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I have to see if $$\int_{0}^{\infty}\frac{1}{\sqrt{x^3+x}}dx$$ is convergent or not.

I know I have two improper points so $$\int_{0}^{\infty}\frac{1}{\sqrt{x^3+x}}dx = \int_{0}^{1}\frac{1}{\sqrt{x^3+x}}dx + \int_{1}^{\infty}\frac{1}{\sqrt{x^3+x}}dx$$

For the integral from 1 to 0 it's used the following comparison $$0\leq \frac{1}{\sqrt{x^3+x}}\leq \frac{1}{\sqrt{x}}$$

and for the integral from 1 to infinity it's used $$0\leq \frac{1}{\sqrt{x^3+x}}\leq \frac{1}{\sqrt{x^3}}$$

Why in one case is used $\sqrt{x}$ and in the other the $\sqrt{x^3}$?

In both cases isn't this valid

$$0\leq \frac{1}{\sqrt{x^3+x}}\leq \frac{1}{\sqrt{x^3}}$$

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up vote 4 down vote accepted

Even though the inequality is true, the fact that $\displaystyle 0\leq \frac{1}{\sqrt{x^3+x}}\leq \frac{1}{\sqrt{x^3}}$ won't tell you much about the integral from $0$ to $1$ since $\displaystyle \int_ 0 ^1 \frac{1}{\sqrt{x^3}}dx$ diverges, because $\displaystyle 1\leq \frac{3}{2}$.

Let $a,b,\alpha$ be real numbers such that $a<b$.

The integrals $\displaystyle \int \limits_a^{b^-} \frac{1}{(b-x)^\alpha}\mathrm dx , \int \limits_{a^+}^b \frac{1}{(x-a)^\alpha}\mathrm dx$ converge if, and only if, $\alpha <1$.

To verify this you just need to compute the antiderivatives and take the limits.

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And could I use this inequality $0\leq \frac{1}{\sqrt{x^3+x}}\leq \frac{1}{\sqrt{x}}$? This way it converges – Favolas Jan 26 '13 at 12:22
Yes, it converges because the power of the denominator $\sqrt{x}$ is $\displaystyle \frac{1}{2}$ and $\frac{1}{2} < 1$. I will add some stuff to my answer to make it more complete. – Git Gud Jan 26 '13 at 12:28
Many thanks foe your help. It's still hard do pick one comparison tent. Got do do more exercises :) – Favolas Jan 26 '13 at 14:19

The last inequality you give is valid, but $1/\sqrt{x^3}=x^{-3/2}$ is not integrable near $x=0$.

Think of it this way: When $x$ is very small, $x^3\ll x$ (in words: $x^3$ is much smaller than $x$), so $x^3+x$ is very close to $x$ (in relative terms).

On the other hand, when $x$ is very large, $x^3\gg x$, so that $x^3+x$ is very close to $x^3$ (again, in relative terms).

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Hi. So what you are saying is that I have to look at the limits. If the limit is defined (lets say 0 to 1) I have to see the smaller exponent in the denominator. If the limit is undefined (lets say 0 to infinity) I have to see the bigger exponent in the denominator. Is this? – Favolas Jan 26 '13 at 12:28
@Favolas: Not quite the way I would have put it, but yes. – Harald Hanche-Olsen Jan 26 '13 at 13:14
Thanks for your help – Favolas Jan 26 '13 at 14:20

First of all, it's your right to choose different comparison functions, since your problem was split and reduced to the analysis of two "subproblems". Moreover, you are also forced to consider different comparison functions, because you're studying convergence on different intervals!

As you know, $\displaystyle \int x^{-\alpha}$ is convergent near 0 iff $\alpha < 1$, while it is convergent near $+\infty$ iff $\alpha>1$. Thus you see that $$ \dfrac{1}{\sqrt{x^3+x}}\leq \dfrac{1}{\sqrt{x^3}}=x^{-\frac32}$$ is unuseful near 0, because $\displaystyle \int_0^1 x^{-\frac32}$ goes to infinity. By the comparison test, you can't say anything about the finiteness of something which is just $< +\infty$.

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