1. ## improper intergal question

Establish convergence/divergence of the following improper integral:

int from 0 to infinity of 1 / ( x^(1/3)*(/x-5/^(1/3))*(1 + sqrt(x))^0.7) )

My attempt at a solution was to break it up into 3 intergrals: 0 to 1, 1 to 5, and 5 to infinity...I showed that the first two of these integrals converges by comparing them to 1/x^(1/3) and 1//x-5/^(1/3) respectively. I can't establish convergence/divergence of last intergral from 5 to infinity.

Thanks, any help is much appreciated.

2. Originally Posted by davido
Establish convergence/divergence of the following improper integral:

$\int_0^\infty\!\!\! \frac1{x^{1/3}|x-5|^{1/3}(1 + \sqrt{x})^{0.7}}\,dx$

My attempt at a solution was to break it up into 3 integrals: 0 to 1, 1 to 5, and 5 to infinity...I showed that the first two of these integrals converges by comparing them to 1/x^(1/3) and 1//x-5/^(1/3) respectively. I can't establish convergence/divergence of last integral from 5 to infinity.
You probably know that $\int^\infty \!\!\!x^{\alpha}dx$ converges if $\alpha<-1$. So you want to estimate the power of x in the denominator of that fraction, as $x\to\infty$. The first factor is $x^{-1/3}$, the second one is also approximately $x^{-1/3}$ (because |x–5| will have the same order of magnitude as x, when x is large). Similarly, the third factor is approximately $x^{-(1/2)0.7} = x^{-7/20}$. So the integrand is approximately $x^{-(1/3)-(1/3)-(7/20)} = x^{-61/60}$.

Since –61/60 < –1, you should conclude that the integral converges on the interval $[5,\infty)$, and you prove it by comparing it with $\int^\infty \!\!\!x^{-61/60}dx$.