$\int_{1}^\infty\frac{\sin(x^2)}{x^p}dx$ for which p values does the integral converge in condition $$\int_1^\infty\frac{\sin(x^2)}{x^p} \, dx$$ 
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 $p&gt;1$, 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 $\frac{p+1}{2}&gt;1 \to p&gt;1$ the integral converges.

In the answers the value for the conditional convergence is for $0&lt;p\le1$ can't understand why
Integrating by parts, (for $n&gt;0$ 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 $n&gt;0$.

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



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

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 $1&lt;p$.
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 $p&gt;0.$