Showing $|\int_{n\pi}^{(n+1)\pi}\frac {\sin x}x\,\mathrm dx| > \frac{1}{2n+1}$ 
Show $$
\left|\int_{n\pi}^{(n+1)\pi}\frac{\sin x}x \,\mathrm{d}x\right| > \frac{1}{2n+1}. \quad n\in \mathbb{N}$$

$$\left|\frac{\sin x}{x}\right|>\left|\frac{\sin x}{(n+1)\pi}\right|,$$ so $$\left|\int_{n\pi}^{(n+1)\pi}\frac{\sin x}x \,\mathrm{d}x\right|>\left|\int_{n\pi}^{(n+1)\pi}\frac{\sin x}{(n+1)\pi} \,\mathrm{d}x\right|=\frac{2}{(n+1)\pi}. $$
I don't know how to get the estimation $\dfrac{1}{2n+1}$.
 A: Hint: ${2\over{\pi(n+1)}}={1\over{{\pi\over 2}n+{\pi\over 2}}}\geq {1\over {2n+1}}$ implies that ${\pi\over 2}n+{\pi\over 2}\leq 2n+1$ and $(2-{\pi\over 2})n\geq {\pi\over 2}-1$ this is true for $n\geq 2$.
Study the cases $n=0,1$ separately.
A: For any $n\geq 1$
$$\int_{n\pi}^{(n+1)\pi}\frac{|\sin x|}{x}\,dx=\int_{0}^{\pi}\frac{\sin x}{x+n\pi}\,dx=\int_{0}^{\pi/2}\sin(x)\underbrace{\left(\frac{1}{n\pi+x}+\frac{1}{(n+1)\pi -x}\right)}_{\text{decreasing over }[0,\pi/2]}\,dx $$
leads to 
$$\int_{n\pi}^{(n+1)\pi}\frac{|\sin x|}{x}\,dx \geq \int_{0}^{\pi/2}\sin(x)\frac{4}{(2n+1)\pi}\,dx = \frac{2}{\pi\left(n+\tfrac{1}{2}\right)}$$
which is stronger than needed.

We may also consider that the Laplace transform of $f(x)=\sin(x)\cdot\mathbb{1}_{(n\pi,(n+1)\pi)}(x)$ is given by
$$\mathcal{L} f(s) = \frac{(-1)^n}{s^2+1}\left(e^{-\pi ns}+e^{-\pi(n+1)s}\right) $$
and the inverse Laplace transform of $\frac{1}{x}$ is simply $1$, hence
$$ \left|\int_{n\pi}^{(n+1)\pi}\frac{\sin x}{x}\,dx\right|=\int_{0}^{+\infty}\frac{e^{-\pi ns}+e^{-\pi(n+1)s}}{s^2+1}\,ds $$
and the RHS is very well suited for providing both upper and lower bounds.
Indeed, the RHS is obviously $\leq \frac{1}{\pi n}+\frac{1}{\pi(n+1)}$, and by the Cauchy-Schwarz inequality
$$\begin{eqnarray*} &&\frac{2}{\pi n}\int_{0}^{+\infty}\frac{e^{-\pi ns}+e^{-\pi(n+1)s}}{s^2+1}\,ds\\ &\geq& \int_{0}^{+\infty}(s^2+1)\left(e^{-\pi ns}+e^{-\pi(n+1)s}\right)\,ds \int_{0}^{+\infty}\frac{e^{-\pi ns}+e^{-\pi(n+1)s}}{s^2+1}\,ds\\&\geq&\left(\int_{0}^{+\infty}e^{-\pi ns}+e^{-\pi(n+1)s}\,ds\right)^2=\left(\frac{1}{\pi n}+\frac{1}{\pi(n+1)}\right)^2 \end{eqnarray*}$$
hence
$$ \int_{0}^{+\infty}\frac{e^{-\pi ns}+e^{-\pi(n+1)s}}{s^2+1}\,ds \geq \frac{(2n+1)^2}{2\pi n (n+1)^2}. $$
