积分不等式的
双阶乘的估计
双阶乘除式的一个估计
(2n)!!(2n−1)!!∼√nπ.
证明：
(2n)!!(2n−1)!!=[(2n)!!]2(2n)!=22n(n!)2(2n)!∼22n[(ne)n√2πn]2(2ne)2n√4πn=22n[(ne)n√2πn]2(2ne)2n√4πn=√nπ.
由此可对如下一个关于sinx的整数次幂在[−π2,π2]上的积分进行数值计算：
∫π20sinnxdx={(2m−1)!!(2m)!!π2∼12√πm,n=2m(2m)!!(2m+1)!!∼√mπ2m+1,n=2m+1,m∈N+.
特殊地，我们有：
0.250037≈I=∫π2−π2sin100xdx=2∫π20sin100xdx=99!!100!!π∼1√50ππ=15√π2=0.25003696.
由上可知，此误差是很小的。
一个级数求解
∞∑n=1lnnn2=−π26(−12lnA+γ+ln2+lnπ).
证明：(Glaisher–Kinkelin constant)lnA=limn→∞[n24−(n22+n2+112)lnn+n∑k=1klnk].
(Riemann zeta函数的导函数)ζ′(s)=−∞∑k=1lnkks.
(Gamma 函数)
Γ(s)=∫∞0xs−1e−xdx.
首先，证明ζ′(−1)=112−lnA.
再证明Γ′(2)=1−γ.
Γ′(2)=∫∞0xe−xlnxdx=∫∞0(e−x+e−xlnx)dx=1−γ.
利用
ζ(s)=2sπs−1sinπs2Γ(1−s)ζ(1−s).
令s=−1，我们有ζ−1=−112.
两边同取对数得
lnζ(s)=sln2+(s−1)lnπ+lnsinπs2+lnΓ(1−s)+lnζ(1−s).
求导，得到ζ′(s)ζ(s)=ln(2π)+π2tanπs2−Γ′(1−s)Γ(1−s)−ζ′(1−s)ζ(1−s).
令s=−1，我们有
ζ′(−1)ζ(−1)=12lnA−1=ln(2π)−1+γ−ζ′(2)ζ(2).
⇒ζ′(2)=π26(ln(2π)−12lnA+γ).
因此我们得到∞∑n=1lnnn2=−π26(−12lnA+γ+ln2+lnπ).
几个欧拉和（Euler sum）的求解
∞∑n=11n2(n∑k=1(−1)k−1k)(n∑k=1(−1)k−1k2);∞∑n=1(−1)k−1n2(n∑k=1(−1)k−1k)(n∑k=11k2);∞∑n=1(∞∑n=11n2)(∞∑n=11n3)n2;∞∑n=11n2Hn,Hn=n∑k=11n.
傅里叶变换求解积分题2
计算
∫∞0cos(mx)x4+x2+1dx.
解：留数理论的一种解答：
注意到
∫∞0cos(mx)x4+x2+1dx=12∫∞−∞cos(mx)x4+x2+1dx.
若令
F(m)=∫∞−∞cos(mx)x4+x2+1dx=∫∞−∞cos(mx)(x2−x+1)(x2+x+1)dx=12∫∞−∞(x+1)cos(mx)x2+x+1dx−12∫∞−∞(x−1)cos(mx)x2−x+1dx.
⇒F′(m)=−12∫∞−∞(x2+x)sin(mx)x2+x+1dx+12∫∞−∞(x2−x)sin(mx)x2−x+1dx=12∫∞−∞sin(mx)x2+x+1dx−12∫∞−∞sin(mx)x2−x+1dx
再令
I=12∫∞−∞sin(mx)x2+x+1dx−12∫∞−∞sin(mx)x2−x+1dx,T=12∫∞−∞ei(mx)x2+x+1dx−12∫∞−∞ei(mx)x2−x+1dx.
则
I=ImT.
即T的虚部为I.因此，为了计算积分I，只需求出积分
∫∞−∞ei(mx)x2+x+1dx−∫∞−∞ei(mx)x2−x+1dx
即可.先求
∫∞−∞ei(mx)x2+x+1dx.
求得辅助函数
P(z)Q(z)ei(mz)=ei(mz)z2+z+1
在上半平面的奇点只有点α=−12+√32i（另一个奇点为β=−12−√32i）.于是我们有
∫∞−∞ei(mx)x2+x+1dx=2πi⋅Res(ei(mz)z2+z+1,−12+√32i).
由于
Res(ei(mz)z2+z+1,−12+√32i)=limz→α(z−α)ei(mz)(z−α)(z−β)=e−√32m−12im√3i.
⇒∫∞−∞ei(mx)x2+x+1dx=2π√3e−√32m−12im=2π√3e−√32m(cosm2−isinm2).
同理亦得
∫∞−∞ei(mx)x2−x+1dx=2π√3e−√32m(cosm2+isinm2).
⇒∫∞−∞ei(mx)x2+x+1dx−∫∞−∞ei(mx)x2−x+1dx=−4πi√3e−√32msinm2
故
F′(m)=I=ImT=Im12(∫∞−∞ei(mx)x2+x+1dx−∫∞−∞ei(mx)x2−x+1dx)=−2π√3e−√32msinm2.
⇒F(m)=−2π√3⋅[e−√32m2(−cosm2−√3sinm2)]=π√3⋅e−√32m(cosm2+√3sinm2).
⇒∫∞0cos(mx)x4+x2+1dx=12∫∞−∞cos(mx)x4+x2+1dx=12F(m)=π2√3⋅e−√32m(cosm2+√3sinm2)=π√3e−√32msin(m2+π6).
另解：由Fourier变换公式，我们有
e−√32m(cosm2+√3sinm2)=2π∫∞0cos(mx)dx∫∞0e−√32u(cosu2+√3sinu2)cos(ux)du=2√3π∫∞0cos(mx)x4+x2+1dx.
立得
∫∞0cos(mx)x4+x2+1dx=π2√3e−√32m(cosm2+√3sinm2)=π√3e−√32msin(m2+π6).
2013年武汉大学数学分析考研真题
数学钟的12个问题
上面是自己在网上看到的一个数学钟，设计者真是别有用心，于是利用周末闲暇时光，对上面的12个问题，一一给出说明.
数字1：Euler公式
eiπ+1=0.
数字2：
对于
Γ(x+1)=∫∞0e−vvxdv.
我们有
Γ′(x+1)=∫∞0e−vvxlnvdv.
另外
Γ′(x+1)Γ(x+1)=−γ+∞∑k=1xk(x+k)⇒Γ′(1)=−γ.
所以我们有
Γ′(1)=∫∞0e−vlnvdv=−γ.
由此得
∫∞0e−x2−e−xxdx=[(e−x2−e−x)lnx]∞0+∫∞02xe−x2lnxdx−∫∞0e−xlnxdx=0+12∫∞0e−tlntdt−∫∞0e−xlnxdx=−12∫∞0e−xlnxdx=γ2.
数字3：
π4=5arctan17+2arctan379.
证明和数字5的大致是一样的，只是计算量偏大了点
(7−i)5(1+i)=(48−14i)2(8+6i)=(2108−1344i)(8+6i)=8(3116+237i)=4(79+3i)2.
由此得
arg(1+i)+arg(7−i)5=arg[4(79+3i)2]=arg(79+3i)2.
若取−π<arg≤π，则由上式可知
π4−5arctan17=2arctan379.
即
π4=5arctan17+2arctan379.
这一结果属于Leonhard Euler.
数字4:
arctanx=∞∑n=0(−1)nx2n+12n+1.
再令x=1即可.
数字5:
应用
π4=4arctan15−arctan1239.
下面是泉叔书上给出的一个较为简便的证明方法
(5−i)4(1+i)=(24−10i)2(1+i)=(476−480i)(1+i)=4(239−i).
于是
arg(1+i)+arg(5−i)4=arg[4(239−i)]=arg(239−i).
若取−π<arg≤π，则由上式可知
π4−4arctan15=−arctan1239.
即
π4=4arctan15−arctan1239.
这一结果归功于 John Machin.
数字6:
∫10lnxx−1dx=∫01ln(1−t)tdt=Li2(1)=π26.
数字7:
tan(2arctan13)=2⋅131−(13)2=34
⇒tan(π4−2arctan13)=1−341+34=17.
数字8:
∞∑k=01(2k+1)2=∞∑k=11k2−∞∑k=11(2k)2=34∞∑k=11k2=π28.
数字9:注意到
m∏k=1tankπ2m+1=√2m+1.
令m=4即可得到.
数字10:
sinπ10=√5−14.
数字11：
首先证明：n−1∑k=1tan2kπ2n=(n−1)(2n−1)3.
由Euler公式
(coskπ2n+isinkπ2n)2n=(−1)k.
由牛顿二项式定理
\sum_{t=0}^{2n}\binom{2n}{t}\left(\cos{\frac{k\pi}{2n}}\right)^t \cdot \left(i\sin{\frac{k\pi}{2n}}\right)^{2n-t}=(-1)^{k}.
只考虑虚部
\sum_{r=0}^{n-1}\binom{2n}{2r+1}\left(\cos{\frac{k\pi}{2n}}\right)^{2r+1} \cdot \left(i\sin{\frac{k\pi}{2n}}\right)^{2n-2r-1}=0.
两边同除 \displaystyle \left(\cos{\frac{k\pi}{2n}}\right)^{2n}
\sum_{r=0}^{n-1}\binom{2n}{2r+1}\left(i\tan{\frac{k\pi}{2n}}\right)^{2n-2r-1}=0.
两边同乘\displaystyle i\tan{\frac{k\pi}{2n}}
\sum_{r=0}^{n-1}\binom{2n}{2r+1}\left(i\tan{\frac{k\pi}{2n}}\right)^{2n-2r}=0.
故 \displaystyle \tan^2{\frac{k\pi}{2n}} 是以下多项式方程的根
\sum_{r=0}^{n-1}\binom{2n}{2r+1}\left(-x\right)^{n-r}=0.
根据 Vieta公式我们有
\frac{\binom{2n}{3}}{\binom{2n}{1}} =\frac{(2n-1)(n-1)}{3}.
进一步，我们有
\sum\limits_{k = 1}^{2n - 1} {{{\csc }^4}\frac{{k\pi }}{{2n}}}= \frac{{\left( {4{n^2} - 1} \right)\left( {4{n^2} + 11} \right)}}{{45}}.
证：
\begin{align*}\sum\limits_{k = 1}^{2n - 1} {{{\csc }^4}\frac{{k\pi }}{{2n}}} &= \sum\limits_{k = 1}^{2n - 1} {\frac{1}{{{{\sin }^4}\frac{{k\pi }}{{2n}}}}} = \sum\limits_{k = 1}^{2n - 1} {\frac{1}{{{{\sin }^4}\frac{{k\pi }}{{2n}}}}} \\&= \sum\limits_{k = 1}^{2n - 1} {\frac{{{{\left( {{{\sin }^2}\frac{{k\pi }}{{2n}} + {{\cos }^2}\frac{{k\pi }}{{2n}}} \right)}^2}}}{{{{\sin }^4}\frac{{k\pi }}{{2n}}}}} \\&= \sum\limits_{k = 1}^{2n - 1} {\left( {1 + 2{{\cot }^2}\frac{{k\pi }}{{2n}} + {{\cot }^4}\frac{{k\pi }}{{2n}}} \right)} \\&= 2n - 1 + 2\sum\limits_{k = 1}^{2n - 1} {{{\cot }^2}\frac{{k\pi }}{{2n}}} + \sum\limits_{k = 1}^{2n - 1} {{{\cot }^4}\frac{{k\pi }}{{2n}}}.\end{align*}
先考虑前者
\begin{align*}\sum\limits_{k = 1}^{2n - 1} {{{\cot }^2}\frac{{k\pi }}{{2n}}} &= \sum\limits_{k = 1}^{n - 1} {{{\cot }^2}\frac{{k\pi }}{{2n}}} + \sum\limits_{k = 1}^{n - 1} {{{\cot }^2}\frac{{\left( {2n - k} \right)\pi }}{{2n}}} + {\cot ^4}\frac{{n\pi }}{{2n}} \\&= 2\sum\limits_{k = 1}^{n - 1} {{{\cot }^2}\frac{{k\pi }}{{2n}}} = 2\sum\limits_{k = 1}^{n - 1} {{{\cot }^2}\frac{{\left( {n - k} \right)\pi }}{{2n}}} \\&= 2\sum\limits_{k = 1}^{n - 1} {{{\tan }^2}\frac{{k\pi }}{{2n}}} = \frac{{2\left( {n - 1} \right)\left( {2n - 1} \right)}}{3}.\end{align*}
再考虑后者
\begin{align*}\sum\limits_{k = 1}^{2n - 1} {{{\cot }^4}\frac{{k\pi }}{{2n}}} &= \sum\limits_{k = 1}^{n - 1} {{{\cot }^4}\frac{{k\pi }}{{2n}}} + \sum\limits_{k = 1}^{n - 1} {{{\cot }^4}\frac{{\left( {2n - k} \right)\pi }}{{2n}}} + {\cot ^4}\frac{{n\pi }}{{2n}} \\&= 2\sum\limits_{k = 1}^{n - 1} {{{\cot }^4}\frac{{k\pi }}{{2n}}} = 2\sum\limits_{k = 1}^{n - 1} {{{\cot }^4}\frac{{\left( {n - k} \right)\pi }}{{2n}}} \\&= 2\sum\limits_{k = 1}^{n - 1} {{{\tan }^4}\frac{{k\pi }}{{2n}}} = 2\left[ {{{\left( {\sum\limits_{k = 1}^{n - 1} {{{\tan }^2}\frac{{k\pi }}{{2n}}} } \right)}^2} - 2\sum\limits_{1 \le i < j \le n - 1} {{{\tan }^2}\frac{{i\pi }}{{2n}}{{\tan }^2}\frac{{j\pi }}{{2n}}} } \right]\\&= 2\left[ {\frac{{{{\left( {n - 1} \right)}^2}{{\left( {2n - 1} \right)}^2}}}{9} - 2\frac{{\left( \begin{array}{c}2n\\5\end{array} \right)}}{{\left( \begin{array}{c}2n\\1\end{array} \right)}}} \right] \\&= \frac{{2{{\left( {n - 1} \right)}^2}{{\left( {2n - 1} \right)}^2}}}{9} - \frac{{2\left( {n - 2} \right)\left( {n - 1} \right)\left( {2n - 3} \right)\left( {2n - 1} \right)}}{{15}} \\&= \frac{{\left( {n - 1} \right)\left( {2n - 1} \right)\left( {24{n^2} + 36n - 78} \right)}}{{135}} = \frac{{2\left( {n - 1} \right)\left( {2n - 1} \right)\left( {4{n^2} + 6n - 13} \right)}}{{45}}.\end{align*}
由此
\begin{align*}\Rightarrow \sum\limits_{k = 1}^{2n - 1} {{{\csc }^4}\frac{{k\pi }}{{2n}}} &= 2n - 1 + \frac{{4\left( {n - 1} \right)\left( {2n - 1} \right)}}{3} + \frac{{2\left( {n - 1} \right)\left( {2n - 1} \right)\left( {4{n^2} + 6n - 13} \right)}}{{45}} \\&= 2n - 1 + \frac{{\left( {n - 1} \right)\left( {2n - 1} \right)\left( {8{n^2} + 12n + 34} \right)}}{{45}} \\&= \frac{{\left( {4{n^2} - 1} \right)\left( {4{n^2} + 11} \right)}}{{45}}.\end{align*}
特别地有
\Rightarrow \sum\limits_{k = 1}^{999} {{{\csc }^4}\frac{{k\pi }}{{1000}}} = 1000009999989 = {10^{12}} + {10^7} - 11.
数字12：
\begin{align*}\int_0^\infty {\frac{{xdx}}{{\sinh \left( {\sqrt 3 x} \right)}}} &= \int_0^\infty {\frac{{xdx}}{{\frac{{{e^{\sqrt 3 x}} - {e^{ - \sqrt 3 x}}}}{2}}}} = \int_0^\infty {\frac{{2xdx}}{{{e^{\sqrt 3 x}} - {e^{ - \sqrt 3 x}}}}} \\&\underline{\underline {t = {e^{ - \sqrt 3 x}}}} \frac{2}{3}\int_0^1 {\frac{{\ln t}}{{{t^2} - 1}}dt} = \frac{1}{3}\int_0^1 {\frac{{\ln t}}{{t - 1}}dt} - \frac{1}{3}\int_0^1 {\frac{{\ln t}}{{t + 1}}dt} \\&= \frac{1}{3}\int_0^1 {\frac{{\ln t}}{{t - 1}}dt} - \frac{1}{3}\left[ {\ln t\ln \left( {t + 1} \right)} \right]_0^1 + \frac{1}{3}\int_0^1 {\frac{{\ln \left( {t + 1} \right)}}{t}dt} \\&= \frac{1}{3}\int_1^0 {\frac{{\ln \left( {1 - t} \right)}}{t}dt} - \frac{1}{3}\int_{ - 1}^0 {\frac{{\ln \left( {1 - t} \right)}}{t}dt} = \frac{1}{3}\mathrm{Li}_2\left( 1 \right) - \frac{1}{3}\mathrm{Li}_2\left( { - 1} \right) \\&= \frac{1}{3}\left( {\frac{{{\pi ^2}}}{6} + \frac{{{\pi ^2}}}{{12}}} \right) = \frac{{{\pi ^2}}}{{12}}\end{align*}
若干常见的傅里叶级数
关于Fourier级数，我们有以下几种常用表达式
\begin{align*}&x = \frac{\pi }{2} - \frac{4}{\pi }\sum\limits_{n = 1}^\infty {\frac{{\cos \left[ {\left( {2n - 1} \right)x} \right]}}{{{{\left( {2n - 1} \right)}^2}}}} = 2\sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^{n - 1}}\frac{{\sin \left( {nx} \right)}}{n}} \qquad 0 \le x < \pi \\&{x^2}=\frac{{{\pi ^2}}}{3}+4\sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^n}\frac{{\cos \left( {nx} \right)}}{{{n^2}}}} \qquad - \pi \le x \le \pi \\&\cos \left( {ax} \right) = \frac{{\sin \left( {a\pi } \right)}}{\pi }\left( {\frac{1}{a} + \sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^n}\frac{{2a}}{{{a^2} - {n^2}}}\cos \left( {nx} \right)} } \right)\qquad - \pi \le x \le \pi \\&\cot x = \frac{1}{x} + \sum\limits_{n = 1}^\infty {\frac{{2x}}{{{x^2} - {n^2}{\pi ^2}}}}\qquad x \ne k\pi ,k = 0, \pm 1, \pm 2, \cdots \\&\frac{1}{{\sin x}} = \frac{1}{x} + \sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^n}\frac{{2x}}{{{x^2} - {n^2}{\pi ^2}}}} \qquad x \ne k\pi ,k = 0, \pm 1, \pm 2, \cdots \\&\left| {\cos x} \right| = \frac{2}{\pi } + \frac{4}{\pi }\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n + 1}}}}{{4{n^2} - 1}}\cos \left[ {\left( {2n} \right)x} \right]} \qquad x \in \left( { - \infty , + \infty } \right)\\&\left| {\sin x} \right| = \frac{2}{\pi } - \frac{4}{\pi }\sum\limits_{n = 1}^\infty {\frac{1}{{4{n^2} - 1}}\cos \left[ {\left( {2n} \right)x} \right]} \qquad x \in \left( { - \infty , + \infty } \right)\\&\sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^{n - 1}}\frac{{\cos \left( {nx} \right)}}{n}} = \ln \left( {2\cos \frac{x}{2}} \right)\qquad - \pi < x < \pi \\&\sum\limits_{n = 1}^\infty {\frac{{\cos \left( {nx} \right)}}{n}} = - \ln \left( {2\sin \frac{x}{2}} \right)\qquad 0 < x < 2\pi \\&{e^{ax}} = \frac{{{e^{2a\pi }} - 1}}{\pi }\left( {\frac{1}{{2a}} + \sum\limits_{k = 1}^\infty {\frac{{a\cos kx - k\sin kx}}{{{k^2} + {a^2}}}} } \right)\qquad x \in \left( {0,2\pi } \right),a \ne 0\end{align*}
根据这些高端级数我们有以下结论：
结论一：\sum\limits_{n = 1}^\infty {\frac{1}{{{n^2} + 1}}} = \frac{{{e^{2\pi }} + 1}}{2} \cdot \frac{\pi }{{{e^{2\pi }} - 1}} - \frac{1}{2} = \frac{\pi }{2}\frac{{{e^{2\pi }} + 1}}{{{e^{2\pi }} - 1}} - \frac{1}{2}.
解：由Fourier级数
{e^{ax}} = \frac{{{e^{2a\pi }} - 1}}{\pi }\left( {\frac{1}{{2a}} + \sum\limits_{n = 1}^\infty {\frac{{a{\mathop{\rm cosn}\nolimits} x - n{\mathop{\rm sinn}\nolimits} x}}{{{n^2} + {a^2}}}} } \right)\quad \;\;\;{\mkern 1mu} x \in \left( {0,2\pi } \right),a \ne 0
将\displaystyle f(x)=e^{ax}(0<x<2\pi)延拓为\displaystyle(-\infty,+\infty)上以\displaystyle 2\pi为周期的函数\displaystyle \widetilde{f}(x)，则必须有
\widetilde{f}(0)=\widetilde{f}(2\pi)=\frac{f(2\pi-0)+f(0+0)}{2}=\frac{e^{2\pi a}+e^{0\cdot a}}{2}=\frac{e^{2\pi a}+1}{2}.
由此得到
\widetilde{f}(x)= \frac{{{e^{2a\pi }} - 1}}{\pi }\left( {\frac{1}{{2a}} + \sum\limits_{n = 1}^\infty{\frac{{a{\mathop{\rm cosn}\nolimits} x - n{\mathop{\rm sinn}\nolimits} x}}{{{n^2} + {a^2}}}} } \right)\qquad {\mkern 1mu} - \infty < x < + \infty
令\displaystyle a=1,x=0，我们立得
\frac{{{e^{2\pi }} + 1}}{2} = \frac{{{e^{2\pi }} - 1}}{\pi }\left( {\frac{1}{2} + \sum\limits_{n = 1}^\infty {\frac{1}{{{n^2} + 1}}} } \right).
即
\sum\limits_{n = 1}^\infty {\frac{1}{{{n^2} + 1}}} = \frac{{{e^{2\pi }} + 1}}{2} \cdot \frac{\pi }{{{e^{2\pi }} - 1}} - \frac{1}{2} = \frac{\pi }{2}\frac{{{e^{2\pi }} + 1}}{{{e^{2\pi }} - 1}} - \frac{1}{2}.
一般地，我们有
\sum\limits_{n = 1}^\infty {\frac{1}{{{n^2} + {a^2}}}} = \frac{\pi }{{2a}}\frac{{{e^{2\pi }} + 1}}{{{e^{2a\pi }} - 1}} - \frac{1}{{2{a^2}}}.
sinn/n和cosn/n类型的若干问题
题一：求极限
\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{k = 1}^n {\left| {\cos k} \right|} .
解：取\displaystyle f\left( x \right) = \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{k = 1}^n {\left| {\cos \left( {k + x} \right)} \right|} ，则
f\left( {x + m} \right) = \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{k = 1}^n {\left| {\cos \left( {k + x + m} \right)} \right|} \left( {m \in {N^*}} \right).
\begin{align*}f\left( x \right) &= \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{k = 1}^n {\left| {\cos \left( {k + x} \right)} \right|} \\&= \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{k = 1}^m {\left| {\cos \left( {k + x} \right)} \right|} + \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{k = m + 1}^n {\left| {\cos \left( {k + x} \right)} \right|} \\&= \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{k = m + 1}^n {\left| {\cos \left( {k + x} \right)} \right|} \\&= \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{s = 1}^n {\left| {\cos \left( {s + m + x} \right)} \right|} = f\left( {x + m} \right).\end{align*}
又
f\left( {x + 2\pi } \right) = \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{k = 1}^n {\left| {\cos \left( {k + x + 2\pi } \right)} \right|} = f\left( x \right).
所以
\begin{align}\label{MA1}f\left( x \right) = f\left( {x + m} \right) = f\left( {x + 2\pi } \right).\qquad(1)\end{align}
对于\displaystyle f\left( x \right) = \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{k = 1}^n {\left| {\cos \left( {k + x} \right)} \right|} ，其中\displaystyle \frac{1}{n}单调且\displaystyle\lim\limits_{n\rightarrow \infty}{\frac{1}{n}}=0,|\cos(k+x)|在x\in[0,1]上一致有界，由狄利克雷判别法可知：函数\displaystyle f(x)一致收敛.同时（1）式说明周期可为任意整数，又可为\displaystyle 2\pi，那么必为常数函数，\displaystyle f(x)\equiv C.所以
\begin{align*}f\left( 0 \right) &= \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{k = 1}^n {\left| {\cos k} \right|} = \int_0^1 {f\left( x \right)dx} \\&= \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\int_0^1 {\sum\limits_{k = 1}^n {\left| {\cos \left( {k + x} \right)} \right|} dx} \\&= \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{k = 1}^n {\int_0^1 {\left| {\cos \left( {k + x} \right)} \right|dx} } \\&= \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{k = 1}^n {\int_k^{k + 1} {\left| {\cos t} \right|dt} } \\&= \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\int_1^{n + 1} {\left| {\cos t} \right|dt} \\&= 2\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\left[ {\frac{n}{\pi }} \right] \\&= \frac{2}{\pi }\mathop {\lim }\limits_{n \to \infty } \frac{\pi }{n}\left[ {\frac{n}{\pi }} \right] = \frac{2}{\pi }\end{align*}
注：在区间[1,n+1]中，含有函数\omega(t)=|\cos t|的周期个数是\left[ {\frac{n}{\pi }} \right]，每个周期的积分值都是2，所以得到上述结果.
\mathop {\lim }\limits_{n \to \infty } \frac{\pi }{n}\left[ {\frac{n}{\pi }} \right] = \mathop {\lim }\limits_{n \to \infty } \frac{{\frac{n}{\pi } - \left( {\frac{n}{\pi }} \right)}}{{\frac{n}{\pi }}} = \mathop {\lim }\limits_{n \to \infty } \left( {1 - \frac{{\left( {\frac{n}{\pi }} \right)}}{{\frac{n}{\pi }}}} \right) = 1
其中\left( {\frac{n}{\pi }} \right)\in(0,1).
另解一：构建|\cos k|,k\in\mathbb{N}^*在|\cos x|上的随机分布，所以|\cos k|的均值就为
\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{k = 1}^n {\left| {\cos k} \right|} = E\left( {\left| {\cos k} \right|} \right) = \frac{{\int_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\left| {\cos x} \right|dx} }}{\pi } = \frac{2}{\pi }.
另解二：令
f(x)=\int_0^x{|\cos t|dt}.
当x\in[n\pi,(n+1)\pi)时，
\int_0^{n\pi } {\left| {\cos t} \right|dt} \le f\left( x \right) < \int_0^{\left( {n + 1} \right)\pi } {\left| {\cos t} \right|dt} .
\Rightarrow 2n\leq f(x)<2(n+1).
由此得
\frac{{2n}}{{\left( {n + 1} \right)\pi }} < \frac{{f\left( x \right)}}{x} < \frac{{2\left( {n + 1} \right)}}{{n\pi }}.
记S_n=\sum\limits_{k = 1}^n {\left| {\cos k} \right|}，则有
\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{k = 1}^n {\left| {\cos k} \right|} = \mathop {\lim }\limits_{n \to \infty } \frac{{{S_n}}}{n} = \mathop {\lim }\limits_{n \to \infty } \frac{{f\left( x \right)}}{x} = \frac{2}{\pi }.
题二：证明
\sum\limits_{n = 1}^\infty {\frac{{\sin n}}{n}} = \frac{{\pi - 1}}{2}.
证明：由Fourier级数
\frac{{\pi-x}}{2} = \sum\limits_{n = 1}^\infty {\frac{{\sin nx}}{n}} \qquad 0 < x < 2\pi
令x=1，我们立得
\sum\limits_{n = 1}^\infty {\frac{{\sin n}}{n}} = \frac{{\pi - 1}}{2}.
题三：证明
\sum\limits_{n = 1}^\infty {\frac{{\cos n}}{n}} = - \frac{1}{2}\ln (2 - 2\cos 1).
证明：由Fourier级数
\sum_{k=1}^\infty \frac{\cos(kx)}{k}=-\frac{1}{2}\ln(2-2\cos x)\qquad x\in\mathbb{R}.
令x=1，我们立得
\sum\limits_{n = 1}^\infty {\frac{{\cos n}}{n}} = - \frac{1}{2}\ln (2 - 2\cos 1).
傅里叶变换求积分函数
来自刚哥的虐心的积分题：
\begin{align*}&\int_0^\infty {\frac{{\cos tx}}{{1 + {t^2}}}} dt;\\&\int_0^\infty {\frac{{\left( {{u^2} + 2} \right)\cos ux}}{{{u^4} + 4}}du} .\end{align*}
解：事实上，由Fourier变换公式
\begin{align*}{e^{ - x}} &= \frac{2}{\pi }\int_0^\infty {cos\left( {\lambda x} \right)d\lambda \int_0^\infty {{e^{ - u}}} cos\left( {\lambda u} \right)} du = \frac{2}{\pi }\int_0^\infty {\frac{{\cos \lambda x}}{{{\lambda ^2} + 1}}d\lambda } ;\\{e^{ - x}}\cos x &= \frac{2}{\pi }\int_0^\infty {\cos \left( {xu} \right)du} \int_0^\infty {{e^{ - t}}\cos t\cos \left( {ut} \right)dt} = \frac{2}{\pi }\int_0^\infty {\frac{{\left( {{u^2} + 2} \right)\cos ux}}{{{u^4} + 4}}du}.\end{align*}
我们得到
\begin{align*}\int_0^\infty {\frac{{\cos \lambda x}}{{{\lambda ^2} + 1}}d\lambda } &= \frac{\pi }{2}{e^{ - x}};\\\int_0^\infty {\frac{{\left( {{u^2} + 2} \right)\cos ux}}{{{u^4} + 4}}du} &= \frac{\pi }{2}{e^{ - x}}\cos x.\end{align*}
一般地，找到以下结论
\begin{align*}\int_0^\infty {\frac{{\cos xu}}{{{\beta ^2} + {u^2}}}du} &= \frac{\pi }{{2\beta }}{e^{ - \beta x}} \qquad x > 0,\beta > 0;\\\int_0^\infty {\frac{{u\sin xu}}{{{\beta ^2} + {u^2}}}du} &= \frac{\pi }{2}{e^{ - \beta x}}\qquad x > 0,\beta > 0;\\\int_0^\infty {\frac{{{x^{\mu - 1}}\sin \left( {ax} \right)}}{{{x^2} + 1}}dx} &= - {a^{2 - \mu }}\mathbf{\Gamma} \left( {\mu - 2} \right){}_1{F_2}\left( {1;\frac{{3 - \mu }}{2},\frac{{4 - \mu }}{2};\frac{{{a^2}}}{4}} \right)\mathrm{sign}\left( a \right)\sin \frac{{\mu \pi }}{2}\\&+ \frac{\pi }{2}\sec \frac{{\mu \pi }}{2}\sinh \left( a \right)\qquad{\mathop{\rm Im}\nolimits} a = 0, - 1 < \mathrm{Re}\mu < 3;\\\int_0^\infty {\frac{{{x^{\mu - 1}}\cos \left( {ax} \right)}}{{{x^2} + 1}}dx} &= \frac{\pi }{{2\sin \frac{{\mu \pi }}{2}}}\cosh a + \frac{1}{2}\cos \frac{{\mu \pi }}{2}\mathbf{\Gamma} \left( \mu \right)\\&\left[ {{e^{ - a + i\pi \left( {1 - \mu } \right)}}\gamma \left( {1 - \mu , - a} \right) - {e^a}\gamma \left( {1 - \mu ,a} \right)} \right]\qquad a > 0,0 < \mathrm{Re}\mu < 3;\\\int_0^\infty {\frac{{{x^{2\mu + 1}}\sin \left( {ax} \right)}}{{{x^2} + {b^2}}}dx} &= - \frac{\pi }{{2\cos \left( {\mu \pi } \right)}}{b^{2\mu }}\mathrm{sinh}\left( {ab} \right) + \frac{{\sin \left( {\mu \pi } \right)}}{{2{a^{2\mu }}}}\mathbf{\Gamma} \left( {2\mu } \right)\\&\left[ {{}_1{F_1}\left( {1;1 - 2\mu ;ab} \right) + {}_1{F_1}\left( {1;1 - 2\mu ; - ab} \right)} \right] \qquad a > 0, - \frac{3}{2} < \mathrm{Re}\mu < \frac{1}{2};\\\int_0^\infty {\frac{{{x^{2\mu + 1}}\cos \left( {ax} \right)}}{{{x^2} + {b^2}}}dx} &= - \frac{\pi }{{2\sin \left[ {\left( {\mu + \frac{1}{2}} \right)\pi } \right]}}{b^{2\mu + 1}}\cosh \left( {ab} \right) + \frac{{\cos \left[ {\left( {\mu + \frac{1}{2}} \right)\pi } \right]}}{{2{a^{2\mu + 1}}}}\mathbf{\Gamma} \left( {2\mu + 1} \right)\\&\left[ {{}_1{F_1}\left( {1; - 2\mu ;ab} \right) + {}_1{F_1}\left( {1; - 2\mu ; - ab} \right)} \right] \qquad a > 0, - 1 < \mathrm{Re}\mu < \frac{1}{2}.\end{align*}
