I want to prove $ \int_0^\infty \frac{e^{-x}}{x} dx = \infty $ How can I prove this integral diverges?
$$ \int_0^\infty \frac{e^{-x}}{x} dx = \infty $$

For $0\lt x\lt 1$, our function is $\gt e^{-1}\frac{1}{x}$.  
Thus $\int_\epsilon^1 \frac{e^{-x}}{x}\,dx \gt -e^{-1}\log(\epsilon)$. But $-\log(\epsilon)$ blows up as $\epsilon$ approaches $0$ from the right.
