The function with Fourier series given by $$f(x)=\sum_{n=1}^\infty \frac{\cos{(nx)}+\sin{(nx)}}n$$ appears to be a curve with vertical asymptotes at $x=2\pi k$ where $k\in\mathbb{Z}$. Is there an elementary closed form for $f(x)$? Wolfram gives us $$f(x)=-\frac12(1+i)(\ln{(1-e^{-ix})}-i\ln{(1 - e^{ix})})$$ but is there a way to simplify the above expression into one which does not involve complex numbers as the function $f(x)$ is clearly real?
Edit: I found that this question may have been asked in different contexts before. I have provided a proof of the related results as one of the answers below.
We can solve each of the sums seperately by considering the sum as the real or imaginary part of $$\sum_{n=1}^\infty \frac{e^{inx}}n=-\ln{(1-e^{ix})},\qquad x\in\mathbb{R}\setminus \{0\}$$ Then we can use the fact that $$\ln{(z)}=\ln{(|z|)}+i\arg{(z)}$$ to seperate the above function into its real and imaginary parts; $$\begin{align} -\ln{(1-e^{ix})} &=-\ln{(1-\cos{(x)}-i\sin{(x)})}\\ &=-\ln{(|1-\cos{(x)}-i\sin{(x)}|)}-i\arg{(1-\cos{(x)}-i\sin{(x)})}\\ &=-\frac12\ln{((1-\cos{(x)})^2+(\sin{(x)})^2)}-i\arctan{\left(\frac{-\sin{(x)}}{1-\cos{(x)}}\right)}\\ &=-\frac12\ln{(2-2\cos{(x)})}+i\arctan{\left(\frac{\sin{(x)}}{1-\cos{(x)}}\right)}\\ &=-\frac12\ln{\left(4\sin^2{\left(\frac{x}2\right)}\right)}+i\arctan{\left(\cot{\left(\frac{x}2\right)}\right)}\\ &=-\ln{\left(2\left|\sin{\left(\frac{x}2\right)}\right|\right)}+i\arctan{\left(\tan{\left(\frac{\pi-x}2\right)}\right)}\\ \end{align}$$ Hence we can evaluate the two sums as $$\sum_{n=1}^\infty \frac{\cos{(nx)}}n=\Re{\left(\sum_{n=1}^\infty \frac{e^{inx}}n\right)}=-\ln{\left(2\left|\sin{\left(\frac{x}2\right)}\right|\right)}$$ $$\sum_{n=1}^\infty \frac{\sin{(nx)}}n=\Im{\left(\sum_{n=1}^\infty \frac{e^{inx}}n\right)}=\arctan{\left(\tan{\left(\frac{\pi-x}2\right)}\right)}$$ For $x\in(0,2\pi)$ this simplifies to the results $$\sum_{n=1}^\infty \frac{\cos{(nx)}}n=-\ln{\left(2\sin{\left(\frac{x}2\right)}\right)}$$ $$\sum_{n=1}^\infty \frac{\sin{(nx)}}n=\frac{\pi-x}2$$