improper (double) integral: $\int_0^\infty\int_x^\infty\frac{1}{\sqrt{t^{3}+1}}\,\mathrm{d}t\,\mathrm{d}x$

Question

I want to determine if the integral $\,\displaystyle\int_0^\infty\displaystyle\int_x^\infty\frac{1}{\sqrt{t^{3}+1}}\,\mathrm{d}t\,\mathrm{d}x$ converges.

I know that $\displaystyle\int_x^\infty\frac{1}{\sqrt{t^{3}+1}}\,\mathrm{d}t$ converges for all $x \geq 0$ and can show this by the comparison theorem. I just am not sure how to use this fact to justify the convergence or divergence of $\displaystyle\int_0^\infty\displaystyle\int_x^\infty\frac{1}{\sqrt{t^{3}+1}}\,\mathrm{d}t\,\mathrm{d}x$.

Could someone point me in the right direction?

Thanks!

Answer 1

Consider the other iterated integral:

$$\int_0^\infty \int_0^t \frac{1}{\sqrt{t^3 + 1}} dx dt = \int_0^\infty \frac{t}{\sqrt{t^3 + 1}} dt$$

This integral is divergent, since the integrand is (asymptotically) about $t/\sqrt{t^3} = 1/\sqrt{t}$. Hence, Fubini's Theorem implies the divergence of the original iterated integral as well.

Answer 2

You could use direct comparison. For $t\in[0,\infty)$, $$\frac{1}{\sqrt{t^3+1}}\geq\frac{1}{\sqrt{(t+1)^3}}=(t+1)^{-3/2}$$ The double integral on this expression is easy to compute; and to see that it is divergent to $\infty$.