Improper Integral Question (exponential integral)

Question

I'm trying to prove that $\int_{0}^{\infty} x^{t} e^{-x} dx$ diverges for $t \leq -1.$

I figured I'd split it up into $\int_{0}^{1} x^{t} e^{-x} dx + \int_{1}^{\infty} x^{t} e^{-x} dx.$

If I can show that the first integral in the sum diverges, then I think I know the original integral diverges.

But how should I go about showing that the first integral diverges? In an earlier problem, I showed that $\int_{0}^{1} x^t dt$ diverges for $t \leq -1$. I was thinking of trying to find an integral that is less than my original integral but also diverges, but I got stuck.

Any ideas?

Answer 1

HINT:

Note that $ x^t e^{-x}\ge e^{-1} x^t$ for $x\in [0,1]$.

Answer 2

The problem is at the lower bound.

If you use Taylor series, $$x^t e^{-x}=\left(1-x+\frac{x^2}{2}+O\left(x^3\right)\right) x^t$$ So, since the first term is $x^t$, using what you already proved, the integral diverges if $t \leq -1$ and converges for $t>-1$.

Just for your curiosity, $$\int x^t\, e^{-x}\,dx=-\Gamma (t+1,x)$$ where appears the incomplete gamma function and, provided $\Re(t)>-1$, $$\int_0^\infty x^t\, e^{-x}\,dx=\Gamma (t+1)$$