Study the convergence of this improper integral

Question

$$ \int_o^\infty t^ae^{bt}dt $$ for a,b reals.

I guess I would have to separate this integral in many cases for different values of a and b.

I know that if b < 0,

$$ \int_o^\infty t^ae^{bt}dt > \int_o^\infty {t^a}dt $$, which diverges for any a. The other cases are more tricky. Any help would be appreciated.

The last inequality is not true, since $e^{bt}<1$ for large $t$ and $b<0$. — mrprottolo, Apr 20 '16 at 09:08
Another point is that $$ \int_o^\infty t^ae^{bt}dt > \int_o^\infty {t^a}dt $$ is not true for $b<0$. — Guilherme Thompson, Apr 20 '16 at 09:18

score 1 · Answer 1 · answered Apr 20 '16 at 09:20

If $b>0$ then $$\lim_{t \to +\infty} t^ae^{bt}=+\infty$$ for any $a$ and the integral diverges. If $b<0$ then the near $0$ you have $1<e^{bt}<2$ hence the integral converges if and only if $a>-1$, since there is no problem at infinity. For $b=0$ the integral never converges.

Study the convergence of this improper integral

1 Answers1