$\int_{1}^\infty\frac{\sin(x^2)}{x^p}dx$ 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 $p>1$, 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
 A: 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 $1<p$.
A: 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?
A: 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 $p>0.$
