# How do you test the improper integral int x^3 dx from (-oo, oo) and evaluate if possible?

Aug 2, 2017

The integral is not convergent as:

${\int}_{- \infty}^{+ \infty} {x}^{3} \mathrm{dx} = {\lim}_{u \to \infty} {\int}_{- u}^{0} {x}^{3} \mathrm{dx} + {\lim}_{v \to \infty} {\int}_{0}^{v} {x}^{3} \mathrm{dx}$

${\int}_{- \infty}^{+ \infty} {x}^{3} \mathrm{dx} = {\lim}_{u \to \infty} - {u}^{4} / 4 + {\lim}_{v \to \infty} {v}^{4} / 4$

The two limits should be finite separately and they are not.

The integral is however convergent in the sense of Cauchy's principal values as ${x}^{3}$ is an odd function, so:

${\int}_{- t}^{t} {x}^{3} \mathrm{dx} = {\left[{x}^{4} / 4\right]}_{- t}^{t} = 0$

and then:

${\lim}_{t \to \infty} {\int}_{- t}^{t} {x}^{3} \mathrm{dx} = 0$