In the continuous Fourier series properties for a periodic continuous-time signal, we have time-integration property:
$$ \int_{-\infty}^t x(\alpha)d\alpha \leftrightarrow \frac{a_k}{jk\omega_0} $$
where $a_k$ is the Fourier series coefficients of the signal $x(t)$. Now I am confused. Is the above integral equivalent to $\int x(t)dt$ ?
For instance, the Fourier series coefficients of $\cos(\omega_0 t)$ are: $a_1=a_{-1}=\frac{1}{2}$, other $a_{k}=0$.
and the Fourier series coefficients of $\sin(\omega_0 t)$ are: $b_1=\frac{1}{2j}, b_{-1}=\frac{-1}{2j}$, other $b_{k}=0$.
and we have: $\int \cos(\omega_0t)dt=\frac{1}{\omega_0}\sin(\omega_0t)$
Then:
$$ \textrm{F.S.}\left\{\int \cos(\omega_0t)dt\right\} = \frac{a_k}{jk\omega_0} = \textrm{F.S.}\left\{\frac{1}{\omega_0}\sin(\omega_0t)\right\}= \frac{1}{\omega_0}b_k $$
Thus:
$$ b_k=\frac{a_k}{jk} $$
as we expected.