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For which values $a,b \in \mathbb{R}$ does the integral

$$ \int_1^{+\infty} x^a e^{bx}\, dx $$ converge?

I bear in mind the case $\int_1^{+\infty} x e^{-x}\, dx$, that clearly converges. By similar arguments, if $a$ is a positive integer and $b<0$, we always have convergence by integral by parts.

But what can I say in general? Does it suffice to take $b<0$ in order to ensure convergence?

Guy Fsone
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TheWanderer
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1 Answers1

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It converges iff $b<0$ or ($b=0$ and $-a>1$)

Note that $$\int_1^\infty\frac{dx}{x^2} =1$$

  • If $b<0$ then $x^{a+2}e^{bx} \to 0$ as $x\to \infty$

Hence for $x$ large enough we have $$x^{a+2}e^{bx} <1\implies x^a e^{bx} \le\frac{1}{x^2} $$ the convergence follows by comparison test.

  • If $b=0$ and $a<-1$ that is $-1-a>0$ then we have $$\int_1^\infty\frac{dx}{x^{-a}} =[\frac{1}{(a+1)x^{-1-a}}]_1^\infty = -\frac{1}{(a+1)}$$
  • Now if $b>0$ then then $x^{a+1}e^{bx} \to \infty$ as $x\to \infty$

Hence for $x$ large enough we have $$x^{a+1}e^{bx} >1\implies x^a e^{bx} >\frac{1}{x} $$ the divergence follows by comparison test.

Guy Fsone
  • 23,903