This is a question from Edwards and Penney 4th edition Differential Equations and Boundary value problems from section 9.3.
Suppose that $f(t)$ is a piecewise continuous period $2L$ funtion.
Define $$F(t) = \int \limits_0^t[f(s) - \frac 12a_0]ds$$. Where ${a_n}$ and ${b_n}$ denote the Fourier coefficients of f(t)
Show that $F(t + 2L) = F(t). $
I feel like this isn't always true unless we impose conditions on $F(2L)$ and $F(0)$.
By definition of $F$:
$$F(t+2L) - F(t) = \int_t^{t+2L} f(s) - \frac{1}{2} a_0\,ds = \int_{t}^{t+2L}f(s)\,ds - a_0 L.$$
By the periodicity of $f$, we have
$$\int_t^{t+2L} f(s)\,ds = \int_0^{2L} f(s)\,ds,$$
and by definition of the Fourier coefficients,
$$a_0 = \frac{1}{L}\int_0^{2L} f(s)\,ds.$$
That means we have indeed
$$F(t+2L) - F(t) = 0$$
for all $t$, i.e. $F$ is $2L$-periodic.