I have been working through problems in Spivak Calculus and in Chapter 14, (Fundamental Theorem of Calculus) Problem 26 it asks if the Integral $\displaystyle\int_0^\infty \frac{\mathrm{d}x}{\sqrt{1+x^3}}$ exists. Now I thought it doesn't because it diverges because the function is unbounded on the interval 0 to 1 by the p-test and comparison, even though from 1 to infinity it converges for the same reason.
But the answer says it exists because it exists from 1 to infinity because it converges on that interval.
I would be grateful if someone could explain this meaning to me.
thank you very much!
By definition an integral is improper when either:
In that case, as already noticed in the comments, we have a bounded function but the integral is said improper because we have $\infty$ as upper limit, that is
$$\int_0^\infty \frac{\mathrm{d}x}{\sqrt{1+x^3}}=\lim_{a\to \infty}\int_0^a \frac{\mathrm{d}x}{\sqrt{1+x^3}}$$
which converges since
$$\frac{1}{\sqrt{1+x^3}} \le \frac{1}{\sqrt{x^3}}$$
and
$$\lim_{a\to \infty}\int_1^a \frac{\mathrm{d}x}{\sqrt{x^3}}=\lim_{a\to \infty} \left(-\frac{2a}{\sqrt{a^3}}+2\right)=2$$