无穷积分 ∫sinx/xdx 的几种巧妙解法

1. Fourier 正弦展开

∫0∞sin⁡xxdx=lim⁡m→∞∫0mπsin⁡xxdx\int_{0}^{\infin}\frac{\sin x}{x}dx=\lim_{m\to\infin}\int_0^{m\pi}\frac{\sin x}{x}dx∫0∞xsinxdx=m→∞lim∫0mπxsinxdx令h=π/k，将区间[0,mπ]分割成km个长度为h的小区间，由黎曼积分的定义\text{令}\,h=\pi/k\text{，将区间}\,[0,m\pi]\,\text{分割成}\,km\,\text{个长度为}\,h\,\text{的小区间，由黎曼积分的定义}令h=π/k，将区间[0,mπ]分割成km个长度为h的小区间，由黎曼积分的定义
∫0mπsin⁡xxdx=lim⁡h→0+∑n=1kmsin⁡nhnhh=lim⁡h→0+∑n=1kmsin⁡nhn\int_0^{m\pi}\frac{\sin x}{x}dx=\lim_{h\to0^+}\sum_{n=1}^{km}\frac{\sin nh}{nh}h=\lim_{h\to0^+}\sum_{n=1}^{km}\frac{\sin nh}{n}∫0mπxsinxdx=h→0+limn=1∑kmnhsinnhh=h→0+limn=1∑kmnsinnh由Fourier正弦展开\text{由}\,\textrm{Fourier}\,\text{正弦展开}由Fourier正弦展开
π−h2=∑n=1∞sin⁡nhn,(0<h<π)\frac{\pi-h}{2}=\sum_{n=1}^{\infin}\frac{\sin nh}{n},(0<h<\pi)2π−h=n=1∑∞nsinnh,(0<h<π)所以\text{所以}所以
∫0∞sin⁡xxdx=lim⁡m→∞∫0mπsin⁡xxdx=lim⁡m→∞lim⁡h→0+∑n=1kmsin⁡nhn=lim⁡h→0+lim⁡m→∞∑n=1kmsin⁡nhn=lim⁡h→0+π−h2=π2\begin{aligned}\int_{0}^{\infin}\frac{\sin x}{x}dx&=\lim_{m\to\infin}\int_0^{m\pi}\frac{\sin x}{x}dx\\&=\lim_{m\to\infin}\lim_{h\to0^+}\sum_{n=1}^{km}\frac{\sin nh}{n}\\&=\lim_{h\to0^+}\lim_{m\to\infin}\sum_{n=1}^{km}\frac{\sin nh}{n}\\&=\lim_{h\to0^+}\frac{\pi-h}{2}\\&=\frac{\pi}{2}\end{aligned}∫0∞xsinxdx=m→∞lim∫0mπxsinxdx=m→∞limh→0+limn=1∑kmnsinnh=h→0+limm→∞limn=1∑kmnsinnh=h→0+lim2π−h=2π

2. 交换积分次序

∫0∞sin⁡xxdx=∫0∞sin⁡x1xdx=∫0∞sin⁡x(∫0∞e−xydy)dx(交换积分次序)=∫0∞(∫0∞e−yxsin⁡xdx)dy∫0∞e−yxsin⁡xdx=∫0∞e−yx12i(eix−e−ix)dx=12i∫0∞(e−(y−i)x−e−(y+i)x)dx=12i(−1y−ie−(y−i)x+1y+ie−(y+i)x)∣0∞=12i(1y−i−1y+i)=1y2+1\begin{aligned}\int_{0}^{\infin}\frac{\sin x}{x}\,dx&=\int_{0}^{\infin}\sin x\,\frac{1}{x}\,dx\\&=\int_{0}^{\infin}\sin x\big(\int_{0}^{\infin}e^{-xy}dy\big)dx \quad(\text{交换积分次序})\\&=\int_{0}^{\infin}\big(\int_{0}^{\infin}e^{-yx}\sin x\,dx\big)dy \\ \\\int_{0}^{\infin}e^{-yx}\sin x\,dx&=\int_{0}^{\infin}e^{-yx}\frac{1}{2i}(e^{ix}-e^{-ix})dx\\&=\frac{1}{2i}\int_{0}^{\infin}(e^{-(y-i)x}-e^{-(y+i)x})dx\\&=\frac{1}{2i}\big(-\frac{1}{y-i}e^{-(y-i)x}+\frac{1}{y+i}e^{-(y+i)x}\big)\Big|_0^\infin\\&=\frac{1}{2i}\big(\frac{1}{y-i}-\frac{1}{y+i}\big)\\&=\frac{1}{y^2+1}\end{aligned}∫0∞xsinxdx∫0∞e−yxsinxdx=∫0∞sinxx1dx=∫0∞sinx(∫0∞e−xydy)dx(交换积分次序)=∫0∞(∫0∞e−yxsinxdx)dy=∫0∞e−yx2i1(eix−e−ix)dx=2i1∫0∞(e−(y−i)x−e−(y+i)x)dx=2i1(−y−i1e−(y−i)x+y+i1e−(y+i)x)∣∣∣0∞=2i1(y−i1−y+i1)=y2+11 ∴原式=∫0∞1y2+1dy=arctan⁡y∣0∞=π2.\begin{aligned} \therefore\,\, \text{原式}=\int_{0}^{\infin}\frac{1}{y^2+1}dy=\arctan y\,\Big|_0^{\infin} =\frac{\pi}{2}\end{aligned}.∴原式=∫0∞y2+11dy=arctany∣∣∣0∞=2π.

3. 构造含参变量函数

记I=∫0∞sin⁡ttdt，构造函数f(x)=∫0∞e−xtsin⁡ttdt，则f(0)=I.\begin{aligned}\text{记}\,I=\int_{0}^{\infin}\frac{\sin t}{t}\,dt\text{，构造函数}\,f(x)=\int_{0}^{\infin}e^{-xt}\frac{\sin t}{t}\,dt\text{，则}\,f(0)=I\end{aligned}.记I=∫0∞tsintdt，构造函数f(x)=∫0∞e−xttsintdt，则f(0)=I.
0≤∣f(x)∣≤∫0∞e−xt∣sin⁡tt∣dt≤∫0∞e−xtdt=1x0\leq\big|f(x)\big|\leq\int_{0}^{\infin}e^{-xt}\bigg|\frac{\sin t}{t}\bigg|dt \leq \int_{0}^{\infin}e^{-xt}dt=\frac{1}{x}0≤∣∣f(x)∣∣≤∫0∞e−xt∣∣∣∣tsint∣∣∣∣dt≤∫0∞e−xtdt=x1

两边取极限，x→∞,f(∞)=0.\text{两边取极限，}x\to\infin,\,f(\infin)=0.两边取极限，x→∞,f(∞)=0.

f′(x)=∫0∞∂∂x(e−xtsin⁡tt)dt=−∫0∞e−xtsin⁡tdt=−1x2+1(由上一方法中的结果)\begin{aligned}f'(x)&=\int_{0}^{\infin}\frac{\partial }{\partial x}\Big( e^{-xt}\frac{\sin t}{t}\Big)dt\\&=-\int_{0}^{\infin}e^{-xt}\sin t \,dt\\&=-\frac{1}{x^2+1} \quad (\text{由上一方法中的结果})\end{aligned}f′(x)=∫0∞∂x∂(e−xttsint)dt=−∫0∞e−xtsintdt=−x2+11(由上一方法中的结果)

由牛顿-莱布尼兹公式\text{由牛顿-莱布尼兹公式}由牛顿-莱布尼兹公式
0−I=f(∞)−f(0)=∫0∞f′(x)dx=−∫0∞1x2+1dx=−arctan⁡x∣0∞=−π20-I=f(\infin)-f(0)=\int_0^{\infin}f'(x)dx=-\int_0^{\infin}\frac{1}{x^2+1} dx =-\arctan x \,\bigg|_0^{\infin}=-\frac{\pi}{2}0−I=f(∞)−f(0)=∫0∞f′(x)dx=−∫0∞x2+11dx=−arctanx∣∣∣∣0∞=−2π所以I=π2.\begin{aligned}\text{所以}\,I=\frac{\pi}{2}\end{aligned}.所以I=2π.

4. Laplace 变换

令f(t)=∫0∞sin⁡txxdx,t>0，对f(t)作拉普拉斯变换，令\begin{aligned}\text{令}\,f(t)=\int_{0}^{\infin}\frac{\sin tx}{x}\,dx,\, t>0\text{，对}\,f(t)\, \text{作拉普拉斯变换，令}\end{aligned}令f(t)=∫0∞xsintxdx,t>0，对f(t)作拉普拉斯变换，令
F(s)=L[f(t)]=L[∫0∞sin⁡txxdx]t=∫0∞1xL[sin⁡tx]tdx=∫0∞1s2+x2dx=1s∫0∞11+(xs)2d(xs)(letu=xs)=1sarctan⁡u∣0∞=1sπ2\begin{aligned}F(s)=\mathscr{L}[f(t)]&=\mathscr{L}\Big[\int_{0}^{\infin}\frac{\sin tx}{x}dx\Big]_t\\&=\int_{0}^{\infin} \frac{1}{x}\mathscr{L}\big[\sin tx \big]_tdx\\&=\int_{0}^{\infin}\frac{1}{s^2+x^2}\,dx\\&=\frac{1}{s}\int_{0}^{\infin}\frac{1}{1+(\displaystyle \frac{x}{s})^2}\,d(\frac{x}{s}) \quad (\textit{let u = $\displaystyle \frac{x}{s}$})\\&=\frac{1}{s}\arctan u\,\Big|_0^{\infin}\\&=\frac{1}{s}\frac{\pi}{2}\end{aligned}F(s)=L[f(t)]=L[∫0∞xsintxdx]t=∫0∞x1L[sintx]tdx=∫0∞s2+x21dx=s1∫0∞1+(sx)21d(sx)(let u = sx)=s1arctanu∣∣∣0∞=s12π 则f(t)=L−1[F(s)]=π2L−1[1s]=π2.\begin{aligned}\text{则}\,\, f(t)=\mathscr{L}^{-1}\big[F(s)\big]=\frac{\pi}{2}\mathscr{L}^{-1}\big[\frac{1}{s}\big]=\frac{\pi}{2}\end{aligned}.则f(t)=L−1[F(s)]=2πL−1[s1]=2π.

令人惊奇的是，f(t)的值竟与t无关，于是我们得到一个更为普遍的结论\begin{aligned}\text{令人惊奇的是，}f(t)\, \text{的值竟与} \,t\, \text{无关，于是我们得到一个更为普遍的结论}\end{aligned}令人惊奇的是，f(t)的值竟与t无关，于是我们得到一个更为普遍的结论 ∫0∞sin⁡txxdx=π2,t>0\int_0^{\infin}\frac{\sin tx}{x}\,dx=\frac{\pi}{2},\,t>0∫0∞xsintxdx=2π,t>0如果你也觉得不可思议的话，不妨看看我的解释\text{如果你也觉得不可思议的话，不妨看看我的解释}如果你也觉得不可思议的话，不妨看看我的解释
f′(t)=∫0∞∂∂t(sin⁡txx)dx=∫0∞cos⁡txdx=lim⁡n→∞∫0nπ/tcos⁡txdx(letu=tx)=lim⁡n→∞1t∫0nπcos⁡udu=1tlim⁡n→∞sin⁡u∣0nπ=0\begin{aligned}f'(t)&=\int_0^{\infin}\frac{\partial }{\partial t}\Big(\frac{\sin tx}{x}\Big)dx \\&=\int_0^{\infin}\cos tx\,dx\\&=\lim_{n\to\infin}\int_0^{n\pi/t}\cos tx\,dx \quad (let\,\, u=tx)\\&=\lim_{n\to\infin}\frac{1}{t}\int_0^{n\pi}\cos u\,du\\&=\frac{1}{t}\lim_{n\to\infin} \sin u\,\Big|_0^{n\pi}\\&=0\end{aligned}f′(t)=∫0∞∂t∂(xsintx)dx=∫0∞costxdx=n→∞lim∫0nπ/tcostxdx(letu=tx)=n→∞limt1∫0nπcosudu=t1n→∞limsinu∣∣∣0nπ=0所以f(t)=C,与t无关.\text{所以}\,f(t)=C,\,\text{与}\,t\,\text{无关}.所以f(t)=C,与t无关.

5. Fourier 变换

令\text{令}令 f(t)={1,∣t∣<10,∣t∣≥1f(t)=\begin{cases}1,\,\,|t|<1 \\ 0,\,\,|t|\geq1\end{cases}f(t)={1,∣t∣<10,∣t∣≥1，对f(t)作傅里叶变换，令\text{，对} \,f(t)\, \text{作傅里叶变换，令}，对f(t)作傅里叶变换，令

F(μ)=F[f(t)]=∫−∞∞f(t)e−iμtdt=∫−11e−iμtdt=−1iμe−iμt∣−11=2μ12i(eiμ−e−iμ)=2sin⁡μμ\begin{aligned}F(\mu)&=\mathscr{F}[f(t)]\\&=\int_{-\infin}^{\infin} f(t)e^{-i\mu t}dt\\&=\int_{-1}^{1} e^{-i\mu t}dt\\&=-\frac{1}{i\mu}e^{-i\mu t}\Big|_{-1}^{1}\\&=\frac{2}{\mu}\frac{1}{2i}(e^{i\mu}-e^{-i\mu})\\&=2\,\frac{\sin \mu}{\mu}\end{aligned}F(μ)=F[f(t)]=∫−∞∞f(t)e−iμtdt=∫−11e−iμtdt=−iμ1e−iμt∣∣∣−11=μ22i1(eiμ−e−iμ)=2μsinμ 则f(t)=F−1[F(μ)]=12π∫−∞∞2sin⁡μμeiμtdμ.\begin{aligned}\text{则}\,f(t)=\mathscr{F}^{-1}\big[F(\mu)\big]=\frac{1}{2\pi}\int_{-\infin}^{\infin}2\,\frac{\sin \mu}{\mu}e^{i\mu t}d\mu.\end{aligned}则f(t)=F−1[F(μ)]=2π1∫−∞∞2μsinμeiμtdμ.
取t=0，则\text{取}\,t=0\text{，则}取t=0，则 1=f(0)=1π∫−∞∞sin⁡μμdμ=2π∫0∞sin⁡μμdμ1=f(0)=\frac{1}{\pi}\int_{-\infin}^{\infin}\frac{\sin \mu}{\mu}\,d\mu=\frac{2}{\pi}\int_{0}^{\infin}\frac{\sin \mu}{\mu}\,d\mu1=f(0)=π1∫−∞∞μsinμdμ=π2∫0∞μsinμdμ 所以∫0∞sin⁡μμdμ=π2.妙哉！.\begin{aligned}\text{所以}\,\int_{0}^{\infin}\frac{\sin \mu}{\mu}\,d\mu=\frac{\pi}{2}.\quad\text{妙哉！}\end{aligned}.所以∫0∞μsinμdμ=2π.妙哉！.

6. 狄拉克函数

首先来介绍一下狄拉克函数(就是 Dirac创造的函数)，也称脉冲函数(比较形象)：\text{首先来介绍一下狄拉克函数(就是 \text{Dirac}\,创造的函数)，也称脉冲函数(比较形象)：}首先来介绍一下狄拉克函数(就是 Dirac创造的函数)，也称脉冲函数(比较形象)：

δ(t)={∞,t=00,t≠0，且满足∫−∞∞δ(t)dt=1.\delta(t)=\begin{cases}\infin,&t=0 \\0,&t\neq 0\end{cases}\,\text{，且满足}\,\displaystyle\int_{-\infin}^{\infin}\delta(t)dt=1.δ(t)={∞,0,t=0t=0，且满足∫−∞∞δ(t)dt=1.

容易验证：∫−∞∞δ(t)f(t)dt=f(0).\text{容易验证：}\displaystyle\int_{-\infin}^{\infin}\delta(t)f(t)dt=f(0).容易验证：∫−∞∞δ(t)f(t)dt=f(0).

取f(t)=e−iμt,f(0)=1.\text{取}\,f(t)=e^{-i\mu t},\,f(0)=1.取f(t)=e−iμt,f(0)=1.

则脉冲函数的傅里叶变换F(μ)=F[δ(t)]=∫−∞∞δ(t)e−iμtdt=1\text{则脉冲函数的傅里叶变换}\,\,\displaystyle F(\mu)=\mathscr{F}[\delta(t)]=\int_{-\infin}^{\infin} \delta(t)e^{-i\mu t}dt=1则脉冲函数的傅里叶变换F(μ)=F[δ(t)]=∫−∞∞δ(t)e−iμtdt=1.

作傅里叶反变换δ(t)=F−1[F(μ)]=12π∫−∞∞eiμtdμ.\text{作傅里叶反变换}\,\,\displaystyle\delta(t)=\mathscr{F}^{-1}\big[F(\mu)\big]=\frac{1}{2\pi}\int_{-\infin}^{\infin} e^{i\mu t}d\mu.作傅里叶反变换δ(t)=F−1[F(μ)]=2π1∫−∞∞eiμtdμ.

准备工作完成，构造函数\text{准备工作完成，构造函数}准备工作完成，构造函数g(λ)=∫−∞∞sin⁡λxxdx，则g(1)=∫−∞∞sin⁡xxdx.\,\displaystyle g(\lambda)=\int_{-\infin}^{\infin}\frac{\sin \lambda x}{x}dx\text{，则}\,g(1)=\int_{-\infin}^{\infin}\frac{\sin x}{x}dx.g(λ)=∫−∞∞xsinλxdx，则g(1)=∫−∞∞xsinxdx.

g′(λ)=∫−∞∞∂∂λ(sin⁡λxx)dx=∫−∞∞cos⁡λxdx+0=∫−∞∞cos⁡λxdx+i∫−∞∞sin⁡λxdx=∫−∞∞eiλxdx=2π(12π∫−∞∞eiλxdx)=2πδ(λ)\begin{aligned}g'(\lambda)&=\int_{-\infin}^{\infin}\frac{\partial }{\partial \lambda}\Big( \frac{\sin \lambda x}{x}\Big)dx\\&=\int_{-\infin}^{\infin}\cos \lambda xdx+0\\&=\int_{-\infin}^{\infin}\cos \lambda x dx+i\int_{-\infin}^{\infin}\sin \lambda x dx\\&=\int_{-\infin}^{\infin}e^{i\lambda x}dx\\&=2\pi \big(\frac{1}{2\pi}\int_{-\infin}^{\infin}e^{i\lambda x}dx\big)\\&=2\pi \delta(\lambda)\end{aligned}g′(λ)=∫−∞∞∂λ∂(xsinλx)dx=∫−∞∞cosλxdx+0=∫−∞∞cosλxdx+i∫−∞∞sinλxdx=∫−∞∞eiλxdx=2π(2π1∫−∞∞eiλxdx)=2πδ(λ)

因为g(λ)是奇函数，所以\text{因为}\,g(\lambda)\,\text{是奇函数，所以}因为g(λ)是奇函数，所以

g(1)=−g(−1)=12(g(1)−g(−1))(由N-L公式)=12∫−11g′(λ)dλ=12∫−112πδ(λ)dλ(δ(λ)=0,λ≠0)=π∫−∞∞δ(λ)dλ=π\begin{aligned}g(1)&=-g(-1)\\&=\frac{1}{2}\big(\,g(1)-g(-1)\,\big) \quad (\text{由}\, N\text{-}L\, \text{公式})\\&=\frac{1}{2} \int_{-1}^{1}g'(\lambda)d\lambda\\&=\frac{1}{2} \int_{-1}^{1} 2\pi \delta(\lambda) d\lambda \quad\,\,\,\, (\delta(\lambda)=0,\,\lambda\neq0)\\&=\pi \int_{-\infin}^{\infin}\delta(\lambda)d\lambda\\&=\pi\end{aligned}g(1)=−g(−1)=21(g(1)−g(−1))(由N-L公式)=21∫−11g′(λ)dλ=21∫−112πδ(λ)dλ(δ(λ)=0,λ=0)=π∫−∞∞δ(λ)dλ=π妙极！\text{妙极！}妙极！

7. 留数定理

定理内容：当被积函数f(x)是x的有理函数(多项式除多项式)，且分母的次数比分子\begin{aligned}\text{定理内容：当被积函数} \,f(x)\, \text{是}\, x \,\text{的有理函数(多项式除多项式)，且分母的次数比分子}\end{aligned}定理内容：当被积函数f(x)是x的有理函数(多项式除多项式)，且分母的次数比分子
的次数至少高一次，f(z)在实轴上除去有限多个一级奇点x1,x2,⋯,xp外处处解析，\begin{aligned}\text{的次数至少高一次，}f(z)\,\text{在实轴上除去有限多个一级奇点}\,x_1,x_2,\cdots,x_p\, \text{外处处解析，}\end{aligned}的次数至少高一次，f(z)在实轴上除去有限多个一级奇点x1,x2,⋯,xp外处处解析，
在上半复平面(Imz>0)除去有限多个奇点z1,z2,⋯,zq外处处解析，则\begin{aligned}\text{在上半复平面}\,(\mathrm{Im}\,z>0)\,\text{除去有限多个奇点}\,z_1,z_2,\cdots,z_q\,\text{外处处解析，则}\end{aligned}在上半复平面(Imz>0)除去有限多个奇点z1,z2,⋯,zq外处处解析，则

∫−∞∞f(x)eimxdx=πi∑k=1pRes[f(z)eimz,xk]+2πi∑k=1qRes[f(z)eimz,zk]\int_{-\infin}^{\infin}f(x)e^{imx}dx=\pi i\sum_{k=1}^{p}\mathrm{Res}[f(z)e^{imz},x_k]+2\pi i\sum_{k=1}^{q}\mathrm{Res}[f(z)e^{imz},z_k]∫−∞∞f(x)eimxdx=πik=1∑pRes[f(z)eimz,xk]+2πik=1∑qRes[f(z)eimz,zk]

其中Res[f(z),z0]为函数f在z0处的留数，定义如下：\begin{aligned}\text{其中}\,\mathrm{Res}[f(z),z_0]\,\text{为函数}\,f\,\text{在}\,z_0\,\text{处的留数，定义如下：}\end{aligned}其中Res[f(z),z0]为函数f在z0处的留数，定义如下：

若z0是f(z)的孤立奇点，f(z)在D={z∣0<∣z−z0∣<R}内解析，C是D内包\begin{aligned}&\text{若}\,z_0\,\text{是}\,f(z)\,\text{的孤立奇点，}f(z)\,\text{在}\,D=\{z\,|\,0<|z-z_0|<R\}\,\text{内解析，}C\,\text{是}\,D\,\text{内包}\end{aligned}若z0是f(z)的孤立奇点，f(z)在D={z∣0<∣z−z0∣<R}内解析，C是D内包
围z0的任一正向简单闭曲线，则称积分\begin{aligned}\text{围}\,z_0\,\text{的任一正向简单闭曲线，则称积分}\end{aligned}围z0的任一正向简单闭曲线，则称积分

12πi∮Cf(z)dz\frac{1}{2\pi i}\oint_C f(z)dz2πi1∮Cf(z)dz为f在z0处的留数，记作Res[f(z),z0].\text{为}\,f\,\text{在}\,z_0\,\text{处的留数，记作}\,\mathrm{Res}[f(z),z_0].为f在z0处的留数，记作Res[f(z),z0].

套用定理，令f(x)=1x，实轴上的一级奇点x1=0，上半复平面内无奇点，则\begin{aligned}\text{套用定理，令}\,f(x)=\frac{1}{x}\text{，实轴上的一级奇点}\,x_1=0\text{，上半复平面内无奇点，则}\end{aligned}套用定理，令f(x)=x1，实轴上的一级奇点x1=0，上半复平面内无奇点，则

∫−∞∞1xeixdx=iπRes[eizz,0]=iπ12πi∮∣z∣=1eizzdz=12∮∣z∣=1eizzdz\int_{-\infin}^{\infin}\frac{1}{x}e^{ix}dx=i\pi\, \mathrm{Res}[\frac{e^{iz}}{z},0] =i\pi \frac{1}{2\pi i}\oint_{|z|=1} \frac{e^{iz}}{z}dz=\frac{1}{2} \oint_{|z|=1} \frac{e^{iz}}{z}dz∫−∞∞x1eixdx=iπRes[zeiz,0]=iπ2πi1∮∣z∣=1zeizdz=21∮∣z∣=1zeizdz

∮∣z∣=1eizzdz=∮∣iz∣=1eizizd(iz)=∮∣z∣=1ezzdz=∮∣z∣=11z(1+z+z22!+⋯+znn!+⋯)dz=∮∣z∣=1(1z+1+z2!+⋯+zn(n+1)!+⋯)dz=∮∣z∣=11zdz+∮∣z∣=1(d(z)+d(z2)2⋅2!+⋯+d(zn+1)(n+1)(n+1)!+⋯)=∮∣z∣=11zdz\begin{aligned}\oint_{|z|=1} \frac{e^{iz}}{z}dz&=\oint_{|iz|=1} \frac{e^{iz}}{iz}d(iz)\\&=\oint_{|z|=1} \frac{e^{z}}{z}dz\\&=\oint_{|z|=1} \frac{1}{z}(1+z+\frac{z^2}{2!}+\cdots+\frac{z^n}{n!}+\cdots) dz \\&=\oint_{|z|=1} (\frac{1}{z}+1+\frac{z}{2!}+\cdots+\frac{z^n}{(n+1)!}+\cdots) dz \\&= \oint_{|z|=1} \frac{1}{z}\,dz\,+\oint_{|z|=1} \big(d(z)+\frac{d(z^2)}{2\cdot2!}+\cdots+\frac{d(z^{n+1})}{(n+1)(n+1)!}+\cdots\big)\\&= \oint_{|z|=1} \frac{1}{z}\,dz\end{aligned}∮∣z∣=1zeizdz=∮∣iz∣=1izeizd(iz)=∮∣z∣=1zezdz=∮∣z∣=1z1(1+z+2!z2+⋯+n!zn+⋯)dz=∮∣z∣=1(z1+1+2!z+⋯+(n+1)!zn+⋯)dz=∮∣z∣=1z1dz+∮∣z∣=1(d(z)+2⋅2!d(z2)+⋯+(n+1)(n+1)!d(zn+1)+⋯)=∮∣z∣=1z1dz

三角换元，令z=eiθ(0≤θ≤2π)，则dzdθ=ieiθ=iz,dzz=idθ.\begin{aligned}\text{三角换元，令}\,z=e^{i\theta}(0\leq\theta\leq 2\pi)\,\text{，则}\,\frac{dz}{d\theta}=ie^{i\theta}=iz,\frac{dz}{z}=id\theta\end{aligned}.三角换元，令z=eiθ(0≤θ≤2π)，则dθdz=ieiθ=iz,zdz=idθ.
∮∣z∣=11zdz=∫02πidθ=2πi\begin{aligned}\oint_{|z|=1} \frac{1}{z}\,dz=\int_0^{2\pi}id\theta=2\pi i\end{aligned}∮∣z∣=1z1dz=∫02πidθ=2πi

于是\text{于是}于是

∫−∞∞1xeixdx=12∮∣z∣=1eizzdz=12∮∣z∣=11zdz=πi\begin{aligned}\int_{-\infin}^{\infin}\frac{1}{x}e^{ix}dx=\frac{1}{2} \oint_{|z|=1} \frac{e^{iz}}{z}dz=\frac{1}{2}\oint_{|z|=1} \frac{1}{z}\,dz=\pi i\end{aligned}∫−∞∞x1eixdx=21∮∣z∣=1zeizdz=21∮∣z∣=1z1dz=πi

又因为\text{又因为}又因为

∫−∞∞1xeixdx=∫−∞∞1x(cos⁡x+isin⁡x)dx=i∫−∞∞sin⁡xxdx\begin{aligned}\int_{-\infin}^{\infin}\frac{1}{x}e^{ix}dx=\int_{-\infin}^{\infin}\frac{1}{x}(\cos x+i\sin x)dx=i\int_{-\infin}^{\infin}\frac{\sin x}{x}dx\end{aligned}∫−∞∞x1eixdx=∫−∞∞x1(cosx+isinx)dx=i∫−∞∞xsinxdx

所以∫−∞∞sin⁡xxdx=π.\begin{aligned}\text{所以}\,\int_{-\infin}^{\infin}\frac{\sin x}{x}dx=\pi\end{aligned}.所以∫−∞∞xsinxdx=π.

8. 黎曼引理

先做些准备工作\text{先做些准备工作}先做些准备工作
sin⁡2n+12x=sin⁡x2+∑k=1n(sin⁡2k+12x−sin⁡2k−12x)\sin \frac{2n+1}{2}x=\sin \frac{x}{2}+\sum_{k=1}^n(\sin \frac{2k+1}{2}x-\sin \frac{2k-1}{2}x)sin22n+1x=sin2x+k=1∑n(sin22k+1x−sin22k−1x)由和差化积公式：sin⁡A−sin⁡B=2sin⁡A−B2cos⁡A+B2.\begin{aligned}\text{由和差化积公式：}\sin A-\sin B=2\sin\frac{A-B}{2}\cos\frac{A+B}{2}\end{aligned}.由和差化积公式：sinA−sinB=2sin2A−Bcos2A+B.
则sin⁡2n+12x=(12+∑k=1ncos⁡kx)2sin⁡x2\begin{aligned}\text{则}\,\displaystyle \sin \frac{2n+1}{2}x=\big(\frac{1}{2}+\sum_{k=1}^n \cos kx\big) 2\sin \frac{x}{2}\end{aligned}则sin22n+1x=(21+k=1∑ncoskx)2sin2x.

x≠2kπ时，有sin⁡2n+12x2sin⁡x2=12+∑k=1ncos⁡kx\begin{aligned}x\neq 2k\pi\,\text{时，有}\,\frac{\sin \displaystyle\frac{2n+1}{2}x}{2\sin\displaystyle \frac{x}{2}}=\frac{1}{2}+\sum_{k=1}^n \cos kx \end{aligned}x=2kπ时，有2sin2xsin22n+1x=21+k=1∑ncoskx.

两边同时积分，得∫0πsin⁡2n+12x2sin⁡x2=π2(n=0,1,2,⋯)\begin{aligned}\text{两边同时积分，得}\int_0^{\pi}\frac{\sin\displaystyle \frac{2n+1}{2}x}{2\sin\displaystyle \frac{x}{2}}=\frac{\pi}{2}\,(n=0,1,2,\cdots)\end{aligned}两边同时积分，得∫0π2sin2xsin22n+1x=2π(n=0,1,2,⋯).

令g(x)=1x−12sin⁡x2=2sin⁡x2−x2xsin⁡x2,0<x≤π.\begin{aligned}\text{令}\,\displaystyle g(x)=\frac{1}{x}-\frac{1}{2\sin\displaystyle \frac{x}{2}}=\frac{2\sin\displaystyle \frac{x}{2}-x}{2x\sin \displaystyle \frac{x}{2}},\,0<x\leq\pi\end{aligned}.令g(x)=x1−2sin2x1=2xsin2x2sin2x−x,0<x≤π.

由洛必达法则，\text{由洛必达法则，}由洛必达法则，

lim⁡x→0+g(x)=lim⁡x→0+cos⁡x2−1xcos⁡x2+2sin⁡x2=lim⁡x→0+−12sin⁡x22cos⁡x2−12xsin⁡x2=0\lim_{x\to0^+}g(x)=\lim_{x\to0^+}\frac{\cos\displaystyle \frac{x}{2}-1}{x\displaystyle\cos \frac{x}{2}+2\sin \frac{x}{2}}=\lim_{x\to0^+}\frac{-\displaystyle\frac{1}{2}\sin\displaystyle \frac{x}{2}}{2\cos\displaystyle \frac{x}{2}-\frac{1}{2}x\sin \frac{x}{2}}=0x→0+limg(x)=x→0+limxcos2x+2sin2xcos2x−1=x→0+lim2cos2x−21xsin2x−21sin2x=0

补充定义g(0)=0，则g在[0,π]上连续.\text{补充定义}\,g(0)=0\text{，则}\,g\,\text{在}\,[0,\pi]\,\text{上连续}.补充定义g(0)=0，则g在[0,π]上连续.

Riemann-Lebesgue(差点儿漏掉s)引理：\textrm{Riemann-Lebesgue}\,(差点儿漏掉 \,\textrm{s})\text{引理：}Riemann-Lebesgue(差点儿漏掉s)引理：若f在[a,b]上连续，则lim⁡p→∞∫abf(x)sin⁡pxdx=0\text{若}\,f\,\text{在}\,[a,b]\,\text{上连续，则}\displaystyle \lim_{p\to \infin}\int_a^bf(x)\sin px \,dx=0若f在[a,b]上连续，则p→∞lim∫abf(x)sinpxdx=0.

令f(x)=g(x),p=n+12，则\text{令}\,\displaystyle f(x)=g(x),\,p=n+\frac{1}{2}\text{，则}令f(x)=g(x),p=n+21，则
lim⁡n→∞∫0π(1x−12sin⁡x2)sin⁡(n+12)xdx=0\lim_{n\to\infin}\int_0^{\pi} \big(\frac{1}{x}-\frac{1}{2\sin \displaystyle \frac{x}{2}}\big)\sin (n+\frac{1}{2})x\,dx=0n→∞lim∫0π(x1−2sin2x1)sin(n+21)xdx=0lim⁡n→∞∫0πsin⁡(n+12)xxdx=lim⁡n→∞∫0πsin⁡(2n+12)x2sin⁡x2dx=lim⁡n→∞π2=π2\lim_{n\to\infin}\int_0^{\pi} \frac{\sin(\displaystyle n+\frac{1}{2})x}{x}\,dx=\lim_{n\to\infin}\int_0^{\pi} \frac{\sin(\displaystyle \frac{2n+1}{2})x}{2\sin\displaystyle \frac{x}{2}}\,dx=\lim_{n\to\infin} \frac{\pi}{2}=\frac{\pi}{2}n→∞lim∫0πxsin(n+21)xdx=n→∞lim∫0π2sin2xsin(22n+1)xdx=n→∞lim2π=2π

令u=(n+12)x，则\begin{aligned}\text{令}\,u=(n+\frac{1}{2})x\text{，则}\end{aligned}令u=(n+21)x，则
lim⁡n→∞∫0πsin⁡(n+12)xxdx=lim⁡n→∞∫0(n+12)πsin⁡uudu=π2\lim_{n\to\infin}\int_0^{\pi} \frac{\sin(n+\displaystyle\frac{1}{2})x}{x}\,dx=\lim_{n\to\infin}\int_0^{(n+\frac{1}{2})\pi}\frac{\sin u}{u}du =\frac{\pi}{2}n→∞lim∫0πxsin(n+21)xdx=n→∞lim∫0(n+21)πusinudu=2π

所以\text{所以}所以 ∫0∞sin⁡uudu=lim⁡n→∞∫0(n+12)πsin⁡uudu=π2\int_0^{\infin}\frac{\sin u}{u}du=\lim_{n\to\infin}\int_0^{(n+\frac{1}{2})\pi}\frac{\sin u}{u}du =\frac{\pi}{2}∫0∞usinudu=n→∞lim∫0(n+21)πusinudu=2π

若读者还有其他巧妙解法，请不吝赐教！\small \text{若读者还有其他巧妙解法，请不吝赐教！}若读者还有其他巧妙解法，请不吝赐教！

文末彩蛋：大家好，这是我的孪生兄弟：\small \textbf{文末彩蛋：大家好，这是我的孪生兄弟：}文末彩蛋：大家好，这是我的孪生兄弟：无穷积分∫e−x2dx的几种巧妙解法！\small \textbf{无穷积分}\,\int e^{-x^2}dx\, \textbf{的几种巧妙解法！}无穷积分∫e−x2dx的几种巧妙解法！

Plus: 如有错误、可以改进的地方、或任何想说的，请在评论区留言！