Theorem. Assume that $f$ is a piecewise continuous function on $[-\pi,\pi]$. For every $x_0$, $x\in [-\pi,\pi]$ there results $ \int_{x_0}^x f(t)\, dt = \int_{x_0}^x \frac{a_0}{2} + \sum_{n=1}^\infty \int_{x_0}^x \left( a_n \cos nt + b_n \sin nt \right)\, dt. $ Given $x_0$, the right-hand side converges uniformly in $[-\pi,\pi]$.
Proof. The function $ F(x)=\int_{x_0}^x \left( f(t)-\frac{a_0}{2} \right)\, dt, \quad x \in [-\pi,\pi] $ is continous, vanishes at $-\pi$ and at $\pi$, and we can extend it as a $2\pi$-periodic function. Moreover its derivative is continuous except at a finite number of points. By a well-know result, its Fourier series converges uniformly. If $A_n$ and $B_n$ are its Fourier coefficients, then $ a_n = n B_n, \quad b_n = -n A_n. $ Hence $ \begin{align} \int_{x_0}^x \left( f(t)- \frac{a_0}{2} \right)dt &= F(x)-F(x_0) =\\ &= \sum_{n=1}^\infty \left\{ A_n (\cos nx -\cos n x_0 ) + B_n (\sin nx - \sin n x_0 ) \right\}\\ &= \sum_{n=1}^\infty \left\{ -b_n \frac{\cos nx - \cos n x_0}{n}+a_n \frac{\sin nx - \sin nx_0}{n} \right\} \\ &=\sum_{n=1}^\infty \left\{ b_n \int_{x_0}^x \sin nt \, dt + a_n \int_{x_0}^x \cos nt \, dt \right\}. \end{align} $