I have to find out wether $$ X_t = \sum_{k=1}^{L} c_k \sin(2\pi v_k t + \phi_k) , t\in \mathbb{R} $$ is a stationary process, where $\phi_1 \ldots \phi_L$ are independent and uniformly distributed in $[-\pi, \pi]$. Also, $c_k$ and $v_k$ are constants.
Yet, I am stuck and can't find out from where to start. I know that:
$$ E(X_t) = \sum_{k=1}^{L} c_k E\left( \sin(2\pi v_k t + \phi_k)\right) $$
Now for every $k \in \{1,2 \ldots k \}$:
$$ E(\sin(2\pi v_k t+ \phi_k) = \frac{1}{2\pi} \int_{-\pi}^{\pi} \sin(2\pi v_k t + \phi_k) d\phi_k =0$$
Since $\int_{-\pi}^{\pi} \sin(u)du = 0 $, because the sine an even function. So the mean is constant for this particular process. I evaluated it and found that the variance is also constant.
For the autocovariance function:
$$ \gamma(h) = E(X_t X_{t+h}) = \sum_k \sum_j c_k c_j E(\sin(2\pi v_k t + \phi_k) \sin(2\pi v_j + \phi_j)) $$
And I know the expected value must have something to do with the ortogonality of sines and cosines. Yet I am stuck. Can you guys give me any advice? Thanks!
Since $\phi_k$ has a uniform distribution on $[-\pi,\pi]$, $$ \mathbb E\left[\sin\left(2\pi v_kt+\phi_k\right)\right]=\frac 1{2\pi}\int_{-\pi}^\pi\sin\left(2\pi v_kt+u\right) \mathrm du. $$ This is not much harder to compute than $\int_{-\pi}^\pi\sin\left( u\right) \mathrm du$; in both cases, a primitive can be found.
For the covariance, let $$ a_{k,j}:= \mathbb E\left[\sin\left(2\pi v_kt+\phi_k\right)\sin\left(2\pi v_j(t+h)+\phi_j\right)\right]. $$ When $k=j$, this reduced to a similar computation as before. When $k\neq j$, the random variable involved in the expectation are independent.