Prove $\int\limits_1^\infty x^a\sin x \, dx$ diverges for $a>1$

Question

Let $a>1$. I need to show that $$ \int_1^\infty x^a\sin x \, dx $$ diverges. I am not sure, but this is my progress

We will look first at intervals $[2m\pi,(2m+2)\pi]$. Then $$ \int_{2m\pi}^{(2m+2)\pi}x^a \sin x\,dx = \int_{2\pi m}^{(2m+1)\pi} x^a\sin x \, dx + \int_{(2m+1)\pi}^{(2m+2)\pi} x^a\sin x \, dx $$ In $[2\pi m,(2m+1)\pi]$, $\sin(x)>0$, and therefore, $$ 2\pi m\leq x\leq (2m+1)\pi \Leftrightarrow (2\pi m)^{a-1}x\sin x \leq x^{a-1}x\sin x \leq \left[(2m+1)\pi\right]^{a-1}x\sin x $$ and in $[(2m+1)\pi,(2m+2)\pi]$, $\sin(x)<0$, and therefore $$ (2m+1)\pi\leq x\leq (2m+2)\pi $$ $$ \Leftrightarrow \left[(2m+2)\pi\right]^{a-1}x\sin x \leq x^{a-1}x\sin x \leq \left[(2m+1)\pi\right]^{a-1}x\sin x $$ From this, I get that $$ \int_{2\pi m}^{(2m+1)\pi}x^a\sin{x}dx + \int_{(2m+1)\pi}^{(2m+2)\pi} x^a\sin x \, dx \geq $$ $$ (2\pi m)^{a-1}\int_{2\pi m}^{(2m+1)\pi}x\sin x \, dx + \left[ (2m+2)\pi \right]^{a-1} \int_{(2m+1)\pi}^{(2m+2)\pi} x\sin x\,dx $$ $$ = (2\pi m)^{a-1}\cdot (4m+1)\pi - \left[(2m+2)\pi\right]^{a-1}\cdot(4m+3)\pi $$ How can I proceed from here?

Answer 1

I would integrate over $[2n\pi, 2n\pi + \pi].$ Then you can say that

$$\int_{2n\pi}^{2n\pi+\pi}x^a\sin x\, dx > (2n\pi)^a\int_{2n\pi}^{2n\pi+\pi}\sin x\, dx = (2n\pi)^a\cdot 2.$$

(This will actually give you divergence for any $a\ge 0.$)

Answer 2

We have for all positive integers $n$,

$$\left| \int_{2n\pi}^{2n\pi + \pi/3}x^a \sin u \, du\right| \geqslant (2n \pi)^a\int_{2n\pi}^{2n\pi + \pi/3} \sin u \, du = \frac{1}{2}(2n \pi)^a.$$

The integral diverges since the Cauchy criterion is not satisfied.

For convergence it is necessary that for all $\epsilon > 0$ there exists $K > 0$ such that for all $b_2 > b_1 >K$ we have

$$\left| \int_{b_1}^{b_2}x^a \sin x \, dx\right| < \epsilon.$$

Answer 3

You can see what happens on intervals $[n\,\pi,(n+1)\,\pi]$. On them $\sin x$ has constant sign. Then $$ \Bigl|\int_{n\pi}^{(n+1)\pi}x^a\sin x\,dx\Bigr|=\int_{n\pi}^{(n+1)\pi}x^a|\sin x|\,dx\ge(n\,\pi)^a\int_{n\pi}^{(n+1)\pi}|\sin x|\,dx=2\,\pi^a\,n^a. $$ If the integral were convergent, we would have hat the following limits exist: $$ \lim_{n\to\infty}\int_\pi^{n\pi}x^a\sin x\,dx=\lim_{n\to\infty}\sum_{k=1}^{n-1}\int_{k\pi}^{(k+1)\pi}x^a\sin x\,dx. $$ This would imply that the series $\displaystyle\sum_{k=1}^\infty\int_{k\pi}^{(k+1)\pi}x^a\sin x\,dx$ converges, which is impossible since the general term does not converge to $0$.

Observe that the proof is valid for $a\ge0$.