Monday, 6 July 2015

calculus - Calculate the following infinite sum in a closed form sumin=1nfty(ntextarccotn1)?



Is it possible to calculate the following infinite sum in a closed form? If yes, please point me to the right direction.
n=1(n arccot n1)


Answer



ncot1(n)1=ntan1(1n)1=n1/n0dx1+x21=n1/n0x2dx1+x2=10x2dxn2+x2
Using formula (9) from this answer and substituting zix, we get
n=11n2+x2=πcoth(πx)2x12x2
Combining (1) and (2) yields
n=1(ncot1(n)1)=1210(1πxcoth(πx))dx=1210(1πx(1+2e2πx1e2πx))dx=2π4π10xe2πx1e2πxdx=2π4π10x(n=1e2πnx)dx=2π4πn=1(1(2πn)2(12πn+1(2πn)2)e2πn)=2π4π2412log(1e2π)+14πLi2(e2π)=12+17π2412log(e2π1)+14πLi2(e2π)


No comments:

Post a Comment

real analysis - How to find limhrightarrow0fracsin(ha)h

How to find limh0sin(ha)h without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...