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I'm quite lost on the following problem:

$$\int_{0}^{\pi/2} \frac{sin^2(x)}{x^{p^2-3p-7}}dx$$

I can't figure out how to work out the given answer. Please help me!

ZellAllon
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  • I was trying Direct comparison test with sin(x) – ZellAllon Jun 24 '14 at 19:12
  • Here's an idea: To start, it converges whenever $p^2 - 3p - 7 \leq 2$, since we have a removable singularity. In fact, $p^2 - 3p - 7 < 3$ shall do the trick (using $\sin(x) \leq x$). So $(p-5)(p+2) < 0$ or $-2 < p < 5$ is a sufficient condition. By the analyticity of $\sin$, we can conclude that this is also necessary. Indeed, you can check this with Wolfram. – Christopher K Jun 24 '14 at 19:20
  • why $p^2-3p-7\leq2$ ? – ZellAllon Jun 24 '14 at 19:27
  • An P-Series converges if P>1 – ZellAllon Jun 24 '14 at 19:30
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    You may have seen $\sin(x) = x\cdot(1-x^2/3! + x^4/5! - ...)$. If $p^2-p-7 \leq 2$, then the integrand becomes $x^{\alpha}\cdot sin^2(x)$ on $(0,\pi/2)$ where $\alpha>0$. So the improper integral converges. But in fact we can do better, by bounding the integrand by $\frac{x^2}{x^{p^2-3p-7}}$ and the integral will converge whenever $2 - (p^2-3p-7) = 9 - p^2 + 3p > -1$ or equivalently, $p^2-3p-10 = (p-5)(p+2) < 0$. – Christopher K Jun 24 '14 at 19:31
  • To tell you the truth, I do not understand. – ZellAllon Jun 24 '14 at 19:33
  • Okay... you do agree that the improper integral $\int_{(0,\pi/2)} 1/x^{\alpha}$ converges for $\alpha < 1$, right? – Christopher K Jun 24 '14 at 19:36
  • Yes! I agree bro – ZellAllon Jun 24 '14 at 19:38
  • Are you familiar with monotonicity of improper integrals - that is $\int_{A} f(x) \leq \int_{A} g(x)$ whenever $\forall x \in A$, $f(x)\leq g(x)$? How about $\sin(x)\leq x$ $\forall x>0$, since $(x-\sin(x))' = 1 - \cos(x) \geq 0$ and $\sin(0) = 0$. – Christopher K Jun 24 '14 at 19:42
  • Yes! yes! $$\int_{a}^{b} f(x)dx$$ Converges if $$\int_{a}^{b} g(x)dx$$ Converges right? – ZellAllon Jun 24 '14 at 19:48
  • Yes, so do you understand my argument on why $$\frac{sin^2(x)}{x^{p^2-3p-7}} \leq \frac{x^2}{x^{p^2-3p-7}} = 1/x^{p^2-3p-9}$$ where we want $p^2-3p-9 < 1$ from before. – Christopher K Jun 24 '14 at 19:51
  • Yes, I understand but what happens if I do this?... $sin(x) \leq 1$ ; $sin^2(x) \leq 1$ ; $ $ $\frac {sin^2(x)}{x^{p^2-3p-7}} \leq \frac {1}{x^{p^2-3p-7}}$ right? – ZellAllon Jun 24 '14 at 20:05
  • And this tells you that whenever $p^2-3p-7<1$ or $p^2-3p-8<0$ that the integral converges. Indeed, if $p^2-3p-8<0$, then $p^2-3p-10<0$. So you've got a solution, but it is not a complete set. By finding a tighter bound, we can show that whenever $p = -1.999$ or $p=4.999$, for example, the integral still converges. But these values do not satisfy $p^2-3p-8 < 0$. – Christopher K Jun 24 '14 at 20:10
  • Thank you Chris K. for providing the requested information. – ZellAllon Jun 24 '14 at 20:37

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