级数化积分

倪举鹏

sum(cos(k*ln(n))/sqrt(n), n = 1 .. infinity)
这个级数可以化为积分再求解么，我想让sum(cos(k*ln(n))/sqrt(n), n = 1 .. infinity)=0  再求解K的值

Buffalo

\(\D\sum_{n=1}^{\infty}\frac{\cos (k \ln n)}{\sqrt{n}}=\sum_{n=1}^{\infty}\frac{e^{ik \ln n}+e^{-ik \ln n}}{2\sqrt{n}}=\sum_{n=1}^{\infty}\frac{n^{ik}+n^{-ik}}{2\sqrt{n}}=\frac12\Bigg(\zeta\bigg(\frac12-ik\bigg)+\zeta\bigg(\frac12+ik\bigg)\Bigg)\)，\(\zeta(z)\)是黎曼\(\zeta\)函数

wayne

楼上的还可以继续化简:

1. 1/2 (Zeta[1/2 + I k ] + Zeta[1/2 - I k]) = Re(Zeta[I k + 1/2]) = sinh(RiemannSiegelTheta[k])

复制代码

\(\frac{1}{2} \left(\zeta \left(i k+\frac{1}{2}\right)+\zeta \left(\frac{1}{2}-i k\right)\right)=\text{Re}\left(\zeta \left(i k+\frac{1}{2}\right)\right)=\sinh(\vartheta (k))\)

也即是RiemannSiegelTheta函数 \(\vartheta (k)\)的零点 ,得到 $k=+-17.845599540410860816826338412519097035693287433696$

wayne

K值还有另一层内涵,第0个gram point
http://mathworld.wolfram.com/GramPoint.html

http://en.wikipedia.org/wiki/Rie ... egel_theta_function