Continuing from an earlier question of mine: Fourier-Series of a part-wise defined function?
I now got a fourier series which I believe is the correct one: $$\frac{\pi(b-a)}{2} + \sum\limits_{n=1}^{\infty} \frac{(a-b)(1-(-1)^n)}{n^2\pi}\cos(nx) + \frac{(-1)^n(b-a)}{n}\sin(nx)$$
Now my next task confused me a little - "What is the sum of this series?". By definition, this should just be $f(x)$ ($\frac{ax+bx}{2}$ if $x$ is a whole multiple of $\pi$) right? So I figured I am probably supposed to find a closed form for this - although given the definition of the function, I find it hard to imagine that it even exists. Am I wrong? If so, what is the closed form of this series?
(This wasn't the correct fourier series after all - the right one is
$$\frac{\pi(b-a)}{2} + \sum\limits_{n=1}^{\infty} \frac{(a-b)(1-(-1)^n)}{n^2\pi}\cos(nx) + \frac{(-1)^{n+1}(a+b)}{n}\sin(nx)$$
Hint1: prove following sums for $x \in (-\pi,\pi)$ and conclude using the appropriate linear combination. $$\tag{1} \sum_{n=1}^\infty \frac {\cos(nx)}{n^2}=\frac{(\pi-|x|)^2}4-\frac{\pi^2}{12}$$ $$\tag{2} \sum_{n=1}^\infty \frac {(-1)^n\cos(nx)}{n^2}=\frac{x^2}4-\frac{\pi^2}{12}$$ $$\tag{3} \sum_{n=1}^\infty \frac {(-1)^n\sin(nx)}n=-\frac x2$$
Hint2: To prove these identities start by computing the Fourier series of $f(x)=\frac x2$ : you should get minus the last identity (it is the classical 'Sawtooth wave').
Setting $x:=y+\pi$ you may get a fourth identity ('sign' is the 'Sign function' that is $\pm$) :
$$\tag{4} \sum_{n=1}^\infty \frac {\sin(nx)}n=\operatorname{sign}(x)\frac {\pi-x}2$$ The integral of $(4)$ will give you minus the first identity (the constant of integration is obtained by considering $x$ at $0$).
The integral of $(3)$ will give you minus the second identity.
Hint3: The result is rather simple (and was probably the starting point!) and it should be easier to separate the cases $x\in (-\pi,0)$ and $x\in (0,\pi)$.