What is the time-integration property in the Fourier series analysis? 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.
Let's consider the complex form of the Fourier series of a $T$-periodic function $x(t)$:

$$x(t)=\sum_{k=-\infty}^{\infty}a_ke^{jk\omega_0t}\tag{1}$$

with $\omega_0=2\pi/T$. The integration property says that the Fourier series of 

$$y(t)=\int x(t)dt\tag{2}$$

is given by

$$y(t)=\sum_{k=-\infty}^{\infty}b_ke^{jk\omega_0t}\tag{3}$$

with

$$b_k=\frac{a_k}{jk\omega_0},\quad k\neq 0\tag{4}$$

where we require that $a_0=0$.

Note that in $(2)$ I used the indefinite integral, and I chose the integration constant to be zero, such that $b_0=0$. In some texts you may find the definite integral instead of the indefinite integral:

$$y(t)=\int_{-\infty}^{t} x(t)dt\tag{5}$$

The problem with $(5)$ is that for certain functions $x(t)$ the definite integral doesn't exist with $-\infty$ as the lower bound. Note that the lower bound in $(5)$ corresponds to the integration constant of the indefinite integral $(2)$. This becomes obvious if we consider the relation between the indefinite and the definite integral:

$$\int x(t)dt=X(t)+C\tag{6}$$

$$\int_{t_0}^tx(\tau)d\tau=X(t)-X(t_0)\tag{7}$$