For which $p>0$ does the following improper integral exist? $$\int^{\infty}_{1} x^{-p}\sin{x} \ dx$$ how do I find the value of p?
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Hint: $$\left|x^{-p}\sin x\right|\le x^{-p};\ldots$$ – DonAntonio Apr 11 '14 at 17:02
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@DonAntonio: this helps prove that the integral converges for all $ p > 1 $, but what about $ p = 1 $? Integrating by parts shows that the integral is convergent even in that case, whereas $ x^{-1} $ obviously isn't integrable on $ (1,+\infty) $... – derpy Apr 11 '14 at 17:04
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1@derpy, that's was one hint. – DonAntonio Apr 11 '14 at 17:07
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I suppose you're right. :P My bad. – derpy Apr 11 '14 at 17:10
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The integral is convergent for any $ p > 0 $; just integrate by parts to get $$ \int_{1}^{+\infty}{\frac{\sin(x)}{x^p}} = -\frac{\cos(x)}{x^p}\Big|_{1}^{+\infty} - p\int_{1}^{+\infty}{\frac{\cos(x)}{x^{p+1}}}, $$ and the integral at the rhs obviously converges since its modulus is no greater than $ 1/x^{p+1} $ (which is summable on $ (1,+\infty) $).
If $ p \le 0 $ the integral is apparently non convergent, since it oscillates without being "tamed".
derpy
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@user2850514: $ 1 / x^p $ is still summable at infinity for $ p > 2 $. What do you mean, it diverges? – derpy Apr 12 '14 at 18:46