女子世界杯_世界杯今日赛事 - fdrwxy.com SPACE


目录:

1. Fourier 正弦展开2. 交换积分次序3. 构造含参变量函数4. Laplace 变换5. Fourier 变换6. 狄拉克函数7. 留数定理8. 黎曼引理

狄利克雷积分

0

sin

x

x

d

x

\,\displaystyle \int_{0}^{\infin}\frac{\sin x}{x}dx\,

∫0∞​xsinx​dx 是一个比较常见的无穷积分,在很多领域有着重要应用。

下面介绍几种巧妙解法。为了您更好的阅读体验,请使用电脑浏览。

1. Fourier 正弦展开

0

sin

x

x

d

x

=

lim

m

0

m

π

sin

x

x

d

x

\int_{0}^{\infin}\frac{\sin x}{x}dx=\lim_{m\to\infin}\int_0^{m\pi}\frac{\sin x}{x}dx

∫0∞​xsinx​dx=m→∞lim​∫0mπ​xsinx​dx

h

=

π

/

k

,将区间

[

0

,

m

π

]

分割成

k

m

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

0

m

π

sin

x

x

d

x

=

lim

h

0

+

n

=

1

k

m

sin

n

h

n

h

h

=

lim

h

0

+

n

=

1

k

m

sin

n

h

n

\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}

∫0mπ​xsinx​dx=h→0+lim​n=1∑km​nhsinnh​h=h→0+lim​n=1∑km​nsinnh​

Fourier

正弦展开

\text{由}\,\textrm{Fourier}\,\text{正弦展开}

由Fourier正弦展开

π

h

2

=

n

=

1

sin

n

h

n

,

(

0

<

h

<

π

)

\frac{\pi-h}{2}=\sum_{n=1}^{\infin}\frac{\sin nh}{n},(0

2π−h​=n=1∑∞​nsinnh​,(0

所以

\text{所以}

所以

0

sin

x

x

d

x

=

lim

m

0

m

π

sin

x

x

d

x

=

lim

m

lim

h

0

+

n

=

1

k

m

sin

n

h

n

=

lim

h

0

+

lim

m

n

=

1

k

m

sin

n

h

n

=

lim

h

0

+

π

h

2

=

π

2

\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∞​xsinx​dx​=m→∞lim​∫0mπ​xsinx​dx=m→∞lim​h→0+lim​n=1∑km​nsinnh​=h→0+lim​m→∞lim​n=1∑km​nsinnh​=h→0+lim​2π−h​=2π​​

2. 交换积分次序

0

sin

x

x

d

x

=

0

sin

x

1

x

d

x

=

0

sin

x

(

0

e

x

y

d

y

)

d

x

(

交换积分次序

)

=

0

(

0

e

y

x

sin

x

d

x

)

d

y

0

e

y

x

sin

x

d

x

=

0

e

y

x

1

2

i

(

e

i

x

e

i

x

)

d

x

=

1

2

i

0

(

e

(

y

i

)

x

e

(

y

+

i

)

x

)

d

x

=

1

2

i

(

1

y

i

e

(

y

i

)

x

+

1

y

+

i

e

(

y

+

i

)

x

)

0

=

1

2

i

(

1

y

i

1

y

+

i

)

=

1

y

2

+

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∞​xsinx​dx∫0∞​e−yxsinxdx​=∫0∞​sinxx1​dx=∫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−i1​e−(y−i)x+y+i1​e−(y+i)x)∣∣∣​0∞​=2i1​(y−i1​−y+i1​)=y2+11​​

原式

=

0

1

y

2

+

1

d

y

=

arctan

y

0

=

π

2

.

\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+11​dy=arctany∣∣∣​0∞​=2π​​.

3. 构造含参变量函数

I

=

0

sin

t

t

d

t

,构造函数

f

(

x

)

=

0

e

x

t

sin

t

t

d

t

,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∞​tsint​dt,构造函数f(x)=∫0∞​e−xttsint​dt,则f(0)=I​.

0

f

(

x

)

0

e

x

t

sin

t

t

d

t

0

e

x

t

d

t

=

1

x

0\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

x

t

sin

t

t

)

d

t

=

0

e

x

t

sin

t

d

t

=

1

x

2

+

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

)

d

x

=

0

1

x

2

+

1

d

x

=

arctan

x

0

=

π

2

0-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+11​dx=−arctanx∣∣∣∣​0∞​=−2π​

所以

I

=

π

2

.

\begin{aligned}\text{所以}\,I=\frac{\pi}{2}\end{aligned}.

所以I=2π​​.

4. Laplace 变换

f

(

t

)

=

0

sin

t

x

x

d

x

,

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∞​xsintx​dx,t>0,对f(t)作拉普拉斯变换,令​

F

(

s

)

=

L

[

f

(

t

)

]

=

L

[

0

sin

t

x

x

d

x

]

t

=

0

1

x

L

[

sin

t

x

]

t

d

x

=

0

1

s

2

+

x

2

d

x

=

1

s

0

1

1

+

(

x

s

)

2

d

(

x

s

)

(

let

u

=

x

s

)

=

1

s

arctan

u

0

=

1

s

π

2

\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∞​xsintx​dx]t​=∫0∞​x1​L[sintx]t​dx=∫0∞​s2+x21​dx=s1​∫0∞​1+(sx​)21​d(sx​)(let u = sx​)=s1​arctanu∣∣∣​0∞​=s1​2π​​

f

(

t

)

=

L

1

[

F

(

s

)

]

=

π

2

L

1

[

1

s

]

=

π

2

.

\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)]=2π​L−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

t

x

x

d

x

=

π

2

,

t

>

0

\int_0^{\infin}\frac{\sin tx}{x}\,dx=\frac{\pi}{2},\,t>0

∫0∞​xsintx​dx=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

t

x

x

)

d

x

=

0

cos

t

x

d

x

=

lim

n

0

n

π

/

t

cos

t

x

d

x

(

l

e

t

u

=

t

x

)

=

lim

n

1

t

0

n

π

cos

u

d

u

=

1

t

lim

n

sin

u

0

n

π

=

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/t​costxdx(letu=tx)=n→∞lim​t1​∫0nπ​cosudu=t1​n→∞lim​sinu∣∣∣​0nπ​=0​

所以

f

(

t

)

=

C

,

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

\text{令}

f

(

t

)

=

{

1

,

t

<

1

0

,

t

1

f(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

i

μ

t

d

t

=

1

1

e

i

μ

t

d

t

=

1

i

μ

e

i

μ

t

1

1

=

2

μ

1

2

i

(

e

i

μ

e

i

μ

)

=

2

sin

μ

μ

\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=∫−11​e−iμtdt=−iμ1​e−iμt∣∣∣​−11​=μ2​2i1​(eiμ−e−iμ)=2μsinμ​​

f

(

t

)

=

F

1

[

F

(

μ

)

]

=

1

2

π

2

sin

μ

μ

e

i

μ

t

d

μ

.

\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​∫−∞∞​2μsinμ​eiμtdμ.​

t

=

0

,则

\text{取}\,t=0\text{,则}

取t=0,则

1

=

f

(

0

)

=

1

π

sin

μ

μ

d

μ

=

2

π

0

sin

μ

μ

d

μ

1=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\mu

1=f(0)=π1​∫−∞∞​μsinμ​dμ=π2​∫0∞​μsinμ​dμ

所以

0

sin

μ

μ

d

μ

=

π

2

.

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∞​μsinμ​dμ=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

=

0

0

,

t

0

,且满足

δ

(

t

)

d

t

=

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

)

d

t

=

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).

f

(

t

)

=

e

i

μ

t

,

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

i

μ

t

d

t

=

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

)

]

=

1

2

π

e

i

μ

t

d

μ

.

\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

λ

x

x

d

x

,则

g

(

1

)

=

sin

x

x

d

x

.

\,\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)=∫−∞∞​xsinλx​dx,则g(1)=∫−∞∞​xsinx​dx.

g

(

λ

)

=

λ

(

sin

λ

x

x

)

d

x

=

cos

λ

x

d

x

+

0

=

cos

λ

x

d

x

+

i

sin

λ

x

d

x

=

e

i

λ

x

d

x

=

2

π

(

1

2

π

e

i

λ

x

d

x

)

=

2

π

δ

(

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

)

=

1

2

(

g

(

1

)

g

(

1

)

)

(

N

-

L

公式

)

=

1

2

1

1

g

(

λ

)

d

λ

=

1

2

1

1

2

π

δ

(

λ

)

d

λ

(

δ

(

λ

)

=

0

,

λ

0

)

=

π

δ

(

λ

)

d

λ

=

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​∫−11​g′(λ)dλ=21​∫−11​2πδ(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

)

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

x

1

,

x

2

,

,

x

p

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

(

I

m

z

>

0

)

除去有限多个奇点

z

1

,

z

2

,

,

z

q

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

)

e

i

m

x

d

x

=

π

i

k

=

1

p

R

e

s

[

f

(

z

)

e

i

m

z

,

x

k

]

+

2

π

i

k

=

1

q

R

e

s

[

f

(

z

)

e

i

m

z

,

z

k

]

\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∑p​Res[f(z)eimz,xk​]+2πik=1∑q​Res[f(z)eimz,zk​]

其中

R

e

s

[

f

(

z

)

,

z

0

]

为函数

f

z

0

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:​

z

0

f

(

z

)

的孤立奇点,

f

(

z

)

D

=

{

z

0

<

z

z

0

<

R

}

内解析,

C

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|

​若z0​是f(z)的孤立奇点,f(z)在D={z∣0<∣z−z0​∣

z

0

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​

1

2

π

i

C

f

(

z

)

d

z

\frac{1}{2\pi i}\oint_C f(z)dz

2πi1​∮C​f(z)dz

f

z

0

处的留数,记作

R

e

s

[

f

(

z

)

,

z

0

]

.

\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

)

=

1

x

,实轴上的一级奇点

x

1

=

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​

1

x

e

i

x

d

x

=

i

π

R

e

s

[

e

i

z

z

,

0

]

=

i

π

1

2

π

i

z

=

1

e

i

z

z

d

z

=

1

2

z

=

1

e

i

z

z

d

z

\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

∫−∞∞​x1​eixdx=iπRes[zeiz​,0]=iπ2πi1​∮∣z∣=1​zeiz​dz=21​∮∣z∣=1​zeiz​dz

z

=

1

e

i

z

z

d

z

=

i

z

=

1

e

i

z

i

z

d

(

i

z

)

=

z

=

1

e

z

z

d

z

=

z

=

1

1

z

(

1

+

z

+

z

2

2

!

+

+

z

n

n

!

+

)

d

z

=

z

=

1

(

1

z

+

1

+

z

2

!

+

+

z

n

(

n

+

1

)

!

+

)

d

z

=

z

=

1

1

z

d

z

+

z

=

1

(

d

(

z

)

+

d

(

z

2

)

2

2

!

+

+

d

(

z

n

+

1

)

(

n

+

1

)

(

n

+

1

)

!

+

)

=

z

=

1

1

z

d

z

\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∣=1​zeiz​dz​=∮∣iz∣=1​izeiz​d(iz)=∮∣z∣=1​zez​dz=∮∣z∣=1​z1​(1+z+2!z2​+⋯+n!zn​+⋯)dz=∮∣z∣=1​(z1​+1+2!z​+⋯+(n+1)!zn​+⋯)dz=∮∣z∣=1​z1​dz+∮∣z∣=1​(d(z)+2⋅2!d(z2)​+⋯+(n+1)(n+1)!d(zn+1)​+⋯)=∮∣z∣=1​z1​dz​

三角换元,令

z

=

e

i

θ

(

0

θ

2

π

)

,则

d

z

d

θ

=

i

e

i

θ

=

i

z

,

d

z

z

=

i

d

θ

.

\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

=

1

1

z

d

z

=

0

2

π

i

d

θ

=

2

π

i

\begin{aligned}\oint_{|z|=1} \frac{1}{z}\,dz=\int_0^{2\pi}id\theta=2\pi i\end{aligned}

∮∣z∣=1​z1​dz=∫02π​idθ=2πi​

于是

\text{于是}

于是

1

x

e

i

x

d

x

=

1

2

z

=

1

e

i

z

z

d

z

=

1

2

z

=

1

1

z

d

z

=

π

i

\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}

∫−∞∞​x1​eixdx=21​∮∣z∣=1​zeiz​dz=21​∮∣z∣=1​z1​dz=πi​

又因为

\text{又因为}

又因为

1

x

e

i

x

d

x

=

1

x

(

cos

x

+

i

sin

x

)

d

x

=

i

sin

x

x

d

x

\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}

∫−∞∞​x1​eixdx=∫−∞∞​x1​(cosx+isinx)dx=i∫−∞∞​xsinx​dx​

所以

sin

x

x

d

x

=

π

.

\begin{aligned}\text{所以}\,\int_{-\infin}^{\infin}\frac{\sin x}{x}dx=\pi\end{aligned}.

所以∫−∞∞​xsinx​dx=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

2

n

+

1

2

x

=

sin

x

2

+

k

=

1

n

(

sin

2

k

+

1

2

x

sin

2

k

1

2

x

)

\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+1​x=sin2x​+k=1∑n​(sin22k+1​x−sin22k−1​x)

由和差化积公式:

sin

A

sin

B

=

2

sin

A

B

2

cos

A

+

B

2

.

\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−B​cos2A+B​​.

sin

2

n

+

1

2

x

=

(

1

2

+

k

=

1

n

cos

k

x

)

2

sin

x

2

\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+1​x=(21​+k=1∑n​coskx)2sin2x​​.

x

2

k

π

时,有

sin

2

n

+

1

2

x

2

sin

x

2

=

1

2

+

k

=

1

n

cos

k

x

\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2sin2x​sin22n+1​x​=21​+k=1∑n​coskx​.

两边同时积分,得

0

π

sin

2

n

+

1

2

x

2

sin

x

2

=

π

2

(

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∫0π​2sin2x​sin22n+1​x​=2π​(n=0,1,2,⋯)​.

g

(

x

)

=

1

x

1

2

sin

x

2

=

2

sin

x

2

x

2

x

sin

x

2

,

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

令g(x)=x1​−2sin2x​1​=2xsin2x​2sin2x​−x​,0

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

x

2

1

x

cos

x

2

+

2

sin

x

2

=

lim

x

0

+

1

2

sin

x

2

2

cos

x

2

1

2

x

sin

x

2

=

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}}=0

x→0+lim​g(x)=x→0+lim​xcos2x​+2sin2x​cos2x​−1​=x→0+lim​2cos2x​−21​xsin2x​−21​sin2x​​=0

补充定义

g

(

0

)

=

0

,则

g

[

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

(

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:

f

[

a

,

b

]

上连续,则

lim

p

a

b

f

(

x

)

sin

p

x

d

x

=

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​∫ab​f(x)sinpxdx=0.

f

(

x

)

=

g

(

x

)

,

p

=

n

+

1

2

,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

0

π

(

1

x

1

2

sin

x

2

)

sin

(

n

+

1

2

)

x

d

x

=

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=0

n→∞lim​∫0π​(x1​−2sin2x​1​)sin(n+21​)xdx=0

lim

n

0

π

sin

(

n

+

1

2

)

x

x

d

x

=

lim

n

0

π

sin

(

2

n

+

1

2

)

x

2

sin

x

2

d

x

=

lim

n

π

2

=

π

2

\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​∫0π​xsin(n+21​)x​dx=n→∞lim​∫0π​2sin2x​sin(22n+1​)x​dx=n→∞lim​2π​=2π​

u

=

(

n

+

1

2

)

x

,则

\begin{aligned}\text{令}\,u=(n+\frac{1}{2})x\text{,则}\end{aligned}

令u=(n+21​)x,则​

lim

n

0

π

sin

(

n

+

1

2

)

x

x

d

x

=

lim

n

0

(

n

+

1

2

)

π

sin

u

u

d

u

=

π

2

\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​∫0π​xsin(n+21​)x​dx=n→∞lim​∫0(n+21​)π​usinu​du=2π​

所以

\text{所以}

所以

0

sin

u

u

d

u

=

lim

n

0

(

n

+

1

2

)

π

sin

u

u

d

u

=

π

2

\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∞​usinu​du=n→∞lim​∫0(n+21​)π​usinu​du=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

x

2

d

x

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!

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