$\operatorname{sinc}x$连乘积的积分

hbghlyj

$$\int_{0}^{\infty} \prod_{k=0}^{n} \operatorname{sinc}\left(\frac{x}{2 k+1}\right)=\frac12$$

1. y=matlabFunction(sinc(x)*sinc(x/3)*sinc(x/5)*sinc(x/7));
2. integral(y,0,inf)

复制代码

0.5000

Borwein integral
These integrals are remarkable for exhibiting apparent patterns that eventually break down. The following is an example.
\begin{align*}
& \int_0^\infty \frac{\sin(x)}{x} \, dx= \frac \pi 2 \\[10pt]
& \int_0^\infty \frac{\sin(x)}{x}\frac{\sin(x/3)}{x/3} \, dx = \frac \pi 2 \\[10pt]
& \int_0^\infty \frac{\sin(x)}{x}\frac{\sin(x/3)}{x/3}\frac{\sin(x/5)}{x/5} \, dx = \frac \pi 2
\end{align*}
This pattern continues up to
\[
\int_0^\infty \frac{\sin(x)}{x}\frac{\sin(x/3)}{x/3}\cdots\frac{\sin(x/13)}{x/13} \, dx = \frac \pi 2.
\]
At the next step the obvious pattern fails,
\begin{align*}
\int_0^\infty \frac{\sin(x)}{x}\frac{\sin(x/3)}{x/3}\cdots\frac{\sin(x/15)}{x/15} \, dx
&= \frac{467807924713440738696537864469}{935615849440640907310521750000}~\pi \\[5pt]
&= \frac \pi 2 - \frac{6879714958723010531}{935615849440640907310521750000}~\pi \\[5pt]
&\approx \frac \pi 2 - 2.31\times 10^{-11}.
\end{align*}

isee

回复 1# hbghlyj


一眼以为多个c,sinc,sin

战巡

回复 1# hbghlyj

不妨试试证明这样一个结论：
对任意$a_1>a_2\ge a_3\ge ...\ge a_n>0$，都有
\[\int_0^{\infty}\prod_{k=1}^n\operatorname{sinc}(a_kx)dx=\frac{1}{2a_1}\]

数量多了不好证明，但至少不难证明这个：
对正数$a\ne b$，会有
\[\int_0^\infty\operatorname{sinc}(ax)\operatorname{sinc}(bx)dx=\frac{1}{2\max\{a,b\}}\]

hbghlyj

假设$\alpha,\beta\in\mathbb{R}$,我们有
$$ I(\alpha,\beta)= \int_{0}^{+\infty}\frac{\sin(\alpha x)\cos(\beta x)}{x}\,dx =\frac{1}{2}\left[\int_{0}^{+\infty}\frac{\sin((\alpha+\beta)x)}{x}\,dx+\int_{0}^{+\infty}\frac{\sin((\alpha-\beta)x)}{x}\,dx\right]=\frac{\pi}{4}\left[\text{sign}(\alpha-\beta)+\text{sign}(\alpha+\beta)\right]$$最后这步用的是$\int_0^{+∞}\operatorname{sinc}x\,dx=\fracπ2$

aops
math.stackexchange.com/questions/441106
math.stackexchange.com/questions/1770005

hbghlyj

对于$a>b>0,$由分部积分法$-\int_{0}^{+\infty}\frac{\sin(a x)\sin(b x)}{x^2}\,dx=\left.\frac{\sin(ax)\sin(bx)}{x}\right|_0^{+∞}-bI(a,b)-aI(b,a)=-bI(a,b)-aI(b,a)$
故$\int_{0}^{+\infty}\frac{\sin(a x)\sin(b x)}{x^2}\,dx=bI(a,b)+aI(b,a)=\frac{πb}2$
故$\int_0^{+\infty}\operatorname{sinc}(ax)\operatorname{sinc}(bx)dx=\fracπ{2a}$

hbghlyj

回复 1# hbghlyj
matlab上面的是normalised sinc,定义为$$\operatorname{sinc} t=\left\{\begin{array}{cc}\frac{\sin \pi t}{\pi t} & t \neq 0 \\ 1 & t=0\end{array}\right.$$而4#,5#是unnormalized sinc,定义为$$\operatorname{sinc} t=\left\{\begin{array}{cc}\frac{\sin t}{t} & t \neq 0 \\ 1 & t=0\end{array}\right.$$所以和1#,3#差个$π$,这样一说就清楚了

hbghlyj

${\displaystyle \int _{-\infty }^{\infty }{\frac {\sin(\theta )}{\theta }}\,d\theta =\int _{-\infty }^{\infty }\left({\frac {\sin(\theta )}{\theta }}\right)^{2}\,d\theta =\pi .}$
${\displaystyle \int _{-\infty }^{\infty }{\frac {\sin ^{3}(\theta )}{\theta ^{3}}}\,d\theta ={\frac {3\pi }{4}}.}$
${\displaystyle \int _{-\infty }^{\infty }{\frac {\sin ^{4}(\theta )}{\theta ^{4}}}\,d\theta ={\frac {2\pi }{3}}.}$

hbghlyj

${\displaystyle \int _{0}^{\infty }{\frac {dx}{x^{n}+1}}=1+2\sum _{k=1}^{\infty }{\frac {(-1)^{k+1}}{(kn)^{2}-1}}={\frac {1}{\operatorname {sinc} ({\frac {\pi }{n}})}}.}$
见这帖

战巡

hbghlyj 发表于 2022-3-17 16:45
${\displaystyle \int _{0}^{\infty }{\frac {dx}{x^{n}+1}}=1+2\sum _{k=1}^{\infty }{\frac {(-1)^{k+1}} ...

搞这么麻烦...

令$y=\frac{1}{x^n+1}$，有$x=(\frac{1}{y}-1)^{\frac{1}{n}}$，而后
\[\int_0^{\infty}\frac{dx}{x^n+1}=-\int_0^1yd(\frac{1}{y}-1)^{\frac{1}{n}}=\int_0^1\frac{1}{ny}\cdot(\frac{1}{y}-1)^{\frac{1}{n}-1}dy\]
\[=\frac{1}{n}\int_0^1y^{-\frac{1}{n}}(1-y)^{\frac{1}{n}-1}dy=\frac{1}{n}B(1-\frac{1}{n},\frac{1}{n})=\frac{1}{n}\cdot\frac{\pi}{\sin(\frac{\pi}{n})}=\frac{1}{\text{sinc}(\frac{\pi}{n})}\]

abababa

战巡 发表于 2022-9-27 11:09
搞这么麻烦...

令$y=\frac{1}{x^n+1}$，有$x=(\frac{1}{y}-1)^{\frac{1}{n}}$，而后

这个可以一般地求$\int_{0}^{\infty}\frac{x^{p-1}}{1+x^q}dx$，其中$0<p<q$。

3楼的那个是不是有问题？比如在Mathematica里
Integrate[Product[Sinc[n x], {n, 1, 5}], {x, 0, Infinity}]
得出的是$\frac{271 \pi }{3072}$，按3楼的公式，结果应该是$\frac{1}{10}$？

abababa

abababa 发表于 2022-10-8 16:26
$\DeclareMathOperator{\sinc}{sinc}$
这个可以一般地求$\int_{0}^{\infty}\frac{x^{p-1}}{1+x^q}dx$，其中$0

应该是要求$a_1>\sum_{k=2}^{n}a_k$且每个$a_k>0$，这样把那个$\sinc$的分子用欧拉公式写成指数形式，最后乘起来时，最低次幂是正数，这样的话$e^{i\alpha\theta}$形式的部分就不再含有极点了，所有的极点全部来自分母的那些$x$的幂，最后取极限时$\frac{\sin ax}{ax}\to1$，所以只留下一个极点，就是$\frac{\sin a_1x}{a_1x}$的那个$x=0$处。然后根据网友以前讲的一个结论（最后图片），就得到
\[\int_{-\infty}^{\infty}f(x)dx=\pi i\text{Res}_{z=0}f(z)\]

其中$f(z)=\frac{e^{ibz}}{z}\prod_{k=1}^{n}\frac{\sin(a_kz)}{z}$，然后一算留数，就得到
\[\int_{-\infty}^{\infty}\frac{e^{ibx}}{x}\prod_{k=1}^{n}\frac{\sin(a_kx)}{x}dx=\pi i\prod_{k=1}^{n}a_k\]
两边再取虚部就得到结果了。

网友讲的结论：

hbghlyj

abababa 发表于 2022-10-8 09:26
应该是要求$a_1>\sum_{k=2}^{n}a_k$且每个$a_k>0$，这样把那个$\sinc$的分子用欧拉公式写成指数形式，最后乘起来时，最低次幂是正数，这样的话$e^{i\alpha\theta}$形式的部分就不再含有极点了，所有的极点全部来自分母的那些$x$的幂，最后取极限时$\frac{\sin ax}{ax}\to1$，所以只留下一个极点，就是$\frac{\sin a_1x}{a_1x}$的那个$x=0$处。

见en.wikipedia.org/wiki/Borwein_integral#General_formula
$$\int _{0}^{\infty }\prod _{k=0}^{n}{\frac {\sin(a_{k}x)}{a_{k}x}}\,dx={\frac {\pi }{2a_{0}}}C_{n}$$
where
$$C_{n}={\frac {1}{2^{n}n!\prod _{k=1}^{n}a_{k}}}\sum _{\gamma \in \{\pm 1\}^{n}}\varepsilon _{\gamma }b_{\gamma }^{n}\operatorname {sgn}(b_{\gamma })$$
In the case when $a_{0}>|a_{1}|+|a_{2}|+\cdots +|a_{n}|$, we have $\displaystyle C_{n}=1$.

hbghlyj

3Blue1Brown Researchers thought this was a bug (Borwein integrals)

0-frequency value in a Fourier Transform
$$\int_{-\infty}^{\infty} f(t) d t=\mathcal{F}[f(t)](0)$$
Fourier transform of $\sinc$ function is the rectangular function$$\mathcal{F}\left[\frac{\sin (\pi t / k)}{\pi t / k}\right](\omega)=k \cdot \operatorname{rect}(k \omega)$$
Convolution theorem$$\mathcal{F}[f(t) \cdot g(t)]=\mathcal{F}[f(t)] * \mathcal{F}[g(t)]$$

从而得到
$$\int_{-\infty}^{\infty} \frac{\sin (x)}{x} \underbrace{\frac{\sin \left(a_{1} x\right)}{a_{1} x} \cdots \frac{\sin \left(a_{n} x\right)}{a_{n} x}}_{a_{1}+\cdots+a_{n}<1} d x=\pi$$

hbghlyj

abababa 发表于 2022-11-2 07:41
其中$f(z)=\frac{e^{ibz}}{z}\prod_{k=1}^{n}\frac{\sin(a_kz)}{z}$

$\frac{e^{ibz}}{z}$是哪里来的呢