Answer
$${\text{No}}$$
Work Step by Step
$$\eqalign{
& \int_0^\infty {{x^n}dx} \cr
& {\text{Use the definition of improper integrals}} \cr
& \int_0^\infty {{x^n}dx} = \mathop {\lim }\limits_{b \to \infty } \int_0^b {{x^n}} dx \cr
& {\text{Integrate}} \cr
& = \mathop {\lim }\limits_{b \to \infty } \left[ {\frac{{{x^{n + 1}}}}{{n + 1}}} \right]_0^b \cr
& = \mathop {\lim }\limits_{b \to \infty } \left[ {\frac{{{b^{n + 1}}}}{{n + 1}} - \frac{{{0^{n + 1}}}}{{n + 1}}} \right] \cr
& = \mathop {\lim }\limits_{b \to \infty } \left[ {\frac{{{b^{n + 1}}}}{{n + 1}}} \right] \cr
& {\text{The limit exists when }}n + 1 < 0,{\text{ then }}n < - 1 \cr
& {\text{For }}n < - 1 \cr
& {x^n} = \frac{1}{{{x^n}}},{\text{ and the lower limit is 0, it is no possible}} \cr
& {\text{the division by 0, then it is not possible to find a number}} \cr
& {\text{such that }}\int_0^\infty {{x^n}dx} {\text{ is convergent}}{\text{.}} \cr} $$
