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Find all the values of $p$ for which the integral converges $$\int_0^{\pi/2} \frac{\sin x}{x^p} dx$$

First, we can note that $$\int_0^{\pi/2} \frac{\sin x}{x^p} dx \leq \int_0^{\pi/2} \frac{1}{x^p} dx$$

Let's find the values $p$ so as to have the upper bound converging:

Case 1: When $p = 1$ $$\int_0^{\pi/2} \frac{1}{x^p} dx = \lim\limits_{t \rightarrow 0^+}[ \ln(x)]_t^{\pi/2}= - \infty $$

Case 2: When $p \neq 1$ $$\int_0^{\pi/2} \frac{1}{x^p} dx = \lim\limits_{t \rightarrow 0^+} \left[\frac{1}{1-p} x^{1-p} \right]_t^{\pi/2}= \lim\limits_{t \rightarrow 0^+} \left( \frac{1}{1-p} \right) \left( \left(\frac{\pi}{2}\right)^{1-p} - t^{1-p} \right) $$

If $p<1$: as $t \rightarrow 0^+$, we have $t^{1-p} \rightarrow 0$ $=>$ the limit exists.

If $p>1$: as $t \rightarrow 0^+$, we have $t^{1-p} \rightarrow \infty $ $=>$ the limit does not exist.

It follows that the upper integral converges when $p<1$, By the comparison test theorem the original integral converges when $p<1$

Not overly confident about my approach. If it is correct, is there a more formal efficient approach to discriminate the values of $p$?

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    $\begingroup$ The integral actually converges for all $p < 2$. $\endgroup$ – Michael L. Aug 2 '17 at 3:07
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    $\begingroup$ Recall that $\lim_{x\to 0} \frac{\sin(x)}{x}=1$, so that the integral is finite for $p=1$. $\endgroup$ – Reveillark Aug 2 '17 at 3:08
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Note that $\frac{2}{\pi}x\leq \sin(x)\leq x$ for all $x\in [0, \pi/2]$, so we have $$\frac{(2/\pi)x}{x^p} = \frac{2}{\pi}x^{1-p}\leq \frac{\sin(x)}{x^p}\leq \frac{x}{x^p} = x^{1-p}$$ on this interval. Therefore, $$\frac{2}{\pi}\int_0^{\pi/2} x^{1-p}\,\mathrm{d}x\leq \int_0^{\pi/2} \frac{\sin(x)}{x^p}\,\mathrm{d}x\leq \int_0^{\pi/2} x^{1-p}\,\mathrm{d}x$$ For $p < 2$, this implies $$\frac{(\pi/2)^{1-p}}{2-p}\leq \int_0^{\pi/2} \frac{\sin(x)}{x^p}\,\mathrm{d}x\leq \frac{(\pi/2)^{2-p}}{2-p}$$ and thus the integral is finite. For $p\geq 2$, $\int_0^{\pi/2} x^{1-p}\,\mathrm{d}x$ diverges, and therefore so does $\int_0^{\pi/2} \frac{\sin(x)}{x^p}\,\mathrm{d}x$.

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  • $\begingroup$ thx for your input. Could you explain me why you can state that $(2/\pi) x \leq sin(x) \leq x, \forall x \in [0, \pi/2]$? $\endgroup$ – gegu Aug 2 '17 at 4:39
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    $\begingroup$ $\sin(x)$ is strictly concave on $[0, \pi/2]$. Therefore, since $x$ is the line tangent to $\sin(x)$ at $x = 0$, $\sin(x)\leq x$ on $[0, \pi/2]$. Furthermore, $\sin(x) = \frac{2}{\pi}x$ at $x = 0$ and $x = \pi/2$, so by concavity $\sin(x)\geq \frac{2}{\pi}x$ on $[0, \pi/2]$. $\endgroup$ – Michael L. Aug 2 '17 at 4:41

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