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I have to show that $\displaystyle\int_{0}^{\pi/2}\displaystyle\frac{x^n}{\sin^mx}dx$ exists iff $m<n+1$

I am not exactly sure, how to go about it, and though I am studying Real Analysis and have a faint idea that perhaps it is related to Darboux Sums, but I am completely ignorant about it. Can you help me on this, and also give me some references where I can learn the theory behind this


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Are there any conditions on $m$ and $n$? – Antonio Vargas Sep 21 '12 at 22:02
I have added the condition, though I was expecting the condition that m<n+1 will automatically pop up. – Soham Sep 21 '12 at 22:06
$\sin^m x\sim x^m$ near $0$, so you want $x^n$ to go to zero faster than $x^m$. This is achieved by looking at $$x^{n-m}$$ It goes to zero for $x\to 0$ iff $n+1>m$. – Pedro Tamaroff Sep 21 '12 at 22:23
up vote 6 down vote accepted

Since $2x/\pi\le\sin x\le x$ for $0\le x\le\pi/2$, this integral exists if and only if

$$ \int_0^{\pi/2}x^{n-m}\mathrm dx $$

exists. This is

$$ \lim_{\epsilon\to0}\int_\epsilon^{\pi/2}x^{n-m}\mathrm dx=\lim_{\epsilon\to0}\left[\frac1{n-m+1}x^{n-m+1}\right]_\epsilon^{\pi/2}\;, $$

and this limit exists iff $n-m+1\gt0$.

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I am chuckling at my own idiocy. Thanks. – Soham Sep 21 '12 at 22:32

Because the integrand in question is continuous on the interval you're integrating over, it typically isn't necessary to go through the technical details of going back to the full definition of the integral (Riemann, Lebesgue, or otherwise) unless the question specifically asks for it.

In this case, and in the case of most integrals of continuous functions, the easiest way to show existence/nonexistence of integrals is to find an inequality that forces the integral to be finite or that forces the integral to be infinite. I would try to do such an estimate on this integral by noticing that the integrand is finite on the inside of the interval, but blows up near the origin, and then try to use the fact that $x/\sin(x) \rightarrow 1$ as $x \rightarrow 0$. Then you can compare the integral in question to a much simpler integral that you can evaluate more directly.

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Oh thank you so very much for the way you put your thoughts. I must admit, though I did understand that we can think about the integrand as $x^{n-m}$*$(x/sinx)^m$ but couldnt foresee how to go from there... much thanks that was really helpful. Plus I am checking jorickis answer – Soham Sep 21 '12 at 22:27

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