I need to show that
$$
\int\limits_1^{\infty}x^\alpha\sin(x^\beta)dx
$$
converges if and only if $\alpha + 1 < \beta$. I used substitution and integration to get
$$
\lim\limits_{t\to\infty}\int\limits_1^tx^\alpha\sin(x^\beta)dx \overset{u = x^\beta}{\underset{du = \beta x^{\beta-1}}{=} }\lim\limits_{t\to\infty}\left(\frac{1}{\beta}\int\limits_1^{t^\beta}u^{\frac{\alpha + 1 – \beta}{\beta}}\sin{u}du\right)
$$
If $\alpha + 1 -\beta < 0$, then we have a product if a function with bounded anti-derivative and a differentiable decreasing function which tend to 0, and from Dirichlet's criterion, the integral converges.
For the converse, I used integration by parts
$$
= \frac{1}{\beta}\lim\limits_{t\to\infty}\left(\left.-u^{\frac{\alpha + 1-\beta}{\beta}}\cos{u}\right|_1^{t^\beta} + \frac{\alpha+1-\beta}{\beta}\int\limits_1^{t^\beta}u^{\frac{\alpha + 1-2\beta}{\beta}}\cos{u}du\right)
$$
For $\alpha + 1 -\beta\geq 0$, I get that the first term indeed diverges, but this is not enough to show total divergence. How can one show it?
Best Answer
For $\beta>0$, $$ \int_1^{+\infty}x^\alpha\sin(x^\beta)\mathrm dx=\frac{1}{\beta}\int_1^{+\infty}u^{\frac{\alpha + 1 - \beta}{\beta}}\sin{u}\mathrm du=\frac{1}{\beta}\int_1^{+\infty}\frac{\sin u}{u^{p}} \mathrm du $$ with $p=1-\frac{\alpha + 1 }{\beta}$.
If $p >1$ the integral $\int_1^{+\infty}\frac{\sin u}{u^{p}}\mathrm du$ converges absolutely; in fact $\left|\frac{\sin u}{u^{p}}\right|\le\frac{1}{u^{p}}$ and $\int_1^{+\infty}\frac{1}{u^{p}}\mathrm du$ converges for $p>1$. From the Comparison Test, $\int_1^{+\infty}\left|\frac{\sin u}{u^{p}}\right|\mathrm du$ converges.
If $0< p\le 1$, the integral $\int_1^{+\infty}\frac{\sin u}{u^{p}}\mathrm du$ converges conditionally. For given $z\ge 1$, $$\left|\int_1^{z}\sin u\mathrm du\right|=\left|\cos 1-\cos z\right|\le 2$$ and $\frac{1}{u^{p}}$ approaches zero monotonically as $u\to\infty$ for $p>0$. Dirichlet's Test implies that $\int_1^{+\infty}\frac{\sin u}{u^{p}}\mathrm du$ converges for $p >0$. On the other hand, it is clear that $$ \left|\frac{\sin u}{u^{p}}\right|\ge \frac{\sin^2 u}{u}=\frac{1}{2u}-\frac{\cos(2u)}{2u} $$ where $\int_1^{+\infty}\frac{\cos(2u)}{2u}\mathrm du$ converges by Dirichlet's Test, and $\int_1^{+\infty}\frac{1}{2u}\mathrm du$ diverges. Hence for $0< p\le 1$, the integral $\int_1^{+\infty}\frac{\sin u}{u^{p}}\mathrm du$ converges conditionally.
In a similar way you can discuss the convergence of the integral for $\beta<0$, $$ \int_1^{+\infty}x^\alpha\sin(x^\beta)\mathrm dx=\frac{-1}{\beta}\int_0^1 u^{\frac{\alpha + 1 - \beta}{\beta}}\sin{u}\mathrm du=\frac{-1}{\beta}\int_0^1\frac{\sin u}{u^{p}} \mathrm du=\frac{1}{\gamma}\int_0^1\frac{\sin u}{u^{p}} \mathrm du $$ with $\gamma=-\beta>0$ and $p=1-\frac{\alpha + 1 }{\beta}=1+\frac{\alpha + 1 }{\gamma}$.
For $u\to 0$, we have $\sin u\sim u$ and then $\frac{\sin u}{u^{p}}\sim \frac{1}{u^{p-1}}$. Thus the integral converges for $p-1<1$, that is $p<2$ or $$ \alpha+1 <\gamma=-\beta$$