Proving the Series Sum of a Trigonometric Function with Exponentials

[arpon]

Homework Statement

Prove that,
$$\sum _{n=1,3,5...} \frac{1}{n} e^{-nx} \sin{ny} = \frac{1}{2}\tan^{-1} (\frac{\sin{y}}{\sinh{x}})$$

Homework Equations

$$\tan^{-1}{x} = x - \frac{x^3}{3} +\frac{x^5}{5} - ... $$

3. The Attempt at a Solution

$$\sum _{n=1,3,5...} \frac{1}{n} e^{-nx} \sin{ny}$$
$$= \sum _{n=1,3,5...} \frac{1}{n} e^{-nx} Im(e^{i ny})$$
$$= \sum _{n=1,3,5...} \frac{ Im(e^{(-x+iy)n})}{n} $$
$$=Im( \sum _{n=1,3,5...} \frac{z^n}{n}) $$
where $z=e^{-x+iy}$.
I am stuck here. Any help will be appreciated.

 

[mfb]

Looking at $\sum_n \frac {(iz)^n}{n}$ looks promising. Note the sum over all n >=1.

 

[Sowmitra Das]

Preliminaries... I'm going to use 2 identities.
Identity 1: $$\displaystyle \sum_{n\,\textrm{odd}}\frac{x^n}{n}=\frac 1 2 \ln \left(\frac {1+x} {1-x} \right)$$
Proof: From the Maclaurin expansion of $\ln (1+x)$ and $\ln(1-x)$, we get,
$$\ln (1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\ldots$$
$$\ln (1-x)=-x-\frac{x^2}{2}-\frac{x^3}{3}-\frac{x^4}{4}-\ldots$$
Subtracting the 2nd expression from the 1st, we get,
$$\ln (1+x)-\ln (1-x)=2\left(x+\frac{x^3}{3}+\frac{x^5}{5}+\ldots\right)=2\sum_{n\,\textrm{odd}}\frac{x^n}{n}$$
$$ \therefore \sum_{n\,\textrm{odd}}{\frac{x^n}{n}} = \frac{1}{2} \ln \left( \frac{1+x}{1-x} \right) $$Identity 2: $$\displaystyle \tan^{-1}x=\frac{i}{2}\ln\left(\frac{i+z}{i-z}\right)$$
Proof: Let, $\tan^{-1}x=w$. So,
$$ x=\tan w=\frac{\sin w}{\cos w}= \frac{\frac{e^{iw}-e^{-iw}}{2i}}{\frac{e^{iw}+e^{-iw}}{2}}= \frac{1}{i}\cdot\frac{e^{iw}-e^{-iw}}{e^{iw}+e^{-iw}} $$
\begin{align}
& \therefore x=\frac{1}{i}\cdot\frac{e^{iw}-e^{-iw}}{e^{iw}+e^{-iw}} \nonumber \\
& \Rightarrow x=-i\cdot\frac{e^{iw}-e^{-iw}}{e^{iw}+e^{-iw}} \nonumber\\
& \Rightarrow x(e^{iw}+e^{-iw})=-i(e^{iw}-e^{-iw}) \nonumber \\
& \Rightarrow e^{iw}(i+x)=e^{-iw}(i-x) \nonumber \\
& \Rightarrow e^{i2w}=\frac{i-x}{i+x} \nonumber \\
& \Rightarrow i2w=\ln\left(\frac{i-x}{i+x}\right) \nonumber \\
& \Rightarrow w=\frac{1}{2i}\cdot\ln\left(\frac{i-x}{i+x}\right)=-\frac{i}{2}\ln\left(\frac{i-x}{i+x}\right)=\frac{i}{2}\ln\left(\frac{i+x}{i-x}\right) \nonumber \\
& \therefore \tan^{-1}x=\frac{i}{2}\ln\left(\frac{i+z}{i-z}\right) \nonumber \\
\end{align}

Now, for the proof of the problem...
\begin{align}
\sum_{n\, \textrm{odd}} \frac{1}{n} e^{-nx} \sin{ny} & = \sum_{n\, \textrm{odd}} \frac{1}{n} e^{-nx}\cdot \frac{e^{iny}-e^{-iny}}{2i} \nonumber\\
& = \sum_{n\, \textrm{odd}} \frac 1 n\cdot \frac{e^{-nx+iny}-e^{-nx-iny}}{2i} \nonumber\\
&= \frac{1}{2i} \left[\sum_{n\, \textrm{odd}} \frac{(e^{-x+iy})^n}{n} -\sum_{n\, \textrm{odd}} \frac{(e^{-x-iy})^n}{n}\right] \nonumber\\
&= \frac{1}{2i} \left[\sum_{n\, \textrm{odd}} \frac{z^n}{n} - \sum_{n\, \textrm{odd}} \frac{{\bar z}^n}{n}\right]\qquad (z=e^{-x+iy},\; \bar z = e^{-x-iy}) \nonumber\\
&= \frac{1}{2i}\left[\frac{1}{2}\ln \left(\frac{1+z}{1-z}\right)-\frac{1}{2}\ln \left(\frac{1+\bar z}{1-\bar z}\right)\right]\quad \textrm{(Identity 1)} \nonumber\\
&= \frac 1 2 \left[\frac{1}{2i}\ln \left(\frac{1+z}{1-z}\right)-\frac{1}{2i}\ln \left(\frac{1+\bar z}{1-\bar z}\right)\right] \nonumber\\
&= \frac 1 2 \left[-\frac{i}{2}\ln \left(\frac{i+iz}{i-iz}\right)+\frac{i}{2}\ln \left(\frac{i+i\bar z}{i-i\bar z}\right)\right] \nonumber\\
&= \frac 1 2 \left[-\tan^{-1}iz+\tan^{-1}i\bar z\right] \qquad \textrm{(Identity 2)} \nonumber\\
&= \frac 1 2 \left[\tan^{-1}i\bar z-\tan^{-1}iz\right] \nonumber\\
&= \frac 1 2 \tan^{-1}\left[\frac{i\bar z-iz}{1+i^2z\bar z}\right] \nonumber\\
&= \frac 1 2 \tan^{-1}\left[\frac{ie^{-x-iy}-ie^{-x+iy}}{1-e^{-x-iy}\cdot e^{-x+iy}}\right] \nonumber\\
&= \frac 1 2 \tan^{-1}\left[\frac{ie^{-x}(e^{-iy}-e^{iy})}{1-e^{-2x}}\right] \nonumber\\
&= \frac 1 2 \tan^{-1}\left[\frac{ie^{-x}(-2i\sin y)}{1-e^{-2x}}\right] \nonumber\\
&= \frac 1 2 \tan^{-1}\left[\frac{2\sin y}{e^x-e^{-x}}\right] \nonumber\\
&= \frac 1 2 \tan^{-1}\left[\frac{\sin y}{\frac{e^x-e^{-x}}{2}}\right] \nonumber\\
&= \frac 1 2 \tan^{-1}\left(\frac{\sin y}{\sinh y}\right) \nonumber\\
\end{align}

And... we're done. :D :D :D