Let the complex exponential Fourier series coefficients of two periodic signals $x_1(t)$ and $x_2(t)$ be $C_{1n}$ and $C_{2n}$, respectively, with $T_0$ being the fundamental time period of both the signals.
Let the signal $x(t)$ be equal to $x_1(t) * x_2(t)$, where $*$ denotes the periodic convolution operation, and its complex exponential Fourier series coefficient be $C_n$.
The aim is to find a relationship between $C_n$, $C_{1n}$ and $C_{2n}$.
$$C_n = \frac1{T_0} \int_{T_0}\left(x(t).e^{-jnw_0t}\right)dt$$ $$= \frac{1}{T_0} \int_0^{T_0}\left((x_1(t) * x_2(t)).e^{-jnw_0t}\right)dt $$$$= \frac{1}{T_0} \int_0^{T_0}\left(\left(\int_0^{T_0}\left(x_1(\tau) . x_2(t - \tau)\right)d\tau\right).e^{-jnw_0t}\right)dt$$
Since $e^{-jnw_0t}$ is constant with respect to the integral $\int_0^{T_0}\left(x_1(\tau) . x_2(t - \tau)\right)d\tau$, therefore we can write
$$C_n = \frac{1}{T_0} \int_0^{T_0}\left(\int_0^{T_0}\left(e^{-jnw_0t}.x_1(\tau) . x_2(t - \tau)\right)d\tau\right)dt $$$$= \frac{1}{T_0} \int_0^{T_0}\left(\int_0^{T_0}\left(\left(x_1(\tau).e^{-jnw_0\tau}\right).\left(x_2(t-\tau).e^{-jnw_0(t-\tau)}\right)\right)d\tau\right)dt$$
Now, I would like to break this double integral into a product of two integrals, but since $x_2(t-\tau).e^{-jnw_0(t-\tau)}$ involves both $t$ and $\tau$, therefore I don't know how to proceed further.
Could someone help me with this?