Bounding $\int_0^{\infty} x^p e^{-x} dx$ to find the values of p for integral convergence.

Question

Find all the values of p for which the integral converges $$\int_0^{\infty} x^p e^{-x} dx$$

I am thinking about bounding the integral using $$\int_0^{\infty} x^p e^{-2x} dx \leq \int_0^{\infty} x^p e^{-x} dx \leq \int_0^{\infty} x^p e^{-0.5x} dx $$

But i do not see those 2 bounds very helpful.

Next option: I was thinking to choose a tangent line as lower bound but can't see one line as upper bound as the original curve runs to infinity.

What could a more adequate approach be here?
Much appreciated

Answer 1

For $x \in [0,1]$ we have $e^{-1} x^p \leqslant x^p e^{-x} \leqslant x^p$.

This should help you determine convergence and divergence of $\int_0^1 x^p e^{-x} \, dx$ by the comparison test.

For $[1, \infty)$ we have

$$\lim_{x \to \infty} \frac{x^p e^{-x}}{x^{-2}} = \lim_{x \to \infty} \frac{x^{p+2}}{ e^{x}} = 0 $$

Since $\int_1^\infty x^{-2} \, dx $ converges, by the limit comparison test $\int_1^\infty x^p e^{-x} \, dx$ converges for all $p$.

Thus, the values of $p$ for which the integral $\int_0^\infty x^p e^{-x} \, dx $ converges/diverges is dictated by the conditions for convergence/divergence of $\int_0^1 x^p e^{-x} \, dx.$ As you will see this is convergence for $p > -1$ and divergence for $p \leqslant -1.$

Answer 2

This is the Gamma Function. Integrating by parts is one of the main tools... For $p\in \mathbb{Z_+}$, $\Gamma (p+1)=p!$. By analytic continuation $\Gamma$ is defined for all complex numbers except the negative integers. ...