Show that the Lebesgue intergral $\int_{1}^{\infty} x^{-b} e^{\sin {x}} \sin {(2x)}\, dx$ exists iff $b>1$ Assume $b>0$. Show that the Lebesgue intergral $\int_{1}^{\infty} x^{-b} e^{\sin {x}} \sin {(2x)}\, dx$ exists iff $b>1$.
We know if $b>1$ the integrand can be bounded and it's just an integral $\int_{1}^{\infty} x^{-b}\,dx$. I don't know how to show the other direction?

Hint: For $0 < b \leqslant 1$
$$\int_1^c |x^{-b}e^{\sin x} \sin 2x| \, dx > e^{-1}\int_1^c \frac{|\sin 2x|}{x} \, dx = \frac{1}{e}\int_2^{2c} \frac{|\sin x|}{x} \, dx $$
Show the integral on the RHS is not finite as $c \to \infty$ using:
$$\int_{n \pi}^{(n+1) \pi} \frac{| \sin x|}{x}\, dx \geqslant \frac{2}{(n+1) \pi}$$
