Show that the Lebesgue intergral $\int_{1}^{\infty} x^{-b} e^{\sin {x}} \sin {(2x)}\, dx$ exists iff $b>1$

Question

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?

Answer 1

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:

$$\int_{n \pi}^{(n+1) \pi} \frac{| \sin x|}{x}\, dx \geqslant \frac{2}{(n+1) \pi}$$

Answer 2

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.