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See this: $$\newcommand{\b}[1]{\left(#1\right)}\left\lfloor\frac{\displaystyle\int_0^{\pi/2}\cos^{\sqrt{13}-1}x{\rm d}x}{\displaystyle\int_0^{\pi/2}\cos^{\sqrt{13}+1}x{\rm d}x}\right\rfloor$$ Well I could only think of Cauchy-Schwarz, but it is also not fitting.

All calculations by hand, no Beta And that implicitly implies not using others like Gamma too.

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  • $\begingroup$ How does ye olde “replace $\cos(x)$ by exponentials” tricke work in this case? $\endgroup$ May 12, 2015 at 13:54
  • $\begingroup$ @Circonflexe can't say, maybe that's overkill sinc eI only want to find the floor. $\endgroup$
    – RE60K
    May 12, 2015 at 13:55
  • $\begingroup$ Use the reduction formula for Wallis' integrals. $\endgroup$
    – Lucian
    May 12, 2015 at 21:13

5 Answers 5

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As Daniel Fischer helped, using Integration by Parts: $$\int_0^{\pi/2}\cos^{\alpha+1}x{\rm d}x=\int_0^{\pi/2}\cos^{\alpha}x\cos x{\rm d}x=\cos^{\alpha}x\sin x\Bigg|_0^{\pi/2}+\alpha\int_0^{\pi/2}\cos^{\alpha-1}x\sin^2 x{\rm d}x=0+\alpha\int_0^{\pi/2}\cos^{\alpha-1}x{\rm d}x-\alpha\int_0^{\pi/2}\cos^{\alpha+1}x{\rm d}x$$ So: $$\frac{\int_0^{\pi/2}\cos^{\alpha-1}x{\rm d}x}{\int_0^{\pi/2}\cos^{\alpha+1}x{\rm d}x}=\frac{\alpha+1}{\alpha}$$

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    $\begingroup$ Could it be that the result is actually $\frac{\alpha + 1}{\alpha}$? $\endgroup$
    – PhoemueX
    May 12, 2015 at 21:30
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We needn't compute the integrals. Because $0\leq \cos x\leq1$ on $[0,\pi/2]$, the higher power of $\cos$ is less then the lower one, hence the floor is at least 1. And it is easy to estimate, that it cannot be 2 or more: $$ \int_0^{\pi/2}\cos^5 x\,dx=\frac8{15},\quad \int_0^{\pi/2}\cos^2 x\,dx=\frac\pi4 $$

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  • $\begingroup$ is all this intutive? I like rigorousity better. $\endgroup$
    – RE60K
    May 12, 2015 at 13:58
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    $\begingroup$ @ADG It can be made into a fully rigorous argument quite directly. $\endgroup$
    – GPerez
    May 12, 2015 at 13:59
  • $\begingroup$ so would you help me eliminating 2 as a possibility, cuz I can't see it now. $\endgroup$
    – RE60K
    May 12, 2015 at 14:00
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    $\begingroup$ @ADG Sign error, it's $\frac{\alpha}{\alpha+1}$. So here, you have $\left\lfloor \frac{\sqrt{13}+1}{\sqrt{13}}\right\rfloor$. $\endgroup$ May 12, 2015 at 14:16
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    $\begingroup$ @ADG I reserve the right to be too lazy for that. You can write it up if you want. $\endgroup$ May 12, 2015 at 14:18
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Recall that $$\int_0^{\pi/2} \cos^a(x)dx = \dfrac12 \beta(1/2,(a+1)/2) = \dfrac12 \dfrac{\Gamma(1/2)\Gamma((a+1)/2)}{\Gamma((a+2)/2)}$$ where $\beta(x,y)$ is the beta function and $\Gamma(z)$ is the gamma function.

Hence, the value of your expression without the floor is $$\dfrac{\Gamma((a+1)/2)}{\Gamma((a+2)/2)} \cdot \dfrac{\Gamma((a+4)/2)}{\Gamma((a+3)/2)} = \dfrac{(a+4)/2-1}{(a+3)/2-1} = \dfrac{a+2}{a+1}=1+\dfrac1{\sqrt{13}}$$ where $a= \sqrt{13}-1$.

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  • $\begingroup$ no beta plz, edited the post. $\endgroup$
    – RE60K
    May 12, 2015 at 13:56
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    $\begingroup$ @ADG What do you mean? $\endgroup$
    – Adhvaitha
    May 12, 2015 at 13:56
  • $\begingroup$ (+1)plz don't use beta or gamma function, that's not my course level yet, sorry for not mentioning it before, but you can kee this here, that'll help other visiters, if you're looking for acceptance, maybe post another answer [sorry times3] $\endgroup$
    – RE60K
    May 12, 2015 at 13:58
  • $\begingroup$ @ADG Done. Below is another answer, which uses nothing but basic trigonometry. $\endgroup$
    – Adhvaitha
    May 12, 2015 at 14:09
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    $\begingroup$ "Recall that..." $\endgroup$
    – djechlin
    May 12, 2015 at 20:42
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Another way is to write $$\cos^{a}(x) = \cos^{a+2}(x) + \sin^2(x)\cos^a(x)$$ Hence, we are after $$1+\dfrac{\int_0^{\pi/2}\sin^2(x)\cos^a(x)dx}{\int_0^{\pi/2}\cos^{a+2}(x)dx}$$ Clearly, $$\dfrac{\int_0^{\pi/2}\sin^2(x)\cos^a(x)dx}{\int_0^{\pi/2}\cos^{a+2}(x)dx} > 0$$ We will now show that $$\dfrac{\int_0^{\pi/2}\sin^2(x)\cos^a(x)dx}{\int_0^{\pi/2}\cos^{a+2}(x)dx} < 1$$ which on rearranging gives us that, it suffices to show $$ \text{ or }\int_0^{\pi/2}\cos(2x) \cos^{a}(x)dx > 0$$ We will show that the last statement is indeed true. \begin{align} \int_0^{\pi/2}\cos(2x) \cos^{a}(x)dx & = \int_0^{\pi/4}\cos(2x) \cos^{a}(x)dx + \int_{\pi/4}^{\pi/2}\cos(2x)\cos^a(x)dx\\ & = \int_0^{\pi/4}\cos(2x) \cos^{a}(x)dx + \int_{0}^{\pi/4}\cos(2(\pi/2-x))\cos^a(\pi/2-x)dx\\ & = \int_0^{\pi/4}\cos(2x) \cos^{a}(x)dx + \int_{0}^{\pi/4}\cos(\pi-2x)\sin^a(x)dx\\ & = \int_0^{\pi/4}\cos(2x) \cos^{a}(x)dx - \int_{0}^{\pi/4}\cos(2x)\sin^a(x)dx\\ & = \int_0^{\pi/4}\cos(2x) \left(\cos^{a}(x) - \sin^a(x) \right)dx > 0 \end{align} where the last inequality is true, since $\cos(x) > \sin(x)$ on the interval $[0,\pi/4)$. Hence, again we have $$1+\dfrac{\int_0^{\pi/2}\sin^2(x)\cos^a(x)dx}{\int_0^{\pi/2}\cos^{a+2}(x)dx} \in (1,2)$$

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  • $\begingroup$ @ADG "Better" is a poor choice of word. My other answer gave you the exact value and this answer rigorously gives you a bound to work with. $\endgroup$
    – Adhvaitha
    May 12, 2015 at 14:17
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Integration by parts gives, for any $\eta>0$: $$\int_{0}^{\pi/2}\cos^{\eta+1}(x)\,dx = \frac{\eta}{\eta+1}\int_{0}^{\pi/2}\cos^{\eta-1}(x)\,dx \tag{1}$$ hence by taking $\eta=\sqrt{13}$ we just need to compute: $$\left\lfloor 1+\frac{1}{\sqrt{13}}\right\rfloor = \color{red}{1}.\tag{2}$$

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