integration property of fourier series Please help me sort this issue out.
The integration property in Fourier series is as follows:

So, for proving the above property, i took this approach:

This is where my doubt is. Some books and websites just put upper limit (ignoring the lower limit) and compare with (1) to conclude that

However, when the lower integral limit is substituted, we get

This value cannot be evaluated. Though some websites say that this corresponds to the &quot;initial value&quot;(?) and that is zero... etc., how does one justify this mathematically?
Any help is appreciated. Thank you.
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.