What is the time-integration property in the Fourier series analysis?
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:

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

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