Improper integral substitution hint

Question

i try solve this improper integral $$\int_0^\infty x^p\sin x^q \ dx$$ I try to compare it with $\displaystyle \int_0^\infty\ \frac{1}{x^p}\ dx$

But I don't know what do when $x\rightarrow\ \infty$ in $\sin x^q$

Any hint I will really appreciate it.

Thanks for your corrections.

Answer 1

If we set $x=z^{\frac{1}{q}}$ we have: $$ I(p,q) = \int_{0}^{+\infty}x^p\sin(x^q)\,dx = \frac{1}{q}\int_{0}^{+\infty}z^{\frac{p+1}{q}}\frac{\sin z}{z}\,dz $$ but: $$ \mathcal{L}\left(\frac{\sin z}{z}\right)=\arctan\frac{1}{s},\qquad \mathcal{L}^{-1}(z^{\alpha}) = \frac{s^{-1-\alpha}}{\Gamma(-\alpha)}$$ hence: $$ I(p,q)=\frac{1}{q\,\Gamma\left(-\frac{p+1}{q}\right)}\int_{0}^{+\infty}s^{-1-\frac{p+1}{q}}\arctan\left(\frac{1}{s}\right)\,ds $$ or: $$ I(p,q)=\frac{1}{q\,\Gamma\left(-\frac{p+1}{q}\right)}\int_{0}^{+\infty}t^{\frac{p+1}{q}}\frac{\arctan t}{t}\,dt $$ so:

$$ I(p,q) = -\frac{\pi}{2(p+1)\,\Gamma\left(-\frac{p+1}{q}\right)\cos\left(\frac{\pi(p+1)}{2q}\right)}$$

provided that $-1<\frac{p+1}{q}<0$.