# Thread: How to show the divergence of integral

1. ## How to show the divergence of integral

Hello, how to show the integral diverges:

$\displaystyle \int_{1}^{\infty}$

2. ## Re: How to show the divergence of integral

Without an integrand and differential, that's going to be a bit tricky.

3. ## Re: How to show the divergence of integral

@MarkFL, the integral is: $\displaystyle \int_{1}^{\infty}\dfrac{\sqrt{x}}{\sqrt[3]{x^2 -1}}dx$...

4. ## Re: How to show the divergence of integral

Originally Posted by lebdim
@MarkFL, the integral is: $\displaystyle \int_{1}^{\infty}\dfrac{\sqrt{x}}{\sqrt[3]{x^2 -1}}dx$...
I would begin with the fact that for $\displaystyle x\in(1,\infty)$ we have:

$\displaystyle \frac{1}{x^{\Large\frac{1}{6}}}< \frac{\sqrt{x}}{\sqrt[3]{x^2-1}}$

Now, let's disregard the singularity at the lower bound, and consider:

$\displaystyle I=\int_2^{\infty} \frac{\sqrt{x}}{\sqrt[3]{x^2-1}}\,dx$

What does the $\displaystyle p$-test tell us about the convergence of $\displaystyle I$?

5. ## Re: How to show the divergence of integral

Ok, thank you. Because $\displaystyle p = \frac{1}{6} < 1$ and then it's divergent.

6. ## Re: How to show the divergence of integral

Because $\displaystyle p$-series are convergent for $\displaystyle p > 1$. Thank you!