Test the following improper integral for convergence:
![\[ \int_0^{\infty} \frac{1}{\sqrt{x^3+1}} \, dx. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-e257a44d24a344601e02635525c1b31c_l3.png "Rendered by QuickLaTeX.com")
The integral converges.
Proof. We know the integral 
converges (example #1 on page 417 of Apostol). Applying the limit comparison test (by the note to Theorem 10.25 on page 418, which says that if 
then the convergence of 
implies the convergence of 
.) we have
![\begin{align*} \lim_{x \to +\infty} \frac{f(x)}{g(x)} &= \lim_{x \to +\infty} \frac{ \frac{1}{\sqrt{x^3+1}}}{\frac{1}{x}} \\[9pt] &= \lim_{x \to +\infty} \frac{x}{\sqrt{x^3+1}} \\[9pt] &= \lim_{x \to +\infty} \frac{1}{x^{\frac{1}{2}} \sqrt{1 + \frac{1}{x}}} \\[9pt] &= 0. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-a2cc908a2d1f75b12821e15072ba42aa_l3.png "Rendered by QuickLaTeX.com")
Since we know converges the theorem establishes the convergence of
![\[ \int_0^{\infty} \frac{1}{\sqrt{x^3+1}} \, dx. \qquad \blacksquare \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-2e964ac669ae72f0bb3233aa58bf7e4d_l3.png "Rendered by QuickLaTeX.com")
Sorry, but integral of q/x^{3/2} do not converge from 0 to infinity, but from 1 to infinity. It seems you need to split the integral in two: one from zero to 1 (converges because its a proper integral), and 1 to infinity (converges because of the comparison limit test).
Is that ok?
However, when you split the integral, the integral of x^(-3/2) between 0 and 1 diverges because this integral is not well defined on 0 and it’s limit diverges.
you can split the question’s integral in two integrals from 0 to 1 and from 1 to inf.The first exists and it is finite because the function inside is continuous from 0 to 1.The second can be comparated with the integral of x^(-3/2) from 1 to inf without the problem of inexistence of this integral at 0.
I think the 0 to 1 part should diverge by example 10.23.E5
I take it back, you can’t compare with 1/x^(-3/2) because 10.25 needs it to exist on 0,infty but it doesn’t.