# Thread: Help with comparison test for improper integrals?

1. ## Help with comparison test for improper integrals?

I have a problem concerning the comparison test for improper integrals. The problem is: Consider the integral of dx/(x^1/3 + x^5/3) from 0 to infinity Determine whether this integral is convergent or divergent and explain why. Do not try to evaluate the integral.

I broke the integral into two parts, one going from 0 to 1 and the other going from 1 to infinity. Both are improper integrals. I reasoned that the one going from 1 to infinity is convergent since there's that x^(5/3) in the denominator, which is greater than 1, causing that integral to converge. However, that doesn't tell me whether or not the entire integral converges or diverges since there's the discontinuity at 0 for the other improper integral.

My reasoning for the second integral is that I can take out the x^(1/3) since in the 0 to 1 bound it's value is between 0 and 1. Then I can compare it to a smaller integral to see if the smaller integral diverges, which if it does, then the larger integral must diverge as well. Can I compare it to 1/(x^2)? I compared it to that and got that 1/(x^2) from 0 to 1 diverges, so the second integral must diverge as well right? So the entire integral diverges? Did I do it right? Thanks.

2. Those fractional powers are annoying, then put $x=u^3$ and the integral becomes $\int_{0}^{\infty }{\frac{dx}{x^{1/3}+x^{5/3}}}=\int_{0}^{\infty }{\frac{3u^{2}}{u+u^{5}}\,du}.$

Now, $\int_{0}^{\infty }{\frac{u^{2}}{u+u^{5}}\,du}=\int_{0}^{1}{\frac{u^ {2}}{u+u^{5}}\,du}+\int_{1}^{\infty }{\frac{u^{2}}{u+u^{5}}\,du}.$

For the first piece, consider the function defined by $f(u)=\left\{\begin{array}{cr}\dfrac{u^{2}}{u+u^{5} },&\text{if }u\ne0,\\[0,3cm]0,&\text{if }u=0,\end{array}\right.$

hence $f$ is a continuous function on $[0,1]$ and thus, integrable on $[0,1].$ From here, the first piece of the integral converges; now, let's check the second one: for $u\ge1$ it's $\frac{u^{2}}{u+u^{5}}\le \frac{u^{2}}{u^{5}}=\frac{1}{u^{3}},$ hence the second piece converges by direct comparison with $\int_1^\infty\frac{du}{u^3}<\infty.$

Both pieces converges, so does the original integral.