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please don't mark this as a replicated post, nobody is answering me on the old one.

Can anyone explain what to look for next:

Find the values of $p>0$ for which the following integral exists:$$I =\int^{\infty}_{1} x^{-p}\sin{(x)} dx$$ which has an infinite bound so let $$I_{t}=\int^{t}_{1} x^{-p}\sin{(x)} dx$$ now if $f(x) = \frac{\sin{(x)}}{x^{p}}$ and let $g(x) = \frac{1}{x^{p}}$ such that $f(x)\le g(x)$. So for $p>1$ we have that $f(x)$ converges.

If $p=1$ we have $$I =\int^{\infty}_{1} \frac{\sin{(x)}}{x} dx$$ which also converges.

So we now know it converges for $p\ge 1$ now how do I analyse $p<1$?

user2850514
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  • What you have is $f \leq g$ in any case. However, for $p>1$ $\int_{1}^{\infty}g < \infty$. So, $I < \infty$. – voldemort Apr 12 '14 at 16:36
  • It was mentioned in comments to the previous question. We have $|\sin x|\le 1$. For $p\gt 1$, $\int_1^\infty \frac{1}{x^p},dx$ converges (say by Integral Test). Therefore by Comparison $\int_1^\infty \frac{|\sin x|}{x^p},dx$ converges. since absolute convergence implies convergence, we conclude that our integral convrges. – André Nicolas Apr 12 '14 at 16:37

1 Answers1

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Your argument is true only for $p \geq 1$, since $\int_0^{\infty} x^{-p}dx$ converges only for $p>1$. For $p < 1$, the integral converges by a generalized version of the alternating test.

This can be seen as follows:

We have $$I = \int_{1}^{\pi}\dfrac{\sin(x)}{x^p}dx + \sum_{n=1}^{\infty} \int_{n \pi}^{(n+1)\pi}\dfrac{\sin(x)}{x^p}dx = \int_{1}^{\pi}\dfrac{\sin(x)}{x^p}dx + \sum_{n=1}^{\infty}(-1)^n a_n$$ where $a_n = \left \vert \displaystyle \int_{n \pi}^{(n+1)\pi}\dfrac{\sin(x)}{x^p}dx\right \vert$.

Show that $a_n$ is monotone decreasing and converges to $0$ and conclude using the generalized alternating test or the Dirichlet's test.

user141421
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  • It seems from your sum you have $a_{0}=\int^{\pi}{0} \frac{\sin{(x)}}{x^{p}} dx$ but clearly the real $a{0} = \int^{\pi}{1} \frac{\sin{(x)}}{x^{p}} dx$. Also how do you find whether $a{n}$ converges? – user2850514 Apr 12 '14 at 17:17
  • @user2850514 Split the integral $$\int_1^{\infty} = \int_1^{\pi} + \int_{\pi}^{2\pi} + \int_{2\pi}^{3\pi} + \int_{3\pi}^{4\pi} + \cdots + \int_{n\pi}^{(n+1)\pi} + \cdots$$ $a_0$ is a typo and I have fixed it now.

    $a_n \to 0$, since $$a_n < \int_{n \pi}^{(n+1)\pi} \dfrac{dx}{x^p} \to 0$$

     – user141421 Apr 12 '14 at 17:19
  • One final thing, to simplify a little, can we let $a_{n} = |\int^{\pi}{0} \frac{\sin{(x)}}{x^{p}} dx|$ since the only thing which changes for each $n$ is the sign which will also occur with that new $a{n}$. – user2850514 Apr 12 '14 at 20:15
  • please read the above comment, thanks – user2850514 Apr 12 '14 at 21:13