How to prove upper bound of this difference of the Sine Integral?

Question

This exercise can be found in Mathematics LibreTexts (bottom of the page) . I have been stuck for about a day and have made minimal progress.

Let $S(x)=\int_0^x\frac{\sin t}{t}$.

Show that for $k \ge 1$ $$|S(2\pi k) - S(2\pi (k+1)| \le \frac{1}{k(2k+1)\pi}$$

They give the additional hint that the identity $\sin (t+\pi) = -\sin t$ should be useful.

The best I have done is bound the difference by $\ln(1 + \frac 1k)$, which I note does approach zero as k tends to infinity, showing that the difference does become arbitrarily small, but this is not what the question asks. This also does not use the hint.

\begin{align} \lvert S(2\pi k) - S(2\pi (k+1)\rvert & = \lvert\int_{2\pi k}^{2\pi(k+1)}\frac{\sin t}{t}dt\rvert \\\\ & \le \int_{2\pi k}^{2\pi (k+1)}\lvert \frac{\sin t}{t}\rvert dt \\\\ & \le \int_{2\pi k}^{2\pi (k+1)}\frac {dt}{t} \\\\ & = \ln (1 + \frac 1k) \end{align}

Any help would be appreciated, thanks!

Answer 1

Try

$$\int_{2 \pi k}^{2 \pi (k+1)} \frac{\sin (t) }{t} \, dt=\int_{-\pi }^{\pi } \frac{\sin (2 \pi k+t+\pi ) }{\pi (2 k+1)+t} \, dt$$

for integer $k$.

$$ \left|-\int _{-\pi }^{\pi } \frac{\sin (t)}{\pi (2 k+1)+t}\cdot dt\right| =\frac{1}{\pi (2 k+1)}\cdot \left|\int _{-\pi }^{\pi }\frac{\sin (t)}{1+ \frac{t}{\pi (2 k+1)}} dt\right|= \frac{1}{\pi (2 k+1)}\cdot \int_0^{\pi } \sin (t) \left| \frac{1}{\frac{t}{\pi (2 k+1)}+1}-\frac{1}{1-\frac{t}{\pi (2 k+1)}}\right| \, dt$$