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Does $$\displaystyle\int\limits^{{\pi}}_{0} \dfrac{3\sin^2\left(2x\right)}{\sqrt{x}}\,\mathrm{d}x$$

converge or diverge?

I'm not even going to evaluate the integral, it's too ugly. I also don't know of a "simpler" function, that can tell me if it will converge or diverge? How can I do this question? I first have to figure out if it converges or diverges, and then find a appropriate function, but I need help figuring that out.

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  • $\begingroup$ Set $\sqrt{x}=t$ and see what happens $\endgroup$ – imranfat Mar 5 '17 at 15:41
  • $\begingroup$ $$\dfrac{3\sin^2(2t^2)}{t}$$ $\endgroup$ – K Split X Mar 5 '17 at 15:43
  • $\begingroup$ Doesnt really help, can u give another hint? $\endgroup$ – K Split X Mar 5 '17 at 15:43
  • $\begingroup$ Sorry, that step isn't right. $\sqrt{x}=t$ implies $x=t^2$ and so $dx=2tdt$. That denominator cancels. You need to do u-sub in the appropriate manner :) $\endgroup$ – imranfat Mar 5 '17 at 15:44
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Hint. The integrand is continuous over $(0,\pi]$ and one has $$ \left|\int\limits^{{\pi}}_{0} \dfrac{3\sin^2\left(2x\right)}{\sqrt{x}}\,\mathrm{d}x\right|\le\int\limits^{{\pi}}_{0} \left|\dfrac{3\sin^2\left(2x\right)}{\sqrt{x}}\right|\,\mathrm{d}x\le3\int\limits^{{\pi}}_{0} \dfrac{1}{\sqrt{x}}\,\mathrm{d}x<\infty $$

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  • $\begingroup$ But how did you know that it converges? Like what was the hint that sparked the idea of convergence? $\endgroup$ – K Split X Mar 5 '17 at 15:48
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    $\begingroup$ @KSplitX Because we have $$\int\limits^{{\pi}}_{0} \dfrac{1}{\sqrt{x}}\,\mathrm{d}x=2\sqrt{\pi}<\infty. $$ $\endgroup$ – Olivier Oloa Mar 5 '17 at 15:49
  • $\begingroup$ No I understand how you reduced the function. What I'm asking is that, given the original function, how did you know whether it would converge or diverge? Only when you knew that it would converge, would you reduce it to the $\dfrac{1}{\sqrt{x}}$ term $\endgroup$ – K Split X Mar 5 '17 at 15:50
  • $\begingroup$ @KSplitX Ok, I 'instantly' saw that the integral converges because the inthegrand was smaller than a known convergent integral. Usually, one is looking for a standard (and simpler) integrand and try to compare the given integrand to it. $\endgroup$ – Olivier Oloa Mar 5 '17 at 15:53
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In particular $$I=3\int_{0}^{\pi}\frac{\sin^{2}\left(2x\right)}{\sqrt{x}}dx=\frac{3}{2}\int_{0}^{\pi}\frac{1-\cos\left(4x\right)}{\sqrt{x}}dx$$ $$=3\sqrt{\pi}-\frac{3}{2}\int_{0}^{\pi}\frac{\cos\left(4x\right)}{\sqrt{x}}dx\stackrel{x=\pi t^{2}/8}{=}3\sqrt{\pi}-\frac{3\sqrt{\pi}}{2\sqrt{2}}\int_{0}^{2\sqrt{2}}\cos\left(\pi t^{2}/2\right)dt$$ $$=\color{red}{3\sqrt{\pi}-\frac{3\sqrt{\pi}}{2\sqrt{2}}C\left(2\sqrt{2}\right)\approx4.3856}$$ where $C\left(x\right)$ is the Fresnel $C$ integral.

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It converges: the only problem can be at $x=0$, and $\sin^2{2x} = 4x^2+o(x^2) $ there, so the whole integrand looks like $x^{3/2}$ near $x=0$, which obviously has a convergent integral since it is bounded.

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  • $\begingroup$ I don't know how you got $\sin^2(2x)=4x^2+o(x^2)$ $\endgroup$ – K Split X Mar 5 '17 at 15:49
  • $\begingroup$ It follows from the Taylor expansion of $\sin{x}$ at $x=0$, $\sin{x} = x+O(x^3)$. $\endgroup$ – Chappers Mar 5 '17 at 16:24
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Ok, Here is a standard approach: We don't know whether or not it converges, so hence my intial step to do a u-sub, which is straight forward: $\sqrt{x}=t$ which gives a new integral of the form $\int{6sin^2{2t}}dt$ with boundaries $0$ and $\sqrt{\pi}$. At this point you can hopefully see that the integral went from improper to proper, and it is going to be convergent. Now you state that the integral is "ugly" to evaluate, but I think this one is quite cute? Set $2t=v$ and you get a $sin^2v$ integral, which is standard. You can tag the limits along. Note, the answer won't be pretty as you are dealing with "ugly" angles. But nonetheless, convergent.

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