Convergence of a double integral

Question

Is the integral $$\int_1^\infty\int_{e^{-x}}^1\frac{\sin y}{x^2y}dy dx$$ convergent or divergent?

Answer 1

Firstly note that the integrand function is positive in your domain. Since $$\int_0^1 \frac{\sin y}{y} dy =L < +\infty$$ you can see that $$\int_1^{+ \infty} \frac{1}{x^2}\int_{e^{-x}}^1 \frac{\sin y}{y} dy \ dx \le \int_1^{+ \infty} \frac{L}{x^2} dx < + \infty$$ so the integral converges.

Answer 2

It is convergent. $\frac{\sin y}{y}$ is an integrable function on it's own.

You could take the lower bound of the $y$ integral as 0 (overestimation), then the y integral will just be a number. And the x integral will be 1, so it will be finite.