Show that if $0\lt L\lt R\lt+\infty$, then $\int_L^R\frac{\sin x}x dx\lt \frac2L$ Show that if $0\lt L\lt R\lt+\infty$, then $\int_L^R\frac{\sin x}x dx\lt \frac2L$
My Work:
a) Integration by parts: $$\int_L^R\frac{\sin x}x dx$$ using the formula $\int a'b=ab-\int ab'$, there is no viable selection for a or b because no amount of integration or derivation will make one of the two functions cancel out
b) Substitution: $u(x)=\frac1x$ $$\int_{u(L)}^{u(R)}u \sin{\frac1u} dx$$ From here we can try IBP again, but I believe that this will also prove to be fruitless because of the term $\sin{\frac1u}$
I don't know of any other method I could use to integrate this function and I'm rather puzzled on what to do next. Any help is appreciated!!
 A: First note that
$$ \left|\int_L^R\frac{\cos x}{x^2} \mathrm{d}x  \right| \le \int_L^R\left|\frac{\cos x}{x^2} \right| \mathrm{d}x \le \int_L^R\frac{1}{x^2}\mathrm{d}x = \frac{1}{L} - \frac{1}{R}$$
Now,
\begin{align}\int_L^R \frac{\sin x}{x} \mathrm{d}x &= \left. \frac{-\cos x}{x}\right|_L^R - \int_L^R\frac{\cos x}{x^2}\mathrm{d}x \\ 
\overset{(a)}\implies \int_L^R \frac{\sin x}{x} \mathrm{d}x &\le \frac{\cos L}{L} - \frac{\cos R}{R} + \frac{1}{L} - \frac{1}{R} \\
&= \frac{1+\cos L}{L} - \underbrace{\frac{1+\cos R}{R}}_{\ge 0} \\
&\le \frac{1+\cos L}{L} \le \frac{2}{L}
\end{align}
where $(a)$ comes from $ x < y-z \implies x < y +|z|$
A: $$
\int_L^R = \int_L^{n\pi} + \overbrace{\int_{n\pi}^{(n+1)\pi} + \int_{(n+1)\pi}^{(n+2)\pi} + \cdots + \int_{(m-1)\pi}^{m\pi}} + \int_{m\pi}^R
$$
where $n\pi$ is the smallest integer multiple of $\pi$ that is $\ge L$ and $m\pi$ is the largest integer multiple of $\pi$ that is $\le R$. If $n$ is even then we have
$$
0 < \int_{n\pi}^{(n+1)\pi} \frac{\sin x} x\,dx < \frac 1 {n\pi} \int_{n\pi}^{(n+1)\pi} \sin x \, dx \le \frac 2 {n\pi} \tag 1
$$
and
$$
0 > \int_{(n+1)\pi}^{(n+2)\pi} \frac{\sin x} x \,dx \ge \frac 1 {(n+1)\pi} \int_{(n+1)\pi}^{(n+2)\pi} \sin x \, dx = \frac{-2}{(n+1)\pi}. \tag 2
$$
Thus the sum under the $\overbrace{\text{overbrace}}$ is less than $2/(n\pi)$ when we've added the first term, then decreases to something $>0$ when we add the second, then increases to something less than first term when we add the third, then decreases to something bigger than the sum of the first two terms when we add the fourth, and so on.  When we add the term after the overbrace, we don't go all the way to the next integer multiple of $\pi$, so we still get something between the last two terms that came before it.  (Still assuming $n$ is even), the integral before the overbrace is negative but somewhere between $0$ and $-2/L$.  Hence the integral from $L$ to $R$ must be between $\pm2/L$.
If $n$ is odd we just reverse some plus and minus signs.
