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?

For $0\leq b\leq 1$ : For $n\in N$ and $y_n=2 n \pi +\pi/6$ we have $x\in (y_n,y_n+\pi/6)\implies$  $f(x)=x^{-b}\exp (\sin x)\sin 2 x\geq$ $(y_n+\pi/6)^{-b} K$ where $K=\exp (\sin \pi/6)\sin \pi/3=\sqrt {3 e}/ 2=K>0$. Therefore $$\int_{y_n}^{y_n+\pi/6}f(x)dx\;\geq \;(\pi/6)(2 n+1/6)^{-b} \pi^{-b} K\;\geq \; (\pi/6)\pi^{-b}K/(2 n+1).$$ $$\text {And we have  }(y_n,y_n+\pi/6)\cap (y_{n+1},y_{n+1}+\pi/6)=\phi.$$ In terms of the Lebesgue integral, if $f^+=\{x :f(x)>0\}$ we have $\int_{f^+} f(x) dx=\infty.$ The case $b<0$ is done similarly.
