# Improper integral substitution hint

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.

• If all you need to do is show that the integral exists then remember that $|\sin t|\leq 1$. That won't cover all cases, but it is a start. – B. S. Thomson Oct 29 '15 at 23:57
• Make a substitution $x^q=u$. – A.S. Oct 30 '15 at 0:40
• Near $0$, use $\sin t\simeq t$. Near $\infty$, see the Riemann-Lebesgue lemma. – Lucian Oct 30 '15 at 4:53
• I have $\int_0^\infty\ u^\frac{p-q+1}q\ sin\ u\ du$ – Camilo Acevedo. Oct 30 '15 at 4:54
• Then you have convergence as long as the power $u$ is raised to is less then $0$ (for convergence around $\infty$) and greater then $-2$ (for convergence around $0$). – A.S. Oct 30 '15 at 11:57

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 $-1<\frac{p+1}{q}<0$.
• Why do you need $\frac {p+1}q<0$? Only $<1$ is needed for convergence. – A.S. Oct 30 '15 at 12:05
• @A.S.: we have to grant integrability in a right neighbourhood of the origin and in a left neighbourhood of $+\infty$. – Jack D'Aurizio Oct 30 '15 at 12:59
• $\sin$ is oscillating so the integral will converge conditionally for $\frac {p+1}q\in [0,1)$. – A.S. Oct 30 '15 at 13:04
• @A.S.: oh, you're right, convergence is granted by $\left|\frac{p+1}{q}\right|<1$, but my manipulations through the Laplace transform are working only for a smaller range. I believe we may enlarge that range by integrating by parts, as usual. – Jack D'Aurizio Oct 30 '15 at 13:09