# How to determine whether an improper integral converges or diverges?

For example: $$\int_{9}^{\infty}\frac{1}{\sqrt{x^3+1}}\,dx$$

The answer is that it converges, but why? I am so confused about this kind of questions. I tried to use the comparison test, so that $$\int_{9}^{\infty}\frac{1}{\sqrt{x^3+1}}\,dx<\int_{9}^{\infty}\frac{1}{\sqrt{x^3}}\,dx<\int_{9}^{\infty}\frac{1}{x}\,dx$$

but $\int_{9}^{\infty}\frac{1}{x}\,dx$ diverges. The thing is, the divergence of one integral doesn't tell us anything about the smaller integral. I have no idea how to solve this kind of questions. Can anyone explain to me how to determine convergence/divergence in general?

$$\int_9^{\infty} \frac{dx}{x^{3/2} }$$ converges, since it is $p$-integral with $p> 1$. Hence, by the comparison test, your integral must converge!

• oh I see I misunderstood \sqrt{x^3} ! Also I have a question, is there any possible if the greater one is divergent, and how to define the original one? – user1234567 Mar 2 '16 at 5:51

$\int_{9}^{\infty } \frac{1}{\sqrt{x^3 + 1}} dx< \int_{9}^{\infty } \frac{1}{\sqrt{x^3}}dx \ \forall x>9$

Now $\int_{9}^{\infty } \frac{1}{\sqrt{x^3}}dx = \frac{-2}{\sqrt{x}}|_{9}^{\infty } = \frac{2}{3}$

So $\int_{9}^{\infty } \frac{1}{\sqrt{x^3 + 1}}dx < \frac{2}{3}$
Hence it is convergent by comparison test.

You should not extend the inequality to $\int_{9}^{\infty } \frac{1}{x}dx$ because it's divergent and a convergent integral is always less than a divergent integral, so it's of no use.

Use the fact that $\int_{1}^{\infty } \frac{1}{x^p}dx$ converges for $p>1$ and diverges for $p\leq1$.