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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.

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    $\begingroup$ Integration by parts $\endgroup$ – mathworker21 Mar 28 at 19:43
  • $\begingroup$ More hint: specifically, the integration by parts creates a sufficient decay in the denominator for you to have a boundary $\endgroup$ – UnbelieveTable Mar 28 at 19:49
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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 $\int_{-\pi}^{\pi} \frac{\sin x}{x}\,dx$.

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  • $\begingroup$ I get a maximum integral of about $3.704$, achieved with bounds $b=\pi$ and $a=-\pi$. $\endgroup$ – jmerry Mar 29 at 8:19
  • $\begingroup$ Yes i saw that too. I plotted the function and verified that the largest possible area is the one between the curve, the $x$-axis and the vertical lines $x=-\pi$ and $x=\pi$. But how can i prove this algebraically? $\endgroup$ – Hussein Eid Mar 29 at 8:26
  • $\begingroup$ I outlined two paths. Which one would you prefer to follow? $\endgroup$ – jmerry Mar 29 at 8:43

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