# Integral test of convergence

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1. Jan 6, 2017

### Rectifier

The problem
I am trying to show that the following integral is convergent
$$\int^{\infty}_{2} \frac{1}{\sqrt{x^3-1}} \ dx$$

The attempt
$x^3 - 1 \approx x^3$ for $x \rightarrow \infty$.

Since $x^3 -1 < x^3$ there is this relation:
$\frac{1}{\sqrt{x^3-1}} > \frac{1}{\sqrt{x^3}}$ (the denominator is smaller for function with $x^3-1$ and hence bigger quotient).

I know since earlier that the integral $\int^{\infty}_{2} \frac{1}{\sqrt{x^3}} \ dx$ is convergent. But the problem is that function $\frac{1}{\sqrt{x^3}}$ is smaller than the other one and therefore always "below" that other function. The other functions integral can diverge and $\int^{\infty}_{2} \frac{1}{\sqrt{x^3}} \ dx$ will still be convergent like nothin' happened.

2. Jan 6, 2017

### Staff: Mentor

Why don't you find an integral that is always "above" that other function?

3. Jan 6, 2017

### Rectifier

Will $\int^{\infty}_{2} \frac{1}{\sqrt{x^3}} + 1$ do the job? Smells fishy though.

4. Jan 6, 2017

### Ray Vickson

For $x > 1$ we have $\sqrt{x^3-1} = x^{3/2} \sqrt{1-x^{-3}}$. Try to find a nice constant lower bound on $\sqrt{1-x^{-3}}$ that is good for $x \geq 2$ (or, at least, for large positive $x$).

5. Jan 6, 2017

### Staff: Mentor

I don't know. I hope the $1$ isn't meant to be part of the integrand. How about $x^{-\frac{5}{4}}$?

6. Jan 6, 2017

### FactChecker

Use a change of variable x = a+1 and (after a little manipulation) compare the integral to the integral of 1/(a3/2)

7. Jan 6, 2017

### Rectifier

On which integral?

8. Jan 6, 2017

### Rectifier

Why do I need to find a constant lower bound? Isn't the bound x=2 already?

9. Jan 6, 2017

### Rectifier

That one is convergent too but I have to show that it is always above the original function.

In other words:

I must show that
$\frac{1}{\sqrt{x^3-1}}< \frac{1}{x^{\frac{5}{4}}}$ for all x:es

Any suggestions on how I can do that?

10. Jan 6, 2017

### Staff: Mentor

Do some algebra. Remove the quotients, the square roots and so on.

11. Jan 6, 2017

### Rectifier

I solved it another way by showing that the quotient of the first function and the second is not infinite for $x \rightarrow \infty$. If you have any alternative solution strategies please share them here.

12. Jan 6, 2017

### Staff: Mentor

I did ($x>0$ ) : $\frac{1}{\sqrt{x^3-1}}<\frac{1}{x^\frac{5}{4}} \Longleftrightarrow x^\frac{5}{4}<\sqrt{x^3-1} \Longleftrightarrow x^\frac{5}{2}<x^\frac{6}{2}-1 \Longleftrightarrow x^6 > x^5 + 2x^3 - 1$ which is true for $x\geq 2$ .

13. Jan 6, 2017

### Ray Vickson

There are other ways of doing your problem, but finding a positive constant $c$ such that $\sqrt{1-x^{-3}} > c$ for large $x>0$ is certainly one way to do it. Your previously-expressed worries about $1/\sqrt{x^3}$ not being an upper bound on $1/\sqrt{x^3-1}$ would then go away.

Last edited: Jan 6, 2017