如何用Fourier系数求两道级数和

青青子衿

\[ \sum\limits_{n=0}^{\infty }\dfrac{1}{n^4+1} \]
\[ \sum\limits_{n=0}^{\infty }\dfrac{1}{n^4+2n^2+1} \]
forum.php?mod=viewthread&tid=1668

战巡

回复 1# 青青子衿

第一个没啥意思

\[\sum_{n=0}^\infty\frac{1}{n^4+1}=\frac{1}{2i}\sum_{n=0}^\infty[\frac{1}{n^2-i}-\frac{1}{n^2+i}]\]
然后套用你引用那个贴里我的办法就行了

战巡

回复 1# 青青子衿

第二个，考虑对
\[f(x)=\sinh(x)x\]
进行展开，会得到
\[\sinh(x)x=\frac{1}{\pi}\int_0^\pi\sinh(t)tdt+\frac{2}{\pi}\sum_{n=1}^\infty\cos(nx)\int_0^{\pi}\sinh(t)t\cos(nt)dt\]
\[=-\frac{1}{\pi}\int_0^\pi\sinh(t)tdt+\frac{2}{\pi}\sum_{n=0}^\infty\cos(nx)\left[\frac{(-1)^n[(n^2+1)\pi\cosh(\pi)+(n^2-1)\sinh(\pi)}{(n^2+1)^2}\right]\]
\[=\frac{\sinh(\pi)}{\pi}-\cosh(\pi)+\frac{2}{\pi}\sum_{n=0}^\infty\cos(nx)\left[\frac{(-1)^n(\pi\cosh(\pi)+\sinh(\pi))}{n^2+1}-\frac{2(-1)^n\sinh(\pi)}{(n^2+1)^2}\right]\]

令$x=\pi$会得到
\[\sinh(\pi)\pi=\frac{\sinh(\pi)}{\pi}-\cosh(\pi)+\frac{2}{\pi}\sum_{n=0}^\infty\left[\frac{(\pi\cosh(\pi)+\sinh(\pi))}{n^2+1}-\frac{2\sinh(\pi)}{(n^2+1)^2}\right]\]
\[\sinh(\pi)\pi^2=\sinh(\pi)-\pi\cosh(\pi)+(1+\pi\coth(\pi))(\pi\cosh(\pi)+\sinh(\pi))-4\sinh(\pi)\sum_{n=0}^\infty\frac{1}{(n^2+1)^2}\]
\[\sum_{n=0}^\infty\frac{1}{(n^2+1)^2}=\frac{1}{4}(\frac{\pi^2}{\sinh(\pi)^2}+\pi\coth(\pi)+2)\]

青青子衿

回复  青青子衿
第一个没啥意思
\[\sum_{n=0}^\infty\frac{1}{n^4+1}=\frac{1}{2i}\sum_{n=0}^\infty[\frac{1}{n^2-i}-\frac{1}{n^2+i}]\] ...
战巡 发表于 2019-4-30 21:56

谢谢战版！
\[ \sum_{n=0}^{\infty} \frac {1}{n^4+1} = \frac{\pi}{2\sqrt 2} \left(\frac {\sinh\left(\sqrt{2}\,\pi\right)+\sin\left(\sqrt{2}\,\pi\right)}{\cosh\left(\sqrt{2}\,\pi\right)-\cos\left(\sqrt{2}\,\pi\right)}\right)+\frac{1}{2} \]
artofproblemsolving.com/community/c7h1843686

青青子衿

回复 4# 青青子衿
\[ \color{black}{\sum_{n=1}^\infty\frac1{n^4+n^2+1}=\frac{\pi\sqrt3}6\tanh\left(\pi\frac{\sqrt3}2\right)-\frac12\tag{7}} \]
math.stackexchange.com/questions/966942/evalu … 11n2n4/966977#966977