$$\int_{0}^{\pi} \frac{\sin{(x)}}{(x+n\pi)^{p}} dx$$ where $p>0$ And $n\in\mathbb{N}$. I understand we can compare this to $$\int_{0}^{\pi} \frac{1}{(x+n\pi)^{p}} dx$$ which tells us it converges for $p\ge1$ but we don't know what happens for $p<1$. How do you determine convergence at this point?
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The integral is over a bounded interval, so the only possible problem (since the integrand is continuous except possibly at one point) is that the integrand blows up. The only point where it can blow up is $0$, and that can only happen for $n = 0$. – Daniel Fischer Apr 15 '14 at 14:08
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How about $$\frac{1}{(x+n\pi)^p} \leq \frac{1}{\pi^p n^p}?$$ Getting rid of $x$ entirely will make your estimations that much easier...
Another way is to simply calculate $$\int_0^\pi \frac{1}{(x+n\pi)^p} dx$$ which equals $$\frac{1}{(x+n\pi)^{p-1}}|_0^\pi = \frac{1}{((n+1)\pi)^{p-1}} - \frac{1}{(n\pi)^{p-1}}$$ for $p\neq 1$ and $$\ln\left(\frac{(n+1)}{n}\right)$$ for $p=1$. The convergence is obvious in all cases.
5xum
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Not really, for $p<1$, the integral will still be a regular integral (the function is bounded and continuous on $[0,\pi]$, therefore the convergence of the integral is not in question. Are you sure you wrote the question down correctly? – 5xum Apr 15 '14 at 18:05