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The original problem is to test the convergence of $\int_{0}^{\infty} \frac{\sin(x)^2}{x^{5/2}}$. It is easy to prove that $\int_{1}^{\infty} \frac{\sin(x)^2}{x^{5/2}}$ is convergent, but I have checked in https://www.integral-calculator.com that $\int_{0}^{1} \frac{\sin(x)^2}{x^{5/2}}$ is not. How could I prove it? Thanks in advance.

egreg
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Andarrkor
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2 Answers2

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Since $\lim_{x\to0}\frac{\sin x}x=1$, you know that $\lim_{x\to0}\frac{\sin^2x}{x^2}=1$. So, near $0$ your function behaves as $\frac1{x^{1/2}}$ and therefore it converges.

José Carlos Santos
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Actually, it does converge. (I'm not sure where you went wrong in using WA.) For a proof, note the integrand approximates $x^{-1/2}$ at small $x$, with antiderivative $\approx 2\sqrt{x}$.

J.G.
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  • Well, I actually checked in https://www.integral-calculator.com and it says it diverges. – Andarrkor Mar 27 '19 at 12:04
  • @Andarrkor I just tried that site; it gets the same exact finite expression as WA did. – J.G. Mar 27 '19 at 12:13
  • @Andarrkor, Huh! It looks like you've found a flaw in that site. I would recommend editing your comment into your question. Why did you say there that you checked Alpha? – Barry Cipra Mar 27 '19 at 12:13
  • @J.G., I checked it too, and it told me the integral diverges! – Barry Cipra Mar 27 '19 at 12:14
  • @BarryCipra Because the OP said they'd checked in WA that the integral doesn't converge. However, both sites agree (for me) that it does. If you're getting the provably wrong "it diverges" result from one or both sites, I'll leave you to report that as a bug, but on my end both resources work fine. – J.G. Mar 27 '19 at 12:15
  • @BarryCipra Because I didn't remember the website and thought that WA would get the same result. My bad. – Andarrkor Mar 27 '19 at 12:16
  • @BarryCipra I have edited the post. – Andarrkor Mar 27 '19 at 12:17
  • @J.G., curiouser and curiouser... Why would the exact same site on the exact same integral give two completely different answers? What value does it give you, anyway? – Barry Cipra Mar 27 '19 at 12:25
  • @BarryCipra $\frac{8\sqrt{\pi}C(2/\sqrt{\pi})-1-4\sin 2+\cos 2}{3}\approx 1.87607$, with $C(x):=\int_0^x\cos t^2dt$. It's possible one or both sites were misused in a way that resulted in incorrect limits being used. it took me a few seconds to work out how to set limits $0,,1$ in integral-calculator. – J.G. Mar 27 '19 at 12:34
  • @J.G., it's weird, the IntegralCalculator site gives me the antiderivative with the constant term and a $2\sin x(\sin x+4x\cos x)/(3x^{3/2})$, but below that it says, in red lettering, that the definite integral from $0$ to $1$ is divergent, as if it's incapable of evaluating $\lim_{x\to0}2\sin x(\sin x+4x\cos x)/(3x^{3/2})=0$. – Barry Cipra Mar 27 '19 at 14:04
  • @BarryCipra There must be something we're missing. Maybe the bug is on mobile or something. (I used a laptop to do this.) – J.G. Mar 27 '19 at 14:11