# Integral representation of a partial sum of a Fourier series using complex exponentials

Consider a periodic function $f: \mathbb{R} \to \mathbb{C}$ of period $T = 2\pi/\omega_0$. What I'd like to know is the integral representation of a partial sum of its Fourier series using complex exponentials.

I have the following guess: let $z_0 := e^{j\omega_0t}$. Then the partial sum is $$S_k(x) = \int_{0}^{T} \left(\frac{f(x+t)+f(x-t)}{2}\right) \left(\frac{z_0^k - 1}{z_0 - 1}\right)\,dt\quad.$$

Please let me know if this expression is correct.

-
Just to clarify (I apologize if my questions are obvious), what is $x$ in your equation ? The variable of $s_k$ ? Your partial sum $s_k$ is taken between $-k$ and $k$ ? Since you seem to have a formula, what is exactly your question ? –  Joel Cohen Jun 8 '11 at 13:03
@Joel Cohen : it is $s_k(x) = \int\limits_{0}^{T} (\frac{f(x+t)+f(x-t)}{2}) (\frac{z_o^k - 1}{z_o - 1})\,dt$ –  Rajesh D Jun 8 '11 at 13:54
@Joel Cohen : I'd like to know if it is correct or anything missing. –  Rajesh D Jun 8 '11 at 13:55

First, denote $f_1*f_2$ for the convolution product of functions, that is

$$f_1*f_2(x) := \int_0^T f_1(t) f_2(x-t) \, dt \underset{_{u=x-t}}{=} \int_0^T f_1(x-u) f_2(u) \, du = f_2*f_1(x)$$

Now for $k \in \mathbb{R}$, write $z^k(t) = e^{j k \omega_0 t}$ (for $k$ integer this is an actual power of $z$, otherwise this is a slight abuse of notation) and $s_k(t) = \sin(k \omega_0 t)$. Your partial sum is

$$s_n = \sum_{k = -n}^n f * z^k = f * \sum_{k = -n}^n z^k$$

Now, we compute the sum (usually called Dirichlet kernel). You can check that

$$\sum_{k = -n}^n z^k = \frac{z^{n+1}-z^{-n}}{z-1} = \frac{z^{n+1/2}-z^{-n-1/2}}{z^{1/2}-z^{-1/2}} = \frac{s_{n+1/2}}{s_{1/2}}$$

In the end, you get

$$s_n(x) = \int_0^T f(x-t) \frac{s_{n+1/2}}{s_{1/2}}(t) \, dt \underset{_{u=-t}}{=} \int_0^T f(x+u) \frac{s_{n+1/2}}{s_{1/2}}(u) \, du$$

And averaging yields

$$s_n(x) = \int_0^T \frac{f(x+t) + f(x-t)}{2} \frac{s_{n+1/2}}{s_{1/2}}(t) \, dt$$

-