How to show $\int_{0}^{\infty}e^{-x}x^{-1} dx = \infty$ How do we obtain result


  $$\int_{0}^{\infty}e^{-x}x^{-1} \;\mathrm dx = \infty$$
How about $$\int_0^\infty e^{-x} x^{-1} \, dx \ge e^{-1} \int_0^1 x^{-1} \, dx = \infty?$$
Notice, 
Laplace transform of a function divided by $t$ is given as $$L\left[\frac{f(t)}{t}\right]=\int_{s}^{\infty}L[f(t)]ds$$

Now, using Laplace transform as follows $$\int_{0}^{\infty}e^{-x}x^{-1}dx$$
$$=\int_{0}^{\infty}e^{-x}\frac{1}{x}dx$$
$$=L\left[\frac{1}{t}\right]_{s=1}$$
$$=\int_{s=1}^{\infty} L\left[1\right] ds$$
$$=\int_{1}^{\infty} \frac{1}{s} ds$$
$$=[\ln |s|]_{1}^{\infty}$$ $$=\lim_{s\to \infty}\ln |s|-\ln|1|$$
$$=\lim_{s\to \infty}\ln |s|=\infty$$