If one expands a function $\phi(x,v(\theta),t)$ to complex Fourier: $\sum_{k=-\infty}^{\infty} \phi_k(x,t) e^{ik\theta}$, then where's $v$?

Question

If one expands a function $\phi(x,v(\theta),t)$ to complex Fourier series marked like: $$\phi(x,v(\theta),t)=\sum_{k=-\infty}^{\infty} \phi_k(x,t) e^{ik\theta}$$

then why/how does argument $v$ "disappear"?

Intuitively,

$$\phi_k(x,t)=\frac{1}{2\pi} \int_{-\pi}^{\pi} e^{-int} \phi(x,v(\theta),t) dsomething$$

Would the notation used perhaps suggest that $dsomething=dv$?

And that by "going backwards" one "recovers" $v$ to $\phi(x,v,t)$, when one computes the complex Fourier series?

Answer 1

As given there are added variables and missing variables.

If given a function of three variables one may expand a function about one of the variables, as in: $$\phi(x,t,\theta) = \sum_{n=0}^{\infty} \phi_{n}(x,t) \, e^{i n \theta}.$$

In an integral case a variable can be removed, or evaluated, as in: $$\int_{a}^{b} \phi(x,t,\theta) \, d\theta = \sigma(x,t),$$ where $\sigma$ is the function of two variables after integration.