Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be $2\pi$-periodic and integrable on $[-\pi, \pi]$. Assume that $f(x)\sim \frac{a_0}{2}+\sum_{n=1}^\infty (a_n \cos nx+b_n \sin nx )$ is its Fourier series.
How to prove that $ \int_0^x \ldots \int_0^x f(x) dx\ldots dx=\int_0^x \ldots \int_0^x \frac{a_0}{2} dx\ldots dx+ \atop \sum_{n=1}^\infty \int_0^x \ldots \int_0^x (a_n \cos nx+ b_n \sin nx) dx \ldots dx, $ where in both sides of equality are $m$-fold integrals.
For $m=1$ the proof is following. Let's consider the function $F(x)=\int_0^x (f(t)-\frac{a_0}{2})dt$ for $x\in \mathbb{R}$, where $a_0=\frac{1}{\pi} \int_{-\pi}^\pi f(t)dt$. Then $F$ is $2\pi$-periodic and absolutely continuous. Assume that $F(x) \sim \frac{A_0}{2}+\sum_{n=1}^\infty (A_n \cos nx+B_n \sin nx)$. Then by Dirichlet-Jordan theorem
$ (*) \ \ F(x) = \frac{A_0}{2}+\sum_{n=1}^\infty (A_n \cos nx+B_n \sin nx) \ \ for \ x\in \mathbb{R}. $ (even uniformly)
By definition of Fourier coefficients and formula of integration by parts we find $A_n=-\frac{b_n}{n}$, $B_n=\frac{a_n}{n}$ for $n \in \mathbb{N}$. Taking in $(*)$ $x=0$ we obtain $\frac{A_0}{2}=-\sum_{n=1}^\infty A_n=\sum_{n=1}^\infty \frac{b_n}{n}$. By $(*)$ we have now $\int_0^x f(t(dt=\int_0^x \frac{a_0}{2}dt +\sum_{n=1}^\infty \int_0^x (a_n \cos nt +b_n \sin nt)dt$.
Thanks.
P.S.
By $m-fold$ integral $\int_0^x... \int_0^x f(x) dx...dx$ I mean
$\int_0^x... \int_0^x f(x) dx...dx:=\int_0^x[ \int_0^{x_m}...(\int_0^{x_3}(\int_0^{x_2} f(x_1) dx_1) dx_2)...dx_{m-1}]dx_m$.
It is equal, by Cauchy's formula,
$\int_0^x... \int_0^x f(x) dx...dx:=\frac{1}{(m-1)!} \int_0^x f(t) (x-t)^{m-1} dt$.