- #1
Rectifier
Gold Member
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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.
Please help me to solve this mystery.
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.
Please help me to solve this mystery.