An upper bound of an integral For any $a<b$, i want to prove that there exists $M_0 \in \mathbb{R}$ such that 
\begin{align}
\left| \int_a^b \frac{\sin x}{x} \text{ d}x \right| \leq M_0.
\end{align}
Here, $M_0$ is independent of $a$ and $b$. How can i prove this? i can find an upper bound which (unfortunately) depends on $a$ and $b$. 
I appreciate if anyone gives me a hint. 
Thanks. 
 A: The proof of this bound should be similar in character to the proof that the improper integral of $\frac{\sin x}{x}$ converges. Over large intervals, the integral behaves like an alternating series, with positive and negative bumps that partially offset. I see two ways to go with this:


*

*Leverage the alternating series estimate. If we're working with full bumps, so that $a$ and $b$ are integer multiples of $\pi$, the bumps alternate in sign and decrease in value as we go to $\infty$. The alternating series estimate tells us that the sum of these is no more than the first term in absolute value.

*Integration by parts. We're not trying to find an antiderivative; we're instead trying to convert this into an absolutely convergent integral, and use this to get an estimate. By the way, the appropriate antiderivative for $\sin x$ is $1-\cos x$. We need this to not blow up at zero.


These are not complete arguments, of course; they're merely enough to get started. I'll also note that the best possible $M_0$ here is $\int_{-\pi}^{\pi} \frac{\sin x}{x}\,dx$.
