(\displaystyle\frac{1}{1+x}\)展开成幂级数, 从而有

\[\begin{array}{l} {\rm{B} }(p,1-p)&=&\displaystyle\lim_{r\to1^-}\int_0^r\frac{x^{p-1}+x^{-p}}{1+x}\mathrm{d}x\\ &=&\displaystyle\lim_{r\to1^-}\int_0^r\left[\sum_{k=0}^\infty(-1)^kx^{k+p-1}+\sum_{k=0}^\infty(-1)^kx^{k-p}\right]\mathrm{d}x\\ &=&\displaystyle\lim_{r\to1^-}\left[\sum_{k=0}^\infty\frac{(-1)^k}{k+p}r^{k+p}+\sum_{k=0}^\infty\frac{(-1)^k}{k-p+1}r^{k-p+1}\right]\\ &=&\displaystyle\sum_{k=0}^\infty\frac{(-1)^k}{k+p}+\sum_{k=0}^\infty\frac{(-1)^k}{k-p+1}\\ &=&\displaystyle\frac{1}{p}+\sum_{k=1}^\infty(-1)^k\left(\frac{1}{k+p}+\frac{1}{p-k}\right)\\ &=&\displaystyle\frac{1}{p}+\sum_{k=1}^\infty(-1)^k\frac{2p}{p^2-k^2}. \end{array}\]

而\(\cos px\)的\(Fourier\)级数为

\[\cos px=\frac{\sin px}{\pi}\left[\frac{1}{p}+\sum_{k=1}^\infty(-1)^k\frac{2p}{p^2-k^2}\cos kx\right]\]

在\(\left|x\right|\leq\pi\)处处收敛, 令\(x=0\)即得

\[{\rm{B}}(p,1-p)=\frac{1}{p}+\sum_{k=1}^\infty(-1)^k\frac{2p}{p^2-k^2}=\frac{\pi}{\sin p\pi}.\]

证毕.

注:令\(p=\displaystyle\frac{1}{2}\), 得 \[\Gamma^2\left(\frac{1}{2}\right)=\frac{\displaystyle\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{1}{2}\right)}{\Gamma(1)}={\rm{B}}(\frac{1}{2},\frac{1}{2})=\pi\] 则\(\displaystyle\int_0^\infty e^{-x^2}\mathrm{d}x=\frac{1}{2}\Gamma\left(\frac{1}{2}\right)=\frac{\sqrt\pi}{2}\), 也是一个得到这个重要等式的方法.

  求\(\Bbb{R}^n\)中单位球体\(D:x_1^2+x_2^2+\cdots x_n^2\leq1\)的体积.

   分析: 考虑用与球面类似的换元求解

   解: 由\(n\)重积分的集合意义, 所求体积为

\[V=\iint\cdots\int_D\mathrm{d}x_1\mathrm{d}x_2\cdots\mathrm{d}x_n.\]

作变换

\[\begin{cases} x_1&=&r\cos\theta_1,\\ x_2&=&r\sin\theta_1\cos\theta_2\\ x_3&=&r\sin\theta_1\sin\theta_2\cos\theta_3\\ \cdots\cdots\\ x_{n-1}&=&r\sin\theta_1\sin\theta_2\cdots\sin\theta_{n-2}\cos\theta_{n-1}\\ x_{n}&=&r\sin\theta_1\sin\theta_2\cdots\sin\theta_{n-2}\sin\theta_{n-1} \end{cases}\]

其中\(\quad 0<r<1, 0<\theta_1,\cdots,\theta_{n-2}<\pi,0<\theta_{n-1}<2\pi\), 则其\(Jacobian\)行列式

\[\displaystyle\frac{\partial(x_1,x_2,\cdots,x_n)}{\partial(r,\theta_1,\theta_2,\cdots,\theta_{n-1})}=r^{n-1}\sin^{n-2}\theta_1\sin^{n-3}\theta_2\cdots\sin\theta_{n-2}\]

由此

\[\begin{array}{l} V &=& \displaystyle\iint\cdots\int_D\mathrm{d}x_1\mathrm{d}x_2\cdots\mathrm{d}x_n\\ &=& \displaystyle\int_0^{2\pi}\mathrm{d}\theta_{n-1}\int_0^\pi\mathrm{d}\theta_{n-2}\cdots\int_0^\pi\mathrm{d}\theta_1\int_0^1r^{n-1}\sin^{n-2}\theta_1\sin^{n-3}\theta_2\cdots\sin\theta_{n-2}\mathrm{d}r\\ &=& \displaystyle\frac{2\pi}{n}\left(\int_0^\pi\sin^{n-2}\theta_1\mathrm{d}\theta_1\right)\left(\int_0^\pi\sin^{n-3}\theta_2\mathrm{d}\theta_2\right)\cdots\left(\int_0^\pi\sin\theta_{n-2}\mathrm{d}\theta_{n-2}\right)\\ &=& \displaystyle\frac{2\pi}{n}{\rm{B}}\left(\frac{1}{2}, \frac{n-1}{2}\right){\rm{B}}\left(\frac{1}{2},\frac{n-2}{2}\right)\cdots{\rm{B}}\left(\frac{1}{2},1\right)\\ &=& \displaystyle\frac{2\pi}{n}\cdot\frac{\displaystyle\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{n-1}{2}\right)}{\displaystyle\Gamma\left(\frac{n}{2}\right)}\cdot\frac{\displaystyle\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{n-2}{2}\right)}{\displaystyle\Gamma\left(\frac{n-1}{2}\right)}\cdots\frac{\displaystyle\Gamma\left(\frac{1}{2}\right)\Gamma(1)}{\displaystyle\Gamma\left(\frac{3}{2}\right)}\\ &=& \displaystyle\frac{2\pi}{n}\cdot\frac{\displaystyle\Gamma^{n-2}\left(\frac{1}{2}\right)}{\displaystyle\Gamma\left(\frac{n}{2}\right)}\\ &=& \displaystyle\frac{\displaystyle\pi^{\frac{n}{2}}}{\displaystyle\Gamma\left(\frac{n}{2}+1\right)}. \end{array}\]

\(\zeta函数\)

  \(Riemann\ zeta\) 函数\(\zeta(s)\), 是一个关于复数\(s\)的函数, 在在复平面上, \(s\)的实数部分\(\sigma=\scr{R}s>1\)时, \(\zeta(s)=\displaystyle\sum_{n=1}^\infty\frac{1}{n^s}\)

特殊点取值

  考虑调和级数\(\zeta(1)=\displaystyle\sum_{i=1}^\infty\frac{1}{n}\), 由\(Cauchy\)判别法, 级数\(S_n\)收敛当且仅当\(\forall \epsilon>0, \exists N, st.\ m,n>N, \left|S_m-S_n\right|<\epsilon\)

\[\left|S_{2n}-S_n\right|=\frac{1}{n+1}+\cdots+\frac{1}{2n}>\frac{1}{2n}+\cdots+\frac{1}{2n}=\frac{1}{2}\]

故调和级数发散

\(Parseval\)等式

\[\frac{1}{\pi}\int_{-\pi}^\pi\left|f(x)\right|^2=\frac{a_0^2}{2}+\sum_{n=1}^\infty(a_n^2+b_n^2)\]

\(\displaystyle\sum_{n=1}^\infty\frac{1}{n^2}=\frac{\pi^2}{6}\)

\(\displaystyle\sum_{n=1}^\infty\frac{1}{n^4}=\frac{\pi^4}{90}\)

  \(zeta\)函数的积分形式 \[\zeta(s)=\sum_{n=1}^\infty\frac{1}{n^{s}}=\frac{1}{\Gamma(s)}\int_0^\infty\frac{x^{s-1}}{e^x-1}\mathrm{d}x,\quad s>1\]

  证: 对\(x\gt0\)有展开式

\[\begin{array}{l} \displaystyle\frac{1}{e^x-1}&=&\displaystyle\frac{e^{-x} }{1-e^{-x} }\\ &=&\displaystyle e^{-x}(1+e^{-x}+e^{-2x}+\cdots)=\sum_{n=1}^\infty e^{-nx} \end{array}\]

于是\(\forall A\gt0\)有

\[\int_0^A\frac{x^{s-1} }{e^x-1}\mathrm{d}x=\int_0^Ax^{s-1}\left(\sum_{n=1}^\infty e^{-nx}\right)\mathrm{d}x\]

对于固定的\(s>1\), 级数\(\sum\limits_{n=1}^\infty x^{s-1}e^{-nx}\)关于\(x\in[0,+\infty)\)是一致收敛的, 所以

\[\begin{array}{l} \displaystyle\int_0^A\frac{x^{s-1} }{e^x-1}\mathrm{d}x&=&\displaystyle\sum_{n=1}^\infty\int_0^A x^{s-1}e^{-nx}\mathrm{d}x\\ &=&\displaystyle\int_0^{nA}\left(\frac{y}{n}\right)^{s-1}\left(\frac{e^{-y} }{n}\right)\mathrm{d}y\\ &=&\displaystyle\sum_{n=1}^\infty\frac{1}{n^s}\int_0^{nA}y^{s-1}e^{-y}\mathrm{d}y \end{array}\]

该级数对于\(A\in[0,\infty)\)是一致收敛的, 于是令\(A\to\infty\), 得

\[\int_0^\infty\frac{x^{s-1} }{e^x-1}\mathrm{d}x=\Gamma(s)\sum_{n=1}^\infty\frac{1}{n^s}\]

  求\(\displaystyle\int_0^1\frac{\ln x}{1-x}\mathrm{d}x\)

   解: 令\(x=e^{-t},\) 则 \[\int_0^1\frac{\ln x}{1-x}\mathrm{d}x=\int_{+\infty}^0\frac{-t}{1-e^{-t}}e^{-t}(-\mathrm{d}t)=-\int_0^{+\infty}\frac{t}{e^t-1}\mathrm{d}t\]

在结论中取\(s=2\), 则

\[\int_0^1\frac{\ln x}{1-x}\mathrm{d}x=-\Gamma(2)\sum_{n=1}^{\infty}\frac{1}{n^2}=-\frac{\pi^2}{6}\]

习题

  1. 计算级数\(\displaystyle\sum_{n=1}^\infty\frac{1}{n\binom{2n}{n}}\)的和。

  解: 利用\({\rm{B} }(p,q)=\frac{\Gamma(p)\Gamma(q)}{\Gamma(p+q)}\)得

\[\begin{array}{l} \displaystyle\sum_{n=1}^\infty\frac{1}{n\binom{2n}{n} }&=&\displaystyle\sum_{n=1}^\infty\frac{1}{n}\frac{n!n!}{(2n)!}\\ &=&\displaystyle\sum_{n=1}^\infty\frac{\Gamma(n)\Gamma(n+1)}{\Gamma(2n+1)}\\ &=&\displaystyle\sum_{n=1}^\infty{\rm{B} }(n+1,n)\\ &=&\displaystyle\sum_{n=1}^\infty\int_0^1t^n(1-t)^{n-1}\mathrm{d}t \end{array}\]

由于当\(0\le t\le 1\)时, \(0\le t(1-t)\le\frac{1}{4}\), 所以

\[0\le t^n(1-t)^{n-1}\le\left(\frac{1}{4}\right)^{n-1}\]

因此级数\(\displaystyle\sum_{n=1}^\infty t^n(1-t)^{n-1}\)在\([0,1]\)一致收敛, 于是有

\[\begin{array}{l} \displaystyle\sum_{n=1}^\infty\frac{1}{n\binom{2n}{n} }&=&\displaystyle\int_0^1\sum_{n=1}^\infty t^n(1-t)^{n-1}\mathrm{d}t\\ &=&\displaystyle\int_0^1t\sum_{n=1}^\infty(t(1-t))^{n-1}\mathrm{d}t\\ &=&\displaystyle\int_0^1\frac{t}{1-t(1-t)}\mathrm{d}t\\ &=&\displaystyle\int_0^1\frac{t}{t^2-t+1}\mathrm{d}t\\ &=&\displaystyle\frac{\pi}{3\sqrt{3}} \end{array}\]