moment of wrapped function (flaw in a proof) I know the moments of a real-valued function $f$:
$$\int_{-\infty}^{\infty}x^{n}f(x)dx=\begin{cases}
1 & ,\ n=0\\
0 & ,\ n=2,...q-1\\
c & ,\ n=q
\end{cases} $$
What can I say about the wrapped function $g$:
$$g(x)=\sum_{j=-\infty}^{\infty}f(x+j) $$
$g(x)$-is a periodic function with Fourier coeffitients:
   $$a(s)=\int_{0}^{1}g(x)\exp(-2\pi ixs)dx=\int_{-\infty}^{\infty}f(x)\exp(-2\pi ixs)dx$$
Can I use inverse Fourier transform on this  step? If so, then 
$$f(x)=\int_{-\infty}^{\infty}\overline{a(s)}\exp(-2\pi isx)ds  $$         $$\int_{-\infty}^{\infty}x{}^{n}f(x)dx=\frac{1}{(2\pi i)^{n}}\frac{\partial^{n}}{\partial s^{n}}\left[\overline{a(s)}\right]_{s=0} $$
(I used Integral with Fourier transform)
From the definition of Fourier coefficients:
$$\frac{\partial^{n}}{\partial s^{n}}\left[\overline{a(s)}\right]=\int_{0}^{1}(2\pi ix)^{n}g(x)\exp(2\pi isx)dx $$
Hence:$$\int_{0}^{1}x^{n}g(x)dx=\int_{-\infty}^{\infty}x{}^{n}f(x)dx $$
But it is not true!(I've plotted it). Where is my mistake?
 A: Since $g(x)$ is periodic, the integral formula for its Fourier coefficients is only correct for integer index $s$. For example, we can write down the same integral, but exponentials with random coefficient in the exponent aren't orthogonal on $[0,1]$, and so on. Thus, differentiation in that parameter doesn't really have the sense we'd might have thought. The coefficient(s) $a(s)$ for $g$ are truly functions of $s\in\mathbb Z$, so differentiation has no natural sense. It is true that the $n$th Fourier coefficient of the $k$th derivative of $g$ is $(2\pi in)^k$ times the $n$th Fourier coefficient of $g$, but that's a different question. 
Edit: in response to further comment... Yes, the Fourier coefficient $a(s)$ of $f$ is defined on the whole real line, and the identity about differentiating it with respect to $s$ is correct. However, to "wind up" (or whatever verb one wants) that identity does not give the integral of $g$ against $x^n$, but, instead $\int_0^1 \sum_{\ell\in\mathbb Z} (x+\ell)^n\,f(x+\ell)\,dx$. That is, the thing inside the integral is not $x^n \sum_\ell f(x+\ell)$, because the $x^n$ has to get summed over translates, too.
Edit 2: In response to second comment... it is not the case that the $n$th derivative $a^{(n)}(s)$ of $a(s)$ is $\int_0^1 x^n\,g(x)\,dx$ (with or without constants). Yes, the corresponding identity does hold with $f$ instead of $g$, but when you try to deduce the $g$-assertion from the $f$-assertion, the winding-up fails, because $x^n$ is not periodic, unlike the exponential.
Just for context, I remember when first learning calculus, thinking that I'd found an easier way to solve high-degree polynomial equations: for example, to solve $x^2=10$, _take_derivatives_ (haha!) of both sides, supposedly getting $2x=0$, so $x=0$??? No. Sure, one can take the derivative of an expression, but that does not guarantee harmony with the context, even if it sorta makes sense. The problem with the pollynomials was that $x$ is just a name for an unknown constant, so it makes no sense to take a derivative with respect to it. There's a similar problem, in part, in this Fourier business. 
A: If you want to compute $\int_{0}^{1}x^{n} g(x)dx$, i.e., consider the wrapped function over one period, you should better consider $h(x) = w(x)   g(x) $  where $w(x)$ is a rectangular window ($w(x)=1$ in $[0,1]$, zero elsewhere).
$$ h(x) = w(x) g(x) = w(x) [ f(x) \star \sum_k \delta(x-k) ]$$
where $\star$ is a convolution. Then, if we call $F(\omega)$ the FT of $f(x)$, we have
$$ H(\omega) =  W(\omega) \star \sum_n \sqrt{2 \pi} \; F(\omega_n) \delta(\omega-\omega_n) =  \sqrt{2 \pi} \; \sum_n F(\omega_n) W(\omega - \omega_n) $$
where $\omega_n = 2 \pi n$ and $W(\omega)$ has a $sinc()$ shape.
We are interested in the $n$ derivative of $H(\omega)$ at $\omega=0$ But
$$ H'(0)= \sqrt{2 \pi} \; \sum_n F(\omega_n) W'(-\omega_n)$$
Hence it's seen that the first derivative (and the higher ones) cannot be expressed in terms of the derivatives of $F(\omega)$ at $\omega=0$.
I think, BTW, that this clarifies the confusion from other comments: how it's that we can derivate with respect to the $frequency$, when at the same time the fourier transform of the (unwindowed) wrapped signal is a fourier series, defined only in discrete values.
