Fourier series of $x-x^2$

Question

Let $f:\mathbb R \to \mathbb R$ be the odd function defined by $f (x) =x-x^2 $ $\ \forall x\in [0,1]$. I must calculate the Fourier series of $f $ and determine if it converges pointwise and uniformly to $f $. After, I must calculate $\sum \frac 1 {n^2}$, probably using Plancherel formula. However I am almost surely making some mistake in the first point: $$\frac 1 2\int^2_0 (x-x^2)e^{-i\pi nx}dx=\frac 1 2\int x e^{-i\pi nx}dx-\frac 1 2\int x^2e^{-i\pi nx}dx=$$ $$=\frac1 {-i\pi n}-\frac1 2\left [ \frac {x^2e^{-i\pi nx}}{-i\pi n}\right ]_0^2+\frac1 {-i\pi n}\int x e^{-i\pi nx}dx = -\frac 1 {i\pi n}+\frac 2 {i\pi n}-\frac 1 {\pi^2 n^2}.$$ I really couldn't find the mistake, and if there isn't one, then I have no idea on how to calculate the sum above. About the pointwise convergence, since $f $ is continuous on $\mathbb R $ and its derivative exists in all points but the integers, the series converges everywhere (pointwise). How could I proceed for uniform convergence? My idea is to show that $S_n =\sum_{-n}^n\tilde {f}e^{i\pi nx} $ is a Cauchy succession (with respect to $||\cdot ||_{\infty}$). Thanks for any help.

Answer 1

Message to OP: the integrations should be in [0,1].

The Fourier series for a function $f(x)$ in the interval [0,1] is given as $$f(x)=\frac{a_0}{2}+\sum_{n=1}^{\infty} [a_n \cos (2\pi nx)+ b_n \sin (2\pi nx)].$$ Here the function is $f(x)=x^2-x$, $$a_n=2 \int_{0}^{1} (x^2-x) \cos (2 \pi nx) dx=\frac{1}{n^2\pi^2}$$ Check that $$b_n=2 \int_{0}^{1} (x^2-x) \sin(2\pi nx) dx=0, a_0=2 \int(x^2-x) dx=-\frac{1}{3}$$ It may be noted that $b_n$ vanishes because $\int_{0}^{z} f(z-x) dx= \int_{0}^{z} f(x) dx$ So finally $$x^2-x=-\frac {1}{6}+\sum_{n=1}^{n} \frac{1}{\pi^2 n^2} \cos(2\pi n x), x \in [0,1].$$