Prove $ | \int_{\nu}^{\delta} \frac{\sin{((2N+1)\pi t)}}{\pi t} \,dt | \leq 2 \sup \limits_{M > 0} | \int_{0}^{M} \frac{\sin{\pi t}}{t} \,dt |$.

Question

I'm trying to understand the proof of Jordan's criterion for the convergence of the Fourier series of a function $f \in L^{1}(\Bbb T)$. At the end of the proof, the following inequalities are used, and I have no idea how to prove they hold.

If $0 < \nu < \delta < \frac{1}{2}$, then:

$\left | \int_{\nu}^{\delta} \frac{\sin{((2N+1)\pi t)}}{\pi t} \,dt \right | \leq 2 \sup \limits_{M > 0} \left | \int_{0}^{M} \frac{\sin{\pi t}}{t} \,dt \right | < C$ for some constant $C$.

Answer 1

$$I=\int_{\nu}^{\delta}\frac{\sin((2N+1)\pi t)}{\pi t}\,dt = \int_{(2N+1)\nu}^{(2N+1)\delta}\frac{\sin(\pi t)}{\pi t}\,dt\\=\int_{0}^{(2N+1)\delta}\frac{\sin(\pi t)}{\pi t}\,dt-\int_{0}^{(2N+1)\nu}\frac{\sin(\pi t)}{\pi t}\,dt$$ so, by considering absolute values, $$ |I|\leq 2\cdot\sup_{M\in\mathbb{R}^+}\left|\int_{0}^{M}\frac{\sin(\pi t)}{\pi t}\,dt\right|.\tag{1}$$ On the other hand, $\operatorname{sinc}(\pi t)$ is a continuous function in a right neighbourhood of the origin, $\sin(\pi t)$ is a function with a bounded primitive and $\frac{1}{\pi t}$ is a continuous and decreasing function on $\mathbb{R}^+$, so the RHS of $(1)$ is finite due to the (continuous version of) Dirichlet's test.

Answer 2

Make the change of variable: $t \to (2N+1)t$,

$\displaystyle \int_{\nu}^{\delta} \frac{\sin{((2N+1)\pi t)}}{\pi t} \,dt = \int_{(2N+1)\nu}^{(2N+1)\delta} \frac{\sin \pi t}{\pi t} \,dt \le \frac{2}{\pi}\int_{0}^{(2N+1)\max\{|\delta|,|\nu|\}} \frac{\sin \pi t}{t} \,dt$