# An improper integral question

everyone who is interested calculus, I wonder ask a question about the value of an improper integral. Here is the integral: $\int_0^\infty \! \frac{e^{-x}}{x} \, \mathrm{d}x$

Is it diverge ( how to proof it's divergence ), or converge ( how to find it's value )?

Thank you guys if you can help me for this! :)

• This diverges due to the divergent behavior of $\frac{1}{x}$ around the origin. – Cameron Williams Jun 14 '14 at 22:18

In fact, $\int_0^1 \frac{e^{-x}}{x}\,dx$ diverges since on $[0,1]$ we have $e^{-x} > \frac{1}{e}$ and $\int_0^1 \frac{1}{x}\,dx$ diverges.
• Well, in that case $\frac{e^x}{x} \ge \frac{1}{x}$, so again it diverges. – rogerl Jun 14 '14 at 22:21