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Prove that: $$\left|\int\limits_a^b\frac{\sin x}{x}dx\right|\le \frac{2}{a}$$

$0<a<b$

I'm not asking for a complete solution to this, just a hint or a push in the right direction. I've been struggling with this and haven't really made any progress.

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    $\begingroup$ Integration by parts? $\endgroup$ Apr 3, 2020 at 22:14

2 Answers 2

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Integrating by parts,

$$\int_a^b\frac{\sin x}x\,dx = \frac{\cos a}a-\frac{\cos b}b -\int_a^b\frac{\cos x}{x^2}\,dx.$$ Then apply the triangle inequality (twice) to get $$\bigg|\int_a^b\frac{\sin x}{x}\,dx\bigg|\leqslant \frac1a+\frac1b+\int_a^b\frac{1}{x^2}\,dx = \frac{2}{a},$$ since $\int_a^b\frac 1{x^2}\,dx = \frac1a-\frac 1b$.

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  • $\begingroup$ Sorry can you explain why $\int_a^b\frac{\sin x}x\,dx$ is $\frac{\cos a}a-\frac{\cos b}b + \int_a^b\frac{\cos x}{x^2}\,dx$ and not $\frac{\cos a}a-\frac{\cos b}b - \int_a^b\frac{\cos x}{x^2}\,dx$? $\endgroup$ Apr 3, 2020 at 22:40
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    $\begingroup$ @Alexander51413 Because I made a mistake :) It doesn't matter anyway, the sign is redundant since I applied the triangle inequality. $\endgroup$ Apr 3, 2020 at 22:44
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\begin{align*} I &= \int_{a}^b \dfrac{\sin x}{x}dx \\ &= \dfrac{-\cos x}{x}|_{a}^b+ \int_{a}^b\dfrac{\cos x}{x^2}dx \\ &= \dfrac{\cos a}{a} - \dfrac{\cos b}{b}+ J \\ &\le \dfrac{1}{a}+\dfrac{1}{b}+ \int_{a}^b \dfrac{1}{x^2}dx \\ &= \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{a}-\dfrac{1}{b} \\ &= \dfrac{2}{a} \text{.} \end{align*} Also: $I \ge -\dfrac{1}{a} - \dfrac{1}{b}- \displaystyle \int_{a}^b \dfrac{1}{x^2}dx = -\dfrac{2}{a}$. Thus $|I| \le \dfrac{2}{a}$ as claimed.

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  • $\begingroup$ Sorry can you explain why $\int_a^b\frac{\sin x}x\,dx$ is $\frac{\cos a}a-\frac{\cos b}b + \int_a^b\frac{\cos x}{x^2}\,dx$ and not $\frac{\cos a}a-\frac{\cos b}b - \int_a^b\frac{\cos x}{x^2}\,dx$? $\endgroup$ Apr 3, 2020 at 22:41
  • $\begingroup$ Ah, it's because $-(-\cos x) = \cos x$. Most students ran into this bump and they couldn't find out why the sign changes... $\endgroup$
    – DeepSea
    Apr 3, 2020 at 22:44
  • $\begingroup$ @DeepSea He's right actually, you're taking $u = \frac1x$ and $dv = \sin x\, dx$, so $du = -\frac1{x^2}\,dx$ and $v=-\cos x$ so applying integration by parts you should get $$\int_a^b u\, dv= uv\big|_a^b - \int_a^b v\,du = -\frac{\cos x}{x^2}\Big|_a^b - \int_a^b\frac{\cos x}{x^2}\,dx.$$ $\endgroup$ Apr 3, 2020 at 22:46
  • $\begingroup$ Yes. The double negative comes from integration by parts and the other from the negative of cosine. $\endgroup$
    – DeepSea
    Apr 3, 2020 at 22:48
  • $\begingroup$ @DeepSea : It's a triple negative, not a double negative. $\endgroup$ Apr 3, 2020 at 22:55

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