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$\def\d{\mathrm{d}}$I would like to compute the following limit, $$\displaystyle{\lim_{n \to \infty} \int_0^{\frac{\pi}{2}} \frac{(\sin(x))^{n}}{1-\sin{(x)}}\,\d x} .$$

I am looking for a high school answer.

I tried writing $$\lim_{n \to \infty} \int_0^{\frac{\pi}{2}}{\frac{(\sin(x))^{n}}{1-\sin{(x)}}\,\d x = \lim_{n \to \infty} \lim_{ε \to \frac{\pi}{2}}\int_0^ε{\frac{(\sin(x))^n}{1-\sin(x)}}\,\d x},$$

but it doesn't help me, since $1 - \sin(x) \leq 1, \forall x \in \left[0, \dfrac{\pi}{2}\right]$.

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  • $\begingroup$ You can write the integrand as $\sum\limits_{k=n}^{\infty}\sin^k x$, if that helps. Don't know if it does. $\endgroup$ – MPW Feb 9 '18 at 22:07
  • $\begingroup$ Is it an exercise? Did you copy the exercise correctly? If it is not an exercise, how you came up with it? The claim is wrong as you can see from the answers below so you cannot prove it. $\endgroup$ – Shashi Feb 10 '18 at 12:07
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Your integral does event convergence, for each $n$ we have $$ \int_0^{\fracπ2}\frac{(\sin x)^n}{1-\sin x}=\infty$$

In fact Since see here $$\frac2πx≤\sin x≤x,~~~~~~\forall x \in \left[0, \displaystyle \frac{\pi}{2}\right]$$ we have

$$\frac{(\frac2πx)^n}{1-\frac2πx}≤\frac{(\sin x)^n}{1-\sin x}≤\frac{x^n}{1-x}\implies \int_0^{\fracπ2}\frac{(\frac2πx)^n}{1-\frac2πx}dx≤\int_0^{\fracπ2}\frac{(\sin x)^n}{1-\sin x}≤\int_0^{\fracπ2}\frac{x^n}{1-x}dx$$ then let $u= \frac2πx$ the we get

$$\infty=\int_0^{1}\frac{x^n}{1-x}dx≤\int_0^{\fracπ2}\frac{(\sin x)^n}{1-\sin x}≤\int_0^{1}\frac{x^n}{1-x}dx+\int_1^{\fracπ2}\frac{x^n}{1-x}dx=\infty$$

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  • $\begingroup$ Please provide a proof for the fact that $\int_0^1{\frac{x^n}{1-x}} \to \infty$, becuase I am pretty sure it tends to 0 as $n \to \infty$... $\endgroup$ – C_M Feb 10 '18 at 6:21
  • $\begingroup$ why a down vote ? may you care to explain? $\endgroup$ – Guy Fsone Feb 10 '18 at 10:52
  • $\begingroup$ $\lim_{n \to \infty}{\int_0^1\frac{x^n}{1-x}dx} \neq \infty$ $\endgroup$ – C_M Feb 10 '18 at 11:12
  • $\begingroup$ @C_M did you compute the integral before you apply the limit? or you are just doing the reverse way $\endgroup$ – Guy Fsone Feb 10 '18 at 11:22
  • $\begingroup$ @C_M for each n that is equals to infinity there is no reason to apply the limit $\endgroup$ – Guy Fsone Feb 10 '18 at 11:23
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The integral fails to converge for all $n$ as $1-\sin x \sim (\pi/2-x)^2$ near $\pi/2.$

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  • $\begingroup$ All too easy. (+1) $\endgroup$ – Mark Viola Feb 9 '18 at 22:20
  • $\begingroup$ Near $\pi/2,$ I get $1 - \sin x \rightarrow 0,$ and $(1-x)^2 \rightarrow (1-\pi/2)^2$ $\endgroup$ – saulspatz Feb 9 '18 at 22:22
  • $\begingroup$ @saulspatz That was a typo, thank you for the correction. $\endgroup$ – zhw. Feb 9 '18 at 22:25
  • $\begingroup$ @C_M Each integral is infinite. The sequence is just $\infty,\infty, \infty,\dots$ There's nothing to do $\endgroup$ – zhw. Feb 14 '18 at 20:00
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just a hint

Write the integral as $$I_1+I_2=$$

$$\int_0^{\frac {\pi}{2}-\epsilon}+\int_{\frac {\pi}{2}-\epsilon}^\frac\pi 2$$ with $$I_1\le \frac {(\cos (\epsilon))^n}{1-\cos (\epsilon) }$$ goes to zero. and

$$I_2\le \epsilon \frac {1}{\cos (\epsilon)} $$

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  • $\begingroup$ I am facing similar issue. but any way what is your point when give the upper bound. ? I dont think the integral converges at all $\endgroup$ – Guy Fsone Feb 10 '18 at 11:36
  • $\begingroup$ @GuyFsone I still cannot delete from mobile. thanks in advance. $\endgroup$ – hamam_Abdallah Feb 10 '18 at 17:15

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