Does the double integral $\int_1^\infty \int_0^x \frac{1}{x^3+y^3} \,dy \,dx$ converge or diverge?

Question

I would need to determine whether the following double integral converges or diverges: $$\int_1^\infty \int_0^x \frac{1}{x^3+y^3}\, dy\, dx$$

I made a change of variable to polar coordinates and therefore I got: $$\int_0^\frac{\pi}{4} \int_1^\infty \frac{r}{r^3\cos^3(\theta)+r^3\sin^3(\theta)} \,dr \,d\theta$$

Which simplifies to: $$\int_0^\frac{\pi}{4} \frac{1}{\cos^3(\theta)+\sin^3(\theta)} d\theta$$

I can't get any further from here. I cannot evaluate this integral by hand and I don't know what I should do next. Is this approach correct in this problem? Thanks!

Answer 1

Another solution using no fancy change of variables. Just note that $\frac{1}{x^3+y^3}\leq \frac{1}{x^3}$ for $y\geq 0$ and thus,

$$ \int_1^{\infty} \int_0^x \frac{1}{x^3+y^3}\textrm{d}y\textrm{d}x\leq \int_1^{\infty} \frac{1}{x^2}\textrm{d}x<\infty $$

Answer 2

Your approach is fine, but not the best one. Hint for the last part: show that for $\theta\in [0,\pi/4]$, $$f(\theta)=\underbrace{\cos^3(\theta)}_{>0}+\underbrace{\sin^3(\theta)}_{\geq 0}$$ is positive and continuous. What may we conclude?

By the way, your integral in polar coordinates is GREATER than the given one (which is OK in order to prove convergence): $$\int_0^\frac{\pi}{4} \int_1^\infty \frac{r}{r^3\cos^3(\theta)+r^3\sin^3(\theta)} dr d\theta=\iint_D\frac{1}{x^3+y^3} dy dx> \int_1^\infty \int_0^x \frac{1}{x^3+y^3} dy dx$$ where $D=\{(x,y): x^2+y^2\geq 1,0\leq y\leq x\}\supset \{(x,y): x\geq 1,0\leq y\leq x\}$.

Answer 3

I believe that if you use polar coordinates your region actually simplifies to the following: $$I=\int_1^\infty\int_0^x\frac{1}{x^3+y^3}dydx=\int_1^\infty\int_{\pi/2-\arctan(x)}^{\pi/2}\frac{rdrd\theta}{r^3\cos^3\theta+r^3\sin^3\theta}$$ However as you may notice it is difficult to get rid of this term involving $x$.

Answer 4

The polar coordinates bounds are incorrect. It should really be

$$\int_0^{\frac{\pi}{4}}\int_{\sec\theta}^\infty \frac{1}{r^2(\cos^3\theta+\sin^3\theta)}dr\:d\theta = \int_0^{\frac{\pi}{4}}\frac{\cos\theta}{\cos^3\theta+\sin^3\theta}\:d\theta$$

$$=\int_0^{\frac{\pi}{4}}\frac{\sec^2\theta}{1+\tan^3\theta}\:d\theta = \int_0^1\frac{1}{1+t^3}\:dt = \frac{\log 2}{3}+\frac{\pi}{3\sqrt{3}}$$

from partial fraction decomposition.