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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 $\int_0^x f(t(dt=\int_0^x \frac{a_0}{2}dt +\sum_{n=1}^\infty \int_0^x (a_n \cos nt +b_n \sin nt)dt$.



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,

$\int_0^x... \int_0^x f(x) dx...dx:=\frac{1}{(m-1)!} \int_0^x f(t) (x-t)^{m-1} dt$.

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If $S_n(t)$ is the $n$'th partial sum of the Fourier series for $f(t)$, you want to show that for any $x \in [0, 2 \pi]$, $\int_0^{2\pi} g(t) S_n(t) \ dt \to \int_0^{2\pi} g(t) f(t) \ dt$ as $n \to \infty$, where $g(t) = (x-t)^{m-1}/(m-1)!$ for $0 < t < x$, 0 otherwise. Now if $T_n(t)$ is the $n$'th partial sum of the Fourier series for $g(t)$, $\int_0^{2 \pi} g(t) S_n(t) = \int_0^{2 \pi} T_n(t) f(t)\ dt$. $g(t)$ is a very nice function, smooth except for jumps at 0 (when identified with $2 \pi$) and $x$, so the partial sums of its Fourier series are bounded and converge to $g(t)$ away from the jumps. Break up $[0,2\pi]$ into small intervals near $0$, $x$ and $2 \pi$, small enough so the integral of $f$ over those intervals is small, and large intervals on which $T_n - g$ is uniformly small...

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Thanks for answer, but there is a thing I dont understand. Does it suffice to show that $\int_0^{2\pi} g(t) S_n(t)dt \rightarrow \int_0^{2\pi} g(t)f(t)dt$ as $n\rightarrow \infty$ for $x \in [0,2 \pi$ ? The function $\int_0^x... \int_0^x f(x) dx...dx=\int_0^{2\pi} g(t)f(t)dt$ is generally not $2\pi$-periodic. – Richard Oct 28 '11 at 6:34
If $x = 2 k \pi + s$ for positive integer $k$, do the change of variables $u = t - 2 j \pi$ on each interval $t \in [2 j \pi, 2 (j+1)\pi]$ for $j = 0, 1, \ldots k$. You'll get a more complicated $g(u)$, but still smooth except for jumps at 0 and $s$. – Robert Israel Oct 28 '11 at 6:59
Thanks for answer. – Richard Oct 28 '11 at 7:05

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