Convergence of improper integral $\int_0^\infty x^{\alpha +1} e^{-x} dx$

Question

Determine the convergence of the following improper integral as $\alpha \in \mathbb{R}$ varies:

$\int_0^\infty x^{\alpha +1} e^{-x} dx$

I tried to do it in this way

$e^{-x}<\frac{1}{x^{\beta}}$ as $x\rightarrow \infty$ for any $\beta \in \mathbb{R}$ Therefore $x^{\alpha +1}e^{-x}<\frac{x^{\alpha +1}}{x^{\beta}}=\frac{1}{x^{\beta-\alpha-1}}$ as $x\rightarrow \infty$

So if $\beta-\alpha-1>1 \rightarrow \alpha<\beta-2$ the integral is convergent

But the solutions say that it is convergent for $\alpha>-2$ and divergent for $\alpha \leq -2$

Where is my mistake?? Besides is this the right way to do this exercise? I mean I used a second arbitrary parameter which I can also delete at the end since $\beta$ can be $0$. Is it wrong to use it?

Thanks for your help

Answer 1

Hint.

$$\int_0^\infty x^{\alpha +1} e^{-x} dx=\int_0^1 x^{\alpha +1} e^{-x} dx+\int_1^\infty x^{\alpha +1} e^{-x} dx.$$

Verify that the second integral is always convergent, no matter what real number $\alpha$ may be, as can be seen by applying the limit test (and using e.g. the convergent reference integral $\int_1^\infty dx/x^2 $).

Additional hint. Notice that since $e^{-x}\rightarrow 1$ as $x\rightarrow 0$ the integrand of $\int_0^1x^{\alpha +1}e^{-x}dx$ behaves like $x^{\alpha +1}$ near $x=0$.

Answer 2

$\displaystyle \int_0^{\infty}e^{-x}x^{\alpha+1}\,dx=\displaystyle \int_0^{\infty}e^{-x}x^{(\alpha+2)-1}\,dx=\Gamma(\alpha+2)$ which is convergent for $\alpha+2>0$, that is $\alpha>-2$.