Test for convergance the integral

Question

1. Test for convergance the integral $$ \int_1^\infty \frac{\sin4x \cos2x}{(x+1)^\frac{1}{3}} \, \mathrm{d}x. $$

I'm confused with this integral because I can't find a comparison. $\frac{\sin x}{x}$ can't help us in solving this problem.

Thank you.

Answer 1

HINT:

Using the addition angle formula for the sine function, write $$\sin(4x)\cos(2x)=\frac12 \left(\sin(6x)+\sin(2x)\right)$$

Then, use the fact that $\int_0^\infty \frac{\sin(x)}{x^a}\,dx$ converges for $0<a$ which is a consequence of Dirichlet's Test.

Answer 2

This integral isn't absolutely convergent, that is $$ \int_1^\infty \left|\frac{\sin 4x\cos 2x}{(x+1)^{1/3}}\right|dx = \infty$$ which means that all comparison tests will be useless., as they can only be used to show absolute convergence. You'll need a completely different method.

Hint: define $$ a_n = (-1)^n\int_{n\pi}^{(n+1)\pi} \frac{\sin 4x\cos 2x}{(x+1)^{1/3}} dx$$ You have $$ \int_{\pi}^\infty \frac{\sin 4x\cos 2x}{(x+1)^{1/3}} dx = \sum_{n=1}^\infty (-1)^n a_n$$ Try to show that $a_n > 0$ and $a_n \rightarrow 0$, and from the Leibniz critrion you'll get the convergence.