improper integral that neither converges nor diverges?

Question

I've been kind of stuck in the following problem.

Determine if the following integral exists: $$\int_0^\infty\frac x{1+x^{3/2}}\,\mathrm dx$$

I think it doesn't but I'm not that convinced. Look, the integral can be written as: $$\int_0^1\frac x{1+x^{3/2}}\,\mathrm dx+\int_1^\infty\frac x{1+x^{3/2}}\,\mathrm dx.$$ The first integral is no problem, since the argument is a continuous function on the compact set $[0,1]$. For the second, notice that the argument is bounded as below: $$\frac1{2x^3}<\frac1{1+x^3}<\frac x{1+x^{3/2}}<\frac1{x^{1/2}},$$ so by the comparison test, there's not enough information to say something about the integral we're concerned with, indeed the integral of the lower bound converges while the integral of the upper bound diverges. Is there any other idea to study its existence?

Thanks!

Answer 1

You're correct the integral doesn't exist. Note for $x \gt 1$ that

$$1 \lt x^{3/2} \implies 1 + x^{3/2} \lt 2x^{3/2} \implies \frac{1}{1 + x^{3/2}} \gt \frac{1}{2x^{3/2}} \tag{1}\label{eq1A}$$

This gives

$$\frac{x}{1+x^{3/2}} \gt \frac{x}{2x^{3/2}} = \frac{1}{2x^{1/2}} \tag{2}\label{eq2A}$$

Since

$$\int_{1}^{\infty} \frac{1}{2x^{1/2}}dx = \left. x^{1/2} \right\rvert_{1}^{\infty} = \infty \tag{3}\label{eq3A}$$

diverges, this means by the direct comparison test the integral

$$\int_{0}^{\infty} \frac{x}{1+x^{3/2}} \tag{4}\label{eq4A}$$

also diverges.

Answer 2

There is another to look at the problem.

For $x$ close to zero, using Taylor for example, the integrand is $x-x^{5/2}+\cdots$; so no problem at the lower bound.

For large $x$ , the integrand is $\frac 1 {\sqrt x}- \frac 1x+\cdots$ and the first term makes a serious problem once integrated trying to use the upper bound.