How can I prove this integral diverges?
$$ \int_0^\infty \frac{e^{-x}}{x} dx = \infty $$
How can I prove this integral diverges?
$$ \int_0^\infty \frac{e^{-x}}{x} dx = \infty $$
$$ \int_{0}^{\infty}\frac{e^{-x}}{x}= \int_{0}^{1}\frac{e^{-x}}{x}+\int_{1}^{\infty}\frac{e^{-x}}{x} \\ > \int_{0}^{1}\frac{e^{-x}}{x} \\ > e^{-1}\int_{0}^{1}\frac{1}{x} $$ which diverges.
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.
$s>0$
$$ \int_0^\infty e^{-sx} dx = \frac{1}{s} $$
$$ \int _1^\infty \int_0^\infty e^{-sx} dx ds = \int _1^\infty \frac{1}{s}ds $$
$$ \int_0^\infty \int _1^\infty e^{-sx} ds dx = \int _1^\infty \frac{1}{s}ds $$
$$ \int_0^\infty (\frac{e^{-sx}}{-x}) |_{s=1}^\infty dx = \int_1^\infty \frac{1}{s}ds $$
$$ \int_0^\infty \frac{e^{-x}}{x} dx = \ln\infty -\ln 1 = \infty$$