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