# Fourier series $\sum_{m=0}^\infty \frac{\cos (2m+1)x}{2m+1}$

Does anyone know the sum of Fourier series $$\sum_{m=0}^\infty \frac{\cos (2m+1)x}{2m+1}?$$ I tried WA; it does not return a function.

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Mathematica returns as $\frac{1}{2} \left(\text{ArcTanh}\left[e^{-i x}\right]+\text{ArcTanh}\left[e^{i x}\right]\right)$ – Mula Ko Saag May 15 '13 at 20:38
Isn't it interesting that the sum $\sum_{n=0}^{\infty}\frac{\sin (2n+1)x}{2n+1}$ is so much easier? (It's just a rectangular pulse train: $f(x)=\frac{\pi}{4}\text{sign} (x)$, $-\pi<x<\pi$) – Matt L. May 15 '13 at 21:02
@MattL.: I disagree: the cosine sum is a lot easier because at least you avoid the multivalued aspect of the logarithm. – Ron Gordon May 15 '13 at 21:06
@RonGordon: OK, agreed. What I meant was the definition of $f(x)$ and its visualization. In this sense the rectangular pulse train is much more basic than the cosine sum (speaking from an engineering viewpoint). – Matt L. May 15 '13 at 21:10

Rewrite the series as

$$\Re{\left [ \sum_{n=0}^{\infty} \frac{\left ( e^{i x}\right )^{2 n+1}}{2 n+1} \right]} = \Re{[\text{arctanh}{(e^{i x})}]} = \frac12 \Re{\left[\log{\left(\frac{1+e^{i x}}{1-e^{i x}}\right)}\right]}$$

With some manipulation,noting that $\log{z} = \log{|z|} + i 2 \pi \arg{z}$, we find that

$$\sum_{n=0}^{\infty} \frac{\cos{(2 n+1) x}}{2 n+1} = \frac12 \log{\left |\cot{\frac{x}{2}}\right |}$$

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Isn't this series for $\text{arctanh}(z)$ only valid for $|z|<1$? – Matt L. May 16 '13 at 7:13
I mean the series $\text{arctanh}(z)=\sum_{n=0}^{\infty}\frac{z^{2n+1}}{2n+1}$. I thought it would not converge for $|z|=1$, will it? – Matt L. May 16 '13 at 7:34
@MattL.: It's a similar issue that I had with respect to the series for $\log{z}$, etc. These series may be summed by the Abel method. See encyclopediaofmath.org/index.php/Abel_summation_method – Ron Gordon May 16 '13 at 8:02
OK, thanks for your explanation. I'm not sure if I entirely understand Abel summation, but I blame it on my lack of knowledge ... – Matt L. May 16 '13 at 9:27