向老师的题目
计算积分
\[\int_0^1{\frac{1}{1+a^2x^2}\left[\left(1-\frac{x}{2}\right)\ln\frac{1+x}{1-x}+\frac{\pi^2x^2}{4}\right]^{-1}\textrm{d}x}.\]
解.先作换元$x\to\tanh x$,可得
\begin{align*}&\hspace{0.5cm}\int_0^1{\frac{1}{1+a^2x^2}\left[\left(1-\frac{x}{2}\right)\ln\frac{1+x}{1-x}+\frac{\pi^2x^2}{4}\right]^{-1}\textrm{d}x}\\&=\int_0^{\infty}{\frac{1}{1+a^2\tanh ^2x}\left(\frac{\coth ^2x-1}{\left(\coth x-x\right)^2+\frac{\pi^2x^2}{4}}\right)\textrm{d}x}\\&=\frac{1}{2}\int_{-\infty}^{\infty}{\frac{1}{1+a^2\tanh ^2x}\left(\frac{\coth ^2x-1}{\left(\coth x-x\right)^2+\frac{\pi^2x^2}{4}}\right)\textrm{d}x}\end{align*}
于是我们考虑函数$$\displaystyle{f\left(z\right)=\frac{1}{1+a^2\coth ^2z}\cdot\frac{\tanh ^2z-1}{\tanh z-z}}$$ 的如下围道积分
注意到$\displaystyle{f\left(z\right)=\frac{1}{1+a^2\coth ^2z}\cdot\frac{\tanh ^2z-1}{\tanh z-z}}$ 的极点为$z=0$,$\displaystyle{z=\pm\frac{\pi}{2}\mathrm{i}}$ （这两个极点在围道边界上）,以及$1+a^2\coth ^2z=0$的根$\displaystyle{z=\pm\mathrm{i}\cdot\mathrm{arccoth}\left(\frac{\mathrm{i}}{a}\right)=\pm\mathrm{i}\cdot\arctan(a)}$, 因此根据留数定理有
\begin{align*}&\int_{ - \infty - \frac{\pi }{2}{\rm{i}}}^{\infty - \frac{\pi }{2}{\rm{i}}} {f\left( z \right){\rm{d}}z} - \int_{ - \infty + \frac{\pi }{2}{\rm{i}}}^{\infty + \frac{\pi }{2}{\rm{i}}} {f\left( z \right){\rm{d}}z} \\&= 2\pi {\rm{i}}\left( {{\rm{Res}}\left[ {f\left( z \right),z = 0} \right] + \left( {{\rm{Res}}\left[ {f\left( z \right),{\rm{i}} \cdot \arctan \left( a \right)} \right] + {\rm{Res}}\left[ {f\left( z \right),z = - {\rm{i}} \cdot \arctan \left( a \right)} \right]} \right)} \right)\\&+ \pi {\rm{i}}\left( {{\rm{Res}}\left[ {f\left( z \right),z = \frac{\pi }{2}{\rm{i}}} \right] + {\rm{Res}}\left[ {f\left( z \right),z = - \frac{\pi }{2}{\rm{i}}} \right]} \right)\\&= 2\pi {\rm{i}}\left( {\frac{3}{{{a^2}}} - \frac{a}{{2\left( {a - \arctan \left( a \right)} \right)}} - \frac{a}{{2\left( {a - \arctan \left( a \right)} \right)}}} \right) + \pi {\rm{i}}\left( {1 + 1} \right)\\&= 2\pi {\rm{i}}\left( {\frac{3}{{{a^2}}} - \frac{{\arctan \left( a \right)}}{{a - \arctan \left( a \right)}}} \right).\end{align*}
注意到$\displaystyle{\tanh \left(z\pm\frac{\pi}{2}\textrm{i}\right)=\coth z,\coth \left(z\pm\frac{\pi}{2}\textrm{i}\right)=\tanh z}$,于是
\begin{align*}&\int_{-\infty -\frac{\pi}{2}\textrm{i}}^{\infty -\frac{\pi}{2}\textrm{i}}{f\left(z\right)\textrm{d}z}-\int_{-\infty +\frac{\pi}{2}\textrm{i}}^{\infty +\frac{\pi}{2}\textrm{i}}{f\left(z\right)\textrm{d}z}=\int_{-\infty}^{\infty}{f\left(x-\frac{\pi}{2}\textrm{i}\right)\textrm{d}x}-\int_{-\infty}^{\infty}{f\left(x+\frac{\pi}{2}\textrm{i}\right)\textrm{d}x}\\&=\int_{-\infty}^{\infty}{\frac{1}{1+a^2\tanh ^2x}\cdot\frac{\coth ^2x-1}{\coth x-\left(x-\frac{\pi}{2}\textrm{i}\right)^2}\textrm{d}x}-\int_{-\infty}^{\infty}{\frac{1}{1+a^2\tanh ^2x}\cdot\frac{\coth ^2x-1}{\coth x-\left(x+\frac{\pi}{2}\textrm{i}\right)^2}\textrm{d}x}\\&=-\pi\textrm{i}\int_{-\infty}^{\infty}{\frac{1}{1+a^2\tanh ^2x}\cdot\frac{\coth ^2x-1}{\left(\coth x-x\right)^2+\frac{\pi^2}{4}}\textrm{d}x}=2\pi\textrm{i}\left(\frac{3}{a^2}-\frac{\arctan\left(a\right)}{a-\arctan\left(a\right)}\right).\end{align*}
因此
\begin{align*}\int_0^{\infty}{\frac{1}{1+a^2\tanh ^2x}\cdot\frac{\coth ^2x-1}{\left(\coth x-x\right)^2+\frac{\pi^2}{4}}\textrm{d}x}&=\frac{1}{2}\int_{-\infty}^{\infty}{\frac{1}{1+a^2\tanh ^2x}\cdot\frac{\coth ^2x-1}{\left(\coth x-x\right)^2+\frac{\pi^2}{4}}\textrm{d}x}\\&=\frac{\arctan\left(a\right)}{a-\arctan\left(a\right)}-\frac{3}{a^2}\end{align*}
在开始本文之前,积分镇楼
\[\int_0^1{x^{20}\left[ \left( 1-\frac{x}{2}\ln \frac{1+x}{1-x} \right) ^2+\frac{\pi ^2x^2}{4} \right] ^{-1}\text{d}x}=\frac{5588512716806912356}{374010621408251953125}\]
下面的这些问题其实最开始来自1998年的美国数学月刊征解题
结果1998年的期刊没有答案,1999年以后的期刊又无处可寻,于是我就自己想办法解决这几个题目.我想了好久,突然想到我在前面某一期公众号上解决过类似的问题,于是灵感来了,解决了这几个征解题的同时还导出了几个额外的式子.
设$\displaystyle{\mathrm{Si}(x)=\int_0^x\frac{\sin t}t\mathrm dt}$表示正弦积分函数,求和
\[\sum_{n=1}^\infty\frac{\mathrm{Si}(n\pi)}{n^3}\]
解.[原创]
首先利用分部积分得
\begin{align*}\text{Si}\left( n\pi \right) &=\int_0^{n\pi}{\frac{\sin t}{t}\text{d}t}=\int_0^{\pi}{\frac{\sin nx}{x}\text{d}x}=\int_0^{\pi}{\sin nx\text{d}\left( \ln x \right)}=-n\int_0^{\pi}{\cos nx\ln x\text{d}x}\\&=-n\int_0^{\pi}{\cos nx\text{d}\left( x\ln x-x \right)}=n\left[ \left( -1 \right) ^{n-1}\left( \pi \ln \pi -\pi \right) -n\int_0^{\pi}{\sin nx\left( x\ln x-x \right) \text{d}x} \right]\end{align*}
于是我们可得
\[\sum_{n=1}^{\infty}{\frac{\text{Si}(n\pi)}{n^3}}=\sum_{n=1}^{\infty}{\frac{\left( -1 \right) ^{n-1}}{n^2}\left( \pi \ln \pi -\pi \right)}-\sum_{n=1}^{\infty}{\frac{1}{n}\int_0^{\pi}{\sin nx\left( x\ln x-x \right) \text{d}x}}\]
而$\displaystyle{\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^2}=\frac{\pi^2}{12}}$,把后一部分式子再分部积分得
\begin{align*}\sum_{n=1}^{\infty}{\frac{1}{n}\int_0^{\pi}{\sin nx\left( x\ln x-x \right) \text{d}x}}&=\sum_{n=1}^{\infty}{\frac{1}{n}\int_0^{\pi}{\sin nx\text{d}\left( \frac{1}{2}x^2\ln x-\frac{3}{4}x^2 \right)}}\\&=\sum_{n=1}^{\infty}{\int_0^{\pi}{\left( \frac{3}{4}x^2-\frac{1}{2}x^2\ln x \right) \cos nx\text{d}x}}.\end{align*}
现在考虑函数\[f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}{\frac{3}{4}{x^2} - \frac{1}{2}{x^2}\ln x,}&{x \in \left( {0,\pi } \right]}\\{0,}&{x = 0}\end{array}} \right..\]作偶对称以后再作$2\pi$周期延拓,则$f(x)$的Fourier余弦级数为
\[\widetilde{f}\left( x \right) =\frac{a_0}{2}+\sum_{n=1}^{\infty}{a_n\cos nx}\]
其中$\displaystyle{a_n=\frac{2}{\pi}\int_0^{\pi}{f\left( x \right) \cos nx\text{d}x}=\frac{2}{\pi}\int_0^{\pi}{\left( \frac{3}{4}x^2-\frac{1}{2}x^2\ln x \right) \cos nx\text{d}x}}$,根据Fourier级数收敛定理可知
\[\frac{a_0}{2}+\sum_{n=1}^{\infty}{a_n}=f(0)=0\]
而$\displaystyle{a_0=\frac2\pi\int_0^{\pi}{\left( \frac{3}{4}x^2-\frac{1}{2}x^2\ln x \right) \text{d}x}=\frac{11}{18}\pi ^2-\frac{1}{3}\pi ^2\ln \pi}$,因此
\[\sum_{n=1}^{\infty}{\int_0^{\pi}{\left( \frac{3}{4}x^2-\frac{1}{2}x^2\ln x \right) \text{d}x}}=\frac{\pi}{2}\sum_{n=1}^{\infty}{a_n}=-\frac{\pi}{4}a_0=\frac{\pi ^3}{12}\ln \pi -\frac{11}{72}\pi ^3\]
于是最后得到
\[\sum_{n=1}^{\infty}{\frac{\text{Si}\left( n\pi \right)}{n^3}}=\frac{\pi ^2}{12}\left( \pi \ln \pi -\pi \right) -\left( \frac{\pi ^3}{12}\ln \pi -\frac{11}{72}\pi ^3 \right) =\frac{5\pi ^3}{72}\]
同样道理我们还能得到
\[\sum_{n=1}^{\infty}{\left( -1 \right) ^n\frac{\text{Si}\left( n\pi \right)}{n^3}}=-\frac{\pi ^2}{6}\left( \pi \ln \pi -\pi \right) -\left( \frac{2\pi ^3}{9}-\frac{\pi ^3}{6}\ln \pi \right) =-\frac{\pi ^3}{18}\]
只不过这时需要利用$\displaystyle{\sum_{n=1}^{\infty}{\frac{1}{n^2}}=\frac{\pi ^2}{6}}$和$\displaystyle{\frac{a_0}{2}+\sum_{n=1}^{\infty}{\left( -1 \right) ^na_n}=f\left( \pi \right) }$即可.在第一步分部积分中我们利用$\ln x$的Fourier展开可得$\displaystyle{\sum_{n=1}^{\infty}{\frac{\left( -1 \right) ^{n-1}\text{Si}\left( n\pi \right)}{n}}=\frac{\pi}{2}}$.
设$\displaystyle{\mathrm{Si}(x)=\int_0^x\frac{\sin t}t\mathrm dt}$表示正弦积分函数,求和
\[\sum_{n=1}^{\infty}{\left( \frac{\text{Si}\left( n\pi \right)}{n} \right) ^2}\]
解.同上,先分部积分得
\[\text{Si}\left( n\pi \right) =-n\int_0^{\pi}{\cos nx\ln x\text{d}x}\]
于是得到
\[\sum_{n=1}^{\infty}{\left( \frac{\text{Si}\left( n\pi \right)}{n} \right) ^2}=\sum_{n=1}^{\infty}{\left( \int_0^{\pi}{\cos nx\ln x\text{d}x} \right) ^2}\]
考虑函数$f(x)=\ln x,x\in(0,\pi)$,作偶函数延拓和$\pi$周期延拓得到的余弦级数为
\[\widetilde{f}\left( x \right) =\frac{a_0}{2}+\sum_{n=1}^{\infty}{a_n\cos nx}\]
其中$\displaystyle{a_n=\frac{2}{\pi}\int_0^{\pi}{\cos nx\ln x\text{d}x}},a_0=2\ln\pi-2$,由Parseval定理得
\[\frac{a_{0}^{2}}{2}+\sum_{n=1}^{\infty}{a_{n}^{2}}=\frac{2}{\pi}\int_0^{\pi}{f^2\left( x \right) \text{d}x}=\frac{2}{\pi}\int_0^{\pi}{\ln ^2x\text{d}x}=4-4\ln \pi +2\ln ^2\pi\]
因此我们最后得到
\[\sum_{n=1}^{\infty}{\left( \frac{\text{Si}\left( n\pi \right)}{n} \right) ^2}=\sum_{n=1}^{\infty}{\left( \int_0^{\pi}{\cos nx\ln x\text{d}x} \right) ^2}=\frac{\pi^2}2\]
好吧,算到这里,虽然解决了这三个美国数学月刊征解题,但是干脆再无聊点,下面我们算两个五次方的级数,其实套路都一样,但是计算量增加很多.
\[\sum_{n=1}^{\infty}{\frac{\text{Si}\left( n\pi \right)}{n^5}}=\frac{269}{43200}\pi ^5,\sum_{n=1}^{\infty}{\frac{\left( -1 \right) ^{n-1}\text{Si}\left( n\pi \right)}{n^5}}=\frac{4}{675}\pi ^5\]
从\begin{align*}\text{Si}\left( n\pi \right) &=\int_0^{n\pi}{\frac{\sin t}{t}\text{d}t}=\int_0^{\pi}{\frac{\sin nx}{x}\text{d}x}=\int_0^{\pi}{\sin nx\text{d}\left( \ln x \right)}=-n\int_0^{\pi}{\cos nx\ln x\text{d}x}\\&=-n\int_0^{\pi}{\cos nx\text{d}\left( x\ln x-x \right)}=n\left[ \left( -1 \right) ^{n-1}\left( \pi \ln \pi -\pi \right) -n\int_0^{\pi}{\sin nx\left( x\ln x-x \right) \text{d}x} \right]\end{align*}
出发,可得
\[\sum_{n=1}^{\infty}{\frac{\text{Si}(n\pi)}{n^5}}=\sum_{n=1}^{\infty}{\frac{\left( -1 \right) ^{n-1}}{n^4}\left( \pi \ln \pi -\pi \right)}-\sum_{n=1}^{\infty}{\frac{1}{n^3}\int_0^{\pi}{\sin nx\left( x\ln x-x \right) \text{d}x}}\]
而$\displaystyle{\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^4}=\frac{7\pi^4}{720}}$(利用$\zeta(4)$即可),把后一部分式子再分部积分得
\begin{align*}\sum_{n=1}^{\infty}{\frac{1}{n^3}\int_0^{\pi}{\sin nx\left( x\ln x-x \right) \text{d}x}}&=\sum_{n=1}^{\infty}{\frac{1}{n^3}\int_0^{\pi}{\sin nx\text{d}\left( \frac{1}{2}x^2\ln x-\frac{3}{4}x^2 \right)}}\\&=\sum_{n=1}^{\infty}\frac1{n^2}{\int_0^{\pi}{\left( \frac{3}{4}x^2-\frac{1}{2}x^2\ln x \right) \cos nx\text{d}x}}\\&=\sum_{n=1}^{\infty}{\frac{1}{n^2}\int_0^{\pi}{\cos nx\text{d}\left( \frac{11}{36}x^3-\frac{1}{6}x^3\ln x \right)}}\\&=\left( \frac{11}{36}\pi ^3-\frac{\pi ^3}{6}\ln \pi \right) \sum_{n=1}^{\infty}{\frac{\left( -1 \right) ^n}{n^2}}+\sum_{n=1}^{\infty}{\frac{1}{n}\int_0^{\pi}{\left( \frac{11}{36}x^3-\frac{x^3}{6}\ln x \right) \sin nx\text{d}x}}\\&=-\left( \frac{11}{36}\pi ^3-\frac{\pi ^3}{6}\ln \pi \right) \frac{\pi ^2}{12}+\sum_{n=1}^{\infty}{\frac{1}{n}\int_0^{\pi}{\left( \frac{11}{36}x^3-\frac{x^3}{6}\ln x \right) \sin nx\text{d}x}}.\end{align*}
而
\begin{align*}\sum_{n=1}^{\infty}{\frac{1}{n}\int_0^{\pi}{\left( \frac{11}{36}x^3-\frac{x^3}{6}\ln x \right) \sin nx\text{d}x}}&=\sum_{n=1}^{\infty}{\frac{1}{n}\int_0^{\pi}{\sin nx\text{d}\left( \frac{25}{288}x^4-\frac{1}{24}x^4\ln x \right)}}\\&=\sum_{n=1}^{\infty}{\int_0^{\pi}{\left( \frac{25}{288}x^4-\frac{1}{24}x^4\ln x \right) \cos nx\text{d}x}}.\end{align*}
下面考虑函数\[f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}{\frac{{25}}{{288}}{x^4} - \frac{1}{{24}}{x^4}\ln x,}&{x \in \left( {0,\pi } \right]}\\{0,}&{x = 0}\end{array}} \right.\]及其余弦级数可得\[\frac{a_0}{2}+\sum_{n=1}^{\infty}{a_n}=f\left( 0 \right) =0\]
其中$\displaystyle{a_n=\frac{2}{\pi}\int_0^{\pi}{\left( \frac{25}{288}x^4-\frac{1}{24}x^4\ln x \right) \cos nx\text{d}x},a_0=\frac{\pi ^4\left( 137-60\ln \pi \right)}{3600}}$,于是$\displaystyle{\sum_{n=1}^{\infty}{a_n}=-\frac{a_0}{2}=-\frac{\pi ^4\left( 137-60\ln \pi \right)}{7200}}$,因此我们最后得到
\[\sum_{n=1}^{\infty}{\frac{\text{Si}\left( n\pi \right)}{n^5}}=\frac{7\pi ^4}{720}\left( \pi \ln \pi -\pi \right) +\frac{\pi ^2}{12}\left( \frac{11\pi ^3}{36}-\frac{\pi ^3}{6}\ln \pi \right) -\frac{\pi}{2}\cdot \frac{\pi ^4\left( 137-60\ln \pi \right)}{7200}=\frac{269}{43200}\pi ^5\]
第二个级数同理.
最后再算一个级数
\[\sum_{n=1}^{\infty}{\frac{\text{Si}^2\left( n\pi \right)}{n^4}}=\frac{\pi ^4}{27}\]
解.首先
\[\frac{\text{Si}\left( n\pi \right)}{n^2}=\frac{\left( -1 \right) ^{n-1}}{n}\left( \pi \ln \pi -\pi \right) -\int_0^{\pi}{\sin nx\left( x\ln x-x \right) \text{d}x}\]
把上式两边乘以$\sin nx$求和,利用Fourier级数的收敛定理得
\begin{align*}\sum_{n=1}^{\infty}{\frac{\text{Si}\left( n\pi \right)}{n^2}\sin nx}&=\left( \pi \ln \pi -\pi \right) \sum_{n=1}^{\infty}{\frac{\left( -1 \right) ^{n-1}}{n}\sin nx}-\sum_{n=1}^{\infty}{\sin nx\int_0^{\pi}{\sin nx\left( x\ln x-x \right) \text{d}x}}\\&=\frac{\left( \pi \ln \pi -\pi \right) x}{2}-\frac{\pi \left( x\ln x-x \right)}{2}.\end{align*}
最后利用Paserval定理得
\[\sum_{n=1}^{\infty}{\frac{\text{Si}^2\left( n\pi \right)}{n^4}}=\frac{2}{\pi}\int_0^{\pi}{\left( \frac{\left( \pi \ln \pi -\pi \right) x}{2}-\frac{\pi \left( x\ln x-x \right)}{2} \right) ^2\text{d}x}=\frac{\pi ^4}{27}\]
注：从这里,其实我们发现了对任意整数$k$,$\displaystyle{\sum_{n=1}^{\infty}{\frac{\text{Si}\left( n\pi \right)}{n^{2k-1}}}(k>1),\sum_{n=1}^{\infty}{\left( -1 \right) ^{n-1}\frac{\text{Si}\left( n\pi \right)}{n^{2k-1}}},\sum_{n=1}^{\infty}{\frac{\text{Si}^2\left( n\pi \right)}{n^{2k}}}}$都可以算出来,只是当次数增加时,这个计算量会越来越大,所以再往上算的话估计只有欧拉和拉马努金这种人去了,不知道这个结果是否可以用类似伯努利数的方法给出通项呢？我也解决不了.
最后再做一点额外补充:我把这些题目放到了数学贴吧,某位吧友也提出了一个不错的方法,就是利用下面的Fourier级数
比如
\begin{align*}\sum_{n=1}^{\infty}{\frac{\text{Si}\left( n\pi \right)}{n^3}}&=\sum_{n=1}^{\infty}{\frac{1}{n^3}\int_0^{\pi}{\frac{\sin nx}{x}\text{d}x}}=\sum_{n=1}^{\infty}{\frac{1}{n^3}\int_0^{\pi}{\sin nx\text{d}x}\int_0^{+\infty}{\text{e}^{-xy}\text{d}y}}\\&=\int_0^{+\infty}{\text{d}y}\int_0^{\pi}{\text{e}^{-xy}}\frac{x^3-3\pi x^2+2\pi ^2x}{12}\text{d}y=\frac{5}{72}\pi ^3.\end{align*}
类似的也可以处理,只不过这些Fourier级数光算出来计算量已经不得了了,如果提高到五次或者更高次,计算量更大了,不过对于有平方的式子,恐怕还只能利用Parseval定理了.好了,本期的问题就到这里了,下一期暂时没想好写什么QAQ.
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