Lebesgue integrability $e^{-\frac{y}{x}}\frac{sen(x)}{y}$

Question

Let's $D := {\{(x,y)\in R^2 : x>0 , x^2<y<2x^2}\}$

Is $f(x,y) = e^{-\frac{y}{x}}\frac{sen(x)}{y}$ Lebesgue integrable on D?

I can see that $f$ is measurable not only on D but also on $R^2$ beacuse it is continue (with the possible exeptions of the two axis. In that case I observe they have Lebesgue measure equals to $0$ so they "do not count" in the measurability). This at least makes sense to the problem.

Then, to be integrable it has to be $\int_D |f(x,y)| d\mu < \infty$.

I tried with this $\Big|e^{-\frac{y}{x}}\frac{sen(x)}{y} \Big| \le e^{-\frac{y}{x}}\frac{|sen(x)|}{y}$ and here I get stuck. I don't know if it's better to see $|sen(x)| < |x|$ or directly $|sen(x)| < 1$, and in any case I'm litte confused on how I can go on.

Answer 1

Hint: by Tonelli's theorem, $$\int_D |f(x,y)| d\mu = \int_{x>0} \bigg( \int_{x^2<y<2x^2} |f(x,y)| \,dy \bigg) dx.$$ For each value of $x$, you can use bounds of the sort you mention to get an upper bound for the inner integral as a function of $x$, and then find an upper bound for the resulting integral over $x$. You might find that $|\sin x|\le 1$ is sufficient when $x$ is not too small, but $|\sin x| \le x$ might be better when $x$ is small.

Of course you could also reverse the variables and do the integral over $x$ first (with appropriate limits).