For what values of p the following integral converges: $\int_{1}^{\infty} \sin(x) \cdot x^p dx$

Question

For what values of p>0 the following integral converges: $$\int_{1}^{\infty} \sin(x) \cdot x^p dx$$

None of the classic convergence tests seem to work.

A hint or direction would be appreciated.

Answer 1

Note that,

$\left | \int\limits_1 ^ \infty x^p\sin x \,dx\right | \le \int\limits_1 ^ \infty |x^p\sin x| \,dx \le \int\limits_1 ^ \infty x^p \,dx$

Answer 2

Since $p>0$, $x^p>1$ so there are intervals where $\sin(x) \cdot x^p>1/2$
and other intervals where $\sin(x)\cdot x^p < -1/2$.

It doesn't converge if $p>0$.

Answer 3

If $p$ is an integer, integrating by parts $p$ times shows that the integral is not convergent. Let's show that the integral is not convergent for all $p>1$. For any $n\in\Bbb N$ $$ \int_{(2n-1)\pi}^{(2n+1)\pi}x^p\sin x\,dx=\int_{(2n-1)\pi}^{2n\pi}(x^p-(x+\pi)^p)\sin x\,dx=\int_{(2n-1)\pi}^{2n\pi}((x+\pi)^p-x^p)|\sin x|\,dx. $$ By the mean value theorem and the fact that $p\ge1$ we have $$ (x+\pi)^p-x^p=p\,\xi^{p-1}\,\pi\ge p\,\pi\,x^{p-1}, $$ where in the above $x\le\xi\le x+\pi$. Then $$ \int_{(2n-1)\pi}^{(2n+1)\pi}x^p\sin x\,dx\ge\int_{(2n-1)\pi}^{2n\pi}p\,\pi\,x^{p-1}|\sin x|\,dx\ge2\,p\,\pi((2\,n-1)\pi)^{p-1}, $$ showing that the integral does not converge.

If $0<p\le1$, we argue in a similar way but use that, since $p-1\le0$, $$ (x+\pi)^p-x^p=p\,\xi^{p-1}\,\pi\ge p\,\pi\,(x+\pi)^{p-1}. $$ Then $$ \int_{(2n-1)\pi}^{(2n+1)\pi}x^p\sin x\,dx\ge\int_{(2n-1)\pi}^{2n\pi}p\,\pi\,(x+\pi)^{p-1}|\sin x|\,dx\ge2\,p\,\pi(2\,n\,\pi)^{p-1}. $$ Since $1-p>0$, the series $\sum n^{p-1}$ diverges, and so does the integral.

Answer 4

Shifting the bottom limit from 1 to $\pi$ doesn't make a difference to convergence, so note that we have $$\begin{align*} \int_\pi^\infty x^p \sin x\, dx &= \int_\pi^{2\pi} x^p \sin x\, dx + \int_{2\pi}^{3\pi} x^p\, \sin x\, dx + \int_{3\pi}^{4\pi} x^p\, \sin x\, dx + \cdots \\ &= -\int_0^\pi (\sin x) ((x+\pi)^p - (x+2\pi)^p + (x+3\pi)^p - (x+4\pi)^p + \cdots)\, dx.\end{align*}$$ The series in the integrand is an alternating series and has a finite value if and only if the terms decrease in absolute value, which they do (for all $x$) if an only if $p < 0$.

Answer 5

If $p=0,$ we already know the integral diverges. If $p>0,$ then

$$\int_{2n\pi + \pi/4}^{2n\pi + \pi/2}\sin x\cdot x^p\,dx \ge \frac{\sqrt 2}{2}\cdot (2n\pi)^p \to \infty.$$

Thus the integral is divergent. If $p<0,$ then $x^p$ decreases on $[1,\infty)$ to $0,$ and so the integral converges by Dirichlet's test.