Solution of a Sum of Cos(m*delta)/m^2 function

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Mukul
Mukul on 8 Feb 2019
I am trying to solve for
m = 2*n - 1;
Sum[Cos[m*d]/m^2, {n, 1, Infinity}]
and end up with a series of LerchPhi function as below:
1/8 e^(-i*d) (LerchPhi[e^(-2*i*d, 2, 1/2] + e^(2*i*d) LerchPhi[e^(2*i*d), 2, 1/2])
I know the simplified answer to this would be
Pi^2/8*[1 - (2 Abs (d))/Pi]
Can anyone help me to find out a way to get the simplified answer. I run the code in Mathematica. Any suggestions from users who are are using Matlab/Maple?

Answers (0)

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