integration property of fourier series
Note that the antiderivative of a function is only defined up to a constant. Furthermore, note that if you integrate a periodic function, the result is not necessarily periodic. Let

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

Now we integrate $(1)$ with an arbitrary lower integration limit $t_0$. I'll say more about that later.

$$\begin{align}y(t)&amp;=\int_{t_0}^tx(\tau)d\tau\\&amp;=\sum_{k=-\infty}^{\infty}a_k\int_{t_0}^te^{jk\omega_0\tau}d\tau\\&amp;=a_0(t-t_0)+\sum_{k=-\infty\\k\neq 0}^{\infty}\frac{a_k}{jk\omega_0}e^{jk\omega_0t}-\sum_{k=-\infty\\k\neq 0}^{\infty}\frac{a_k}{jk\omega_0}e^{jk\omega_0t_0}\tag{2}\end{align}$$

Clearly, if $a_0\neq 0$, $y(t)$ isn't periodic, so we can't hope to obtain a formula for its Fourier coefficients. So in order to end up with a periodic function $y(t)$, we require that $a_0=0$.

If $a_0=0$, we get from $(2)$

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

with

$$b_k=\begin{cases}\displaystyle\frac{a_k}{jk\omega_0},&amp;k\neq 0\\\displaystyle-\sum_{l=-\infty\\l\neq 0}^{\infty}\frac{a_l}{jl\omega_0}e^{jl\omega_0t_0},&amp;k=0\end{cases}\tag{4}$$

So in general the antiderivative $y(t)$ has a non-zero DC component, which depends on the choice of the lower integration limit $t_0$.

The antiderivative

$$\tilde{y}(t)=\sum_{k=-\infty\\k\neq 0}^{\infty}\frac{a_k}{jk\omega_0}e^{jk\omega_0t}\tag{5}$$

is the specific antiderivative of $x(t)$ that has a zero DC Fourier coefficient, and this is the one that is meant by the formula that is usually given in textbooks.

Consequently, the expression

$$y(t)=\int_{-\infty}^tx(\tau)d\tau\tag{6}$$

for the antiderivative of $x(t)$ with Fourier coefficients $b_k=a_k/(jk\omega_0)$, $k\neq 0$, and $b_0=0$,
is at least inaccurate; it is in fact a sloppy way of expressing the specific antiderivative $\tilde{y}(t)$ given by $(5)$, i.e., the one with a zero DC coefficient.