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$$\int_1^\infty\frac{\sin(x^2)}{x^p} \, dx$$ for which p values does the integral converge in condition? and for which values does it converge absolutely?

I managed to find the values of $p$ which will make the integral to converge absolutely , they are $p>1$, but i could not manage to solve for the condition values of $p.$

$$\int_1^\infty\frac{\sin(x^2)}{x^p} \, dx $$ assign $t=x^2$ you will get : $$\frac{1}{2}\int_1^\infty\frac{\sin(t)}{t^\frac{p+1}{2}} \, dt$$ $$\int_1^\infty \left|\frac{\sin(t)}{t^\frac{p+1}{2}}\,dt\right| \le \int_1^\infty\frac{1}{t^\frac{p+1}{2}}\,dt$$ hence if $\frac{p+1}{2}>1 \to p>1$ the integral converges.

In the answers the value for the conditional convergence is for $0<p\le1$ can't understand why

Michael Hardy
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Maor Rocky
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  • Conditional convergence follows the well-known Abel-Dirichlet test for improper integrals. – xbh Sep 19 '18 at 15:40

3 Answers3

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Integrating by parts, (for $n>0$ only (why?)) $$\int^\infty_1 \frac{\sin x}{x^n}dx=\cos 1-(1+n)\int^\infty_1\frac{\cos x}{x^{n+1}}dx$$

By integrating by parts repeatedly, you can always make the power in the denominator as large as possible to obtain absolute convergence, so the only condition is $n>0$.

Correspondingly, $p>-1$ (which is quite counter-intuitive to me).

To show the type of convergence (absolute or conditional), we may consider $$\begin{align} &~~~~\int^\infty_1\frac{|\sin x|}{x^n}dx \\ &\ge\int^\infty_1\frac{\sin^2 x}{x^n}dx \\ &=\frac12\int^\infty_1\frac{1-\cos 2x}{x^n}dx \\ &=\frac12\int^\infty_1\frac1{x^n}dx-\frac12\int^\infty_1\frac{\cos 2x}{x^n}dx \end{align} $$

We have shown that the second term always converges. Therefore, absolute convergence of the original integral is equivalent to the convergence of the first term, which can be examined by p-test.

The result is the convergence is absolute for $1<p$.

Szeto
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  • By using your approach, you have shown only that the integral converges for $p>-1$. You have not shown that while the integral converges conditionally for $-1<p$, it converges absolutely only if $p>1$? And while Abel-Dirichet might be unnecessary to show conditional convergence, I would not call its application herein "overkill." That is an opinion, which I've taken the liberty to delete. – Mark Viola Sep 19 '18 at 16:58
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As the OP showed, we have

$$\int_1^L \frac{\sin(x^2)}{x^p}\,dx=\frac12\int_1^{L^2} \frac{\sin(x)}{x^{(p+1)/2}}\,dx\tag1$$

Using the Abel-Dirichlet test, the integral on the right-hand side of $(1)$ converges (conditionally) whenever $\frac12(p+1)>0$. So, for $p>-1$, the integral in $(1)$ is convergent.

It can be shown that it is absolutely convergent only when $\frac12(p+1)>1$ or $p>1$. To do so, express the integral on the right-hand side of $(1)$ into the sum of integrals

$$\int_1^{L^2} \frac{\sin(x)}{x^{(p+1)/2}}\,dx=\int_1^\pi \frac{\sin(x)}{x^{(p+1)/2}}\,dx+\sum_{k=1}^{\lfloor L^2/\pi\rfloor -1}\int_{k\pi}^{(k+1)\pi} \frac{\sin(x)}{x^{(p+1)/2}}\,dx+\int_{\lfloor L^2/\pi\rfloor \pi}^{L^2} \frac{\sin(x)}{x^{(p+1)/2}}\,dx$$

Can you finish now?

Mark Viola
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  • This covers all aspects well (+1) – RRL Sep 19 '18 at 18:16
  • @RRL Much appreciated! How are you? It's nice to see you've returned. – Mark Viola Sep 19 '18 at 19:09
  • Hi Maor. Please let me know how I can improve my answer. I really want to give you the best answer I can. If this was not useful, I am happy to delete it. If you would, please let me know either way. Thank you. – Mark Viola Oct 10 '18 at 14:38
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If you graph the integrand and consider the sequence of positive and negative areas, you'll see that you have an alternating series with terms of decreasing absolute value, for any $p>0.$

B. Goddard
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