While preparing the Calculus final exam, I came across this improper integral from a past exam:

$\int_{0}^{\infty}{x\ln\left(x\right)\over \sqrt{x^5+1}}\,\mathrm{d}x$

I used some online calculators to see if it converges and it does (and it totally makes sense when you look at the graph of the function). The problem is that I'm supposed to prove that it converges using only comparison and quotient tests. This is what I've done:

1) Work with the absolute value to be able to make sure the function is always positive and be able to use the comparison and quotient tests:

$\int_{0}^{\infty}{\left|{x\ln\left(x\right)\over \sqrt{x^5+1}}\right|}\,\mathrm{d}x = \int_{0}^{\infty}{{x\,\mid ln\left(x\right)\mid\over \sqrt{x^5+1}}}\,\mathrm{d}x $

2) Split the integral:

$\int_{0}^{1}{x\ln\left(x\right)\over \sqrt{x^5+1}}\,\mathrm{d}x + \int_{1}^{\infty}{x\ln\left(x\right)\over \sqrt{x^5+1}}\,\mathrm{d}x$

3) For the integral from 0 to 1 I used the comparison test:

$\int_{0}^{1}{x\ln\left(x\right)\over \sqrt{x^5+1}}\,\mathrm{d}x \leq \int_{1}^{0}{1\over \sqrt{x^5+1}}\,\mathrm{d}x$.

Since ${1\over \sqrt{x^5+1}}$ is continuous and bounded in (0,1], it converges and thus the original integral converges in (0,1].

4) For the integral from 1 to infinity, I used the quotient test with $\int_{1}^{\infty}{\mid\ln\left(x\right)\mid\over x^{3/2}}\,\mathrm{d}x$. The limit of the quotient is 1, so the original function converges iff $\int_{1}^{\infty}{\mid\ln\left(x\right)\mid\over x^{3/2}}\,\mathrm{d}x$ converges. And then I calculate the integral using integration by parts and it is equal to 4. Therefore, it converges.

I'm not very confident about how I solved it, so I would really appreciate any comments. Also, calculating the last integral was quite nasty and I don't think I would have time to do it in an exam, so I think there must be a better way to do this that I'm missing. Any ideas?


The basic steps are all correct and well conceived. A few smallish comments:

First split the integral at $x=1$, then consider absolute values. The integrand is negative on $(0,1)$ and positive on $(1, \infty)$. So the integrand needs to estimated a bit differently in step 3.

The integral may be expressed in terms of the $\Gamma$ function. It's value is approximately 3.74431.

  • $\begingroup$ Thank you! I'll give it a try. $\endgroup$ – muripic Aug 3 '19 at 17:46

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