# Computing $\sum_{n=1}^{\infty} \frac{e^{-An}sin(Bn)}{n}$, where $A,B$ are constants.

Problem

Compute: $$S = \sum_{n=1}^{\infty} \frac{e^{-An}\sin(Bn)}{n}$$

My attempt, let $$I = \sum_{n=1}^{\infty} \frac{e^{-An}}{n} \left( \cos(Bn) + i \sin(Bn) \right) = \sum_{n=1}^{\infty} \frac{e^{n(-A + iB)}}{n}$$

Using the series expansion,

$$\sum_{x=1}^{\infty} \frac{e^{zx}}{x} = \log{\left( \frac{1}{1-e^{iz}} \right)}$$

we obtain

$$I = \sum_{n=1}^{\infty} \frac{e^{n(-A + iB)}}{n} = \log{\left( \frac{1}{1-e^{i(-A + iB)}} \right)}$$

Which gives, $$S = \Im(I) = \Im\left( \log{\left( \frac{1}{1-e^{i(-A + iB)}} \right)} \right).$$

However, I am not sure of this answer. This series originated while doing a problem of Physics, whose solution I do not possess. Any methods to do it more precisely are appreciated.

It seems that you have an extra $$i$$ $$I = \sum_{n=1}^{\infty} \frac{e^{n(-A + iB)}}{n} =-\log \left(1-e^{-A+i B}\right)$$
Expanding the complex numbers $$\Im(I)=\tan ^{-1}\left(\frac{\sin (B)}{e^A-\cos (B)}\right)$$ $$\Re(I)=\frac{1}{2} (A-\log (2 \cosh (A)-2 \cos (B)))$$
• Yes I accidentally put an $i$ n the Taylor expansion. Oct 1, 2023 at 12:21