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

REPLY [2 votes]: 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 $\infty$.
{"set_name": "stack_exchange", "score": 1, "question_id": 461450}