Let $f,g$ be functions such that $f(x)=x^2 \ , \ g(x)=x^3+\pi^2 x$.
Use Fourier series of $f$ to evaluate Fourier series of $g$. Domain is $[-\pi,\pi]$.
My try:
The Fourier series of $f(x)$ is $\displaystyle f(x)\sim \frac {\pi^2}{3}+4\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}\cos(nx)$.
Define $\displaystyle h(x)=f(x)+\frac{\pi^2}{3}$. $h(x)$ is piecewise continuous, hence we can integrate term by term.
We get $$\int\limits_{0}^{x}h(t)\text{d}t=\int\limits_{0}^{x}\left(x^2+\frac{\pi^2}{3}\right)\text{d}t=\frac{x^3+\pi^2x}{3}=\frac{g(x)}{3}=\frac{2\pi^2}{3}x+4\sum_{n=1}^{\infty}\int\frac{(-1)^n}{n^2}\cos(nx) \\ \therefore g(x)=2\pi^2 x+12\sum_{n=1}^{\infty}\frac{(-1)^n}{n^3}\sin(nx)$$ My problem is that I don't get Fourier series. How should I tackle it?
Let us write $$f(x)=x^2=\frac{a_0}2+\sum_{n=1}^\infty a_n\,\cos(nx)$$ the Fourier series of $f$ on $[-\pi,\,\pi]$. As $f$ is even, we only have cosines. The derivative of $f$ is $$f'(x)=2x=-\sum_{n=1}^\infty na_n\sin(nx)\tag{1}$$ and the antiderivative of $f$ is $$\int_0^xf(y)\mathrm dy=\frac{x^3}3=\frac{a_0}2x+\sum_{n=1}^\infty \frac{a_n}n\sin(nx).\tag{2}$$ Using (1) in (2) we get $$\int_0^xf(y)\mathrm dy=\frac{x^3}3=-\frac{a_0}4\sum_{n=1}^\infty na_n\sin(nx)+\sum_{n=1}^\infty \frac{a_n}n\sin(nx).\tag{3}$$ Let us use (1) and (3) to express $$g(x)=x^3+\pi^2x=3\int_0^xf(y)\mathrm dy+\frac{\pi^2}2f'(x)= \sum_{n=1}^\infty\left(3\frac{a_n}n-\frac{\pi^2}2na_n-\frac{3a_0}4na_n\right)\sin(nx).$$ The Fourier coefficients of $g$, such that $g(x)=\sum_{n=1}^\infty b_n\sin(nx)$ are therefore $$b_n=\left(\frac{3}n-n\frac{2\pi^2+3a_0}4\right)a_n.$$