Improper integral

[Eden20]

Can someone help me determine if this improper integral is converges or diverges?

 

[Deleted member 4993]

Can someone help me determine if this improper integral is converges or diverges?

Please show us what you have tried and exactly where you are stuck.

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Please share your work/thoughts about this problem

 

[blamocur]

View attachment 33280

Can someone help me determine if this improper integral is converges or diverges?

I'll read this in bed

 

[BigBeachBanana]

 

[nasi112]

I would guess that it converges. There is no easy way to show that.

My guess is based on a similar problem.

$\displaystyle \int_2^{\infty}\ln(x^3 + 1) - \ln(x^3 - 1) \ dx = 0.2503272784470577$

The difference between the two integrals is that there is a way to solve this integral.

 

[limiTS]

Turn $$\int_{1}^{\infty}\sqrt{x^3+1}-\sqrt{x^3-1}\,\text{d}x$$ into:
$$\int_{1}^{\infty}\sqrt{x^3}\sqrt{1+\frac{1}{x^3}}-\sqrt{x^3}\sqrt{1-\frac{1}{x^3}}\,\text{d}x$$ and then:
$$\int_{1}^{\infty}\sqrt{x^3}\left[\sqrt{1+\frac{1}{x^3}}-\sqrt{1-\frac{1}{x^3}}\right]\text{d}x$$ and eventually:
$$\int_{1}^{\infty}x^{3/2}\left[\left(1+\frac{1}{x^3}\right)^{1/2}-\left(1-\frac{1}{x^3}\right)^{1/2}\right]\text{d}x$$
As $x\to\infty$, the $\left(1+\frac{1}{x^3}\right)^{1/2}$ and the $\left(1-\frac{1}{x^3}\right)^{1/2}$ can both be expanded using the binomial expansion.

The dominant $1'\text{s}$ are subtracted away, and then the next most dominant terms become a $p$-integral of the form:
$$\int_{1}^{\infty}\frac{1}{x^p}\text{d}x$$The value of $p$ determines if the integral converges or diverges.

 

[pka]

View attachment 33280

Can someone help me determine if this improper integral is converges or diverges?

This is anything but an elementary problem. Yes it does converge to $\approx 2.02336$

To find that number must use the $\Gamma (x)$ function.


$$$$

 

[lookagain]

Can someone help me determine if this improper integral is converges or diverges?

Here is a suggestion/hint:

Rewrite the integrand by using the conjugate:

$\displaystyle \dfrac{2}{ \ \sqrt{x^3 + 1} \ + \ \sqrt{x^3 - 1} \ }$

Then come up with an appropriate related expression of which to compare it.