Evaluating ∑n=1∞sin⁡(nx)n without integrating ∑n=1∞enx My solution proceeds by integrating ∑n=1∞enx=eix1−eix

Lizeth Jackson

Lizeth Jackson

Answered

2022-01-26

Evaluating n=1sin(nx)n without integrating n=1enx
My solution proceeds by integrating
n=1enx=eix1eix

Answer & Explanation

Emilie Booker

Emilie Booker

Expert

2022-01-27Added 14 answers

Note that
n=1sin(nx)n=Im(n=1exn)
=Im(n=1(eix)nn)
If xR {0}, then eixS1 {1} and therefore
n=1(eix)nn=log(11eix)
(where log is the principal branch of the logarithm), since
log(1+x)=xx22+x33x4x+log(1x)=(x+x22+x33+x44+)
log(11x)=x+x22+x33+x44+
Therefore
n=1sin(nx)n=arctan(sinx1cos(x))
=arctan(cot(x2))
Of course, if x=0, then this equality also holds.

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