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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 $1 - \sin(x) \leq 1, \forall x \in \left[0, \dfrac{\pi}{2}\right]$.

Guy Fsone
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C_M
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  • You can write the integrand as $\sum\limits_{k=n}^{\infty}\sin^k x$, if that helps. Don't know if it does. – MPW Feb 09 '18 at 22:07
  • 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. – Shashi Feb 10 '18 at 12:07

3 Answers3

2

The integral fails to converge for all $n$ as $1-\sin x \sim (\pi/2-x)^2$ near $\pi/2.$

zhw.
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2

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

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

Guy Fsone
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  • 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 $n \to \infty$... – C_M Feb 10 '18 at 06:21
  • why a down vote ? may you care to explain? – Guy Fsone Feb 10 '18 at 10:52
  • $\lim_{n \to \infty}{\int_0^1\frac{x^n}{1-x}dx} \neq \infty$ – C_M Feb 10 '18 at 11:12
  • @C_M did you compute the integral before you apply the limit? or you are just doing the reverse way – Guy Fsone Feb 10 '18 at 11:22
  • @C_M for each n that is equals to infinity there is no reason to apply the limit – Guy Fsone Feb 10 '18 at 11:23
  • From my point of view, you did it the reverse way. By all means, prove me wrong, I want to see how you solve the integral. – C_M Feb 10 '18 at 11:26
  • @C_M I have solve it ? that integral is obviously infinity since the singularity is at x=1. then at x= 1 we have $$\frac{x^n}{1-x}\sim \frac{1}{1-x}$$ this is blatantly obvious I don't see how I can prove something obvious – Guy Fsone Feb 10 '18 at 11:29
  • You can switch the limit with the integral only if the integral converges uniformly. I want to see how you prove that it converges uniformly. – C_M Feb 10 '18 at 11:30
  • @C_M my friend where did I use the limit in my text? nowhere. please read at least the first sentence I have added for you . for all n we have this $$\int_0^{\fracπ2}\frac{(\sin x)^n}{1-\sin x}=\infty$$ then why do you want to use the limit here ? it is non sense to use the limit there – Guy Fsone Feb 10 '18 at 11:32
  • My apologies, but I just don't see why. This is not "obvious". Open any graphing calculator, use every approximation for $\frac{\pi}{2}$, make $n \to \infty$ and you will see that integral doesn't diverge to $\infty$ – C_M Feb 10 '18 at 11:44
  • @C_M I have added the last line why it is infinity if you dont get that one then I am sorry tooo. there is nothing I can do for you all details a re there – Guy Fsone Feb 10 '18 at 11:45
  • @C_M thanks for removing your down vote – Guy Fsone Feb 10 '18 at 12:31
  • You are right. Also, your idea to use $\frac{2}{\pi}x \leq \sin{x} \leq x$ helped me solve the problem from which I got the integral my question. – C_M Feb 10 '18 at 12:33
  • @C_M. that is indeed a powerfull inequality. you should keep it in mind – Guy Fsone Feb 10 '18 at 12:34
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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

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

hamam_Abdallah
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