I've been struggling to prove that partial sum of Fourier series can be defined in the following form:
$$ S_{n}(f)(x) = \frac{1}{\pi} \int_{-\pi}^{\pi} f(t-x)D_{n}(t)dt, $$ where:
$$ D_{n}(t) = \frac{1}{2} + \sum_{k=1}^n \cos(kt)=\frac{\sin(t(n+\frac{1}{2}))}{2\sin(\frac{t}{2})}.$$ And $f$ is a $2\pi$-periodic function. I've managed to show that $$ S_{n}(f)(x) = \frac{1}{\pi} \int_{-\pi}^{\pi} f(t)D_{n}(t-x)dt $$, but I don't know how to use the fact of $f$ being $2\pi$-periodic.
Any ideas?
Note that $D_n$ is even, so we have $$S_{n}(f)(x) = \frac{1}{\pi} \int_{-\pi}^{\pi} f(t)D_{n}(t-x)dt=\frac{1}{\pi} \int_{-\pi}^{\pi} f(t)D_{n}(x-t)dt$$
Write $s=x-t$, so that $t=x-s$, and $$\frac{1}{\pi} \int_{-\pi}^{\pi} f(t)D_{n}(t-x)dt=\frac{1}{\pi}\int_{x-\pi}^{x+\pi}f(x-s)D_n(s)ds=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x-s)D_n(s)ds.$$