I'm studying Fourier series out of Rudin's "Principals of Mathematical Analysis". In the proof that the Fourier series $s_N(f;x)$ converges pointwise to $f$, it assumes that at a point $x$, there is some $\delta,M>0$ such that $\vert f(x+t)-f(x)\vert \leq M\vert t\vert$ for any $t$ with $\vert t\vert<\delta$. It then goes on to show that for the Fourier series: $$\begin{split} s_N(f;x)-f(x)&=\int_{-\pi}^{\pi}(f(x-t)-f(t))D_N(t)dt\\ &=\int_{-\pi}^{\pi}\frac{f(x-t)-f(t)}{\sin(t/2)}\sin((N+1/2)t)dt\\ &=\int_{-\pi}^{\pi}\frac{f(x-t)-f(t)}{\sin(t/2)}\cos(t/2)\sin(Nt)dt+\int_{-\pi}^{\pi}(f(x-t)-f(t))\cos(Nt)dt \end{split}$$ The terms that aren't $\cos(Nt)$ or $\sin(Nt)$ are bounded because of the assumption about $f$, so then the book claims that the integrals tend to $0$ as $N\rightarrow\infty$. This makes total sense to me intuitively, but I have no idea how to prove it; I can't use the boundedness of the left terms in the integral without taking the absolute value, but if I do that then the trig terms of $Nt$ don't tend to 0, they tend to 1.
Really I'd just like to show that $\lim_{N\rightarrow\infty}\int_{-\pi}^{\pi}f(x)D_N(x)dx=f(0)$, but it's basically the same problem.
A much simpler proof goes as follows:
By the Riemann-Lebesgue lemma, we know that the Fourier coefficient of $f$, defined as $$\int_{-\pi}^{\pi} f(x)e^{-inx} dx$$ goes to $0$ as $n$ goes to $\infty$. Using Euler's equation, given by $$e^{ix} = cos(x) + isin(x)$$ you can conclude that the imaginary part of the integral also goes to $0$, which is precisely what you need.