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! :)

  • 1
    $\begingroup$ This diverges due to the divergent behavior of $\frac{1}{x}$ around the origin. $\endgroup$ – 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.

  • $\begingroup$ Wow, get it~! Thanks! How about... if I change e^{-x} to e^{x} ? $\endgroup$ – Richard Jun 14 '14 at 22:20
  • $\begingroup$ Well, in that case $\frac{e^x}{x} \ge \frac{1}{x}$, so again it diverges. $\endgroup$ – rogerl Jun 14 '14 at 22:21
  • $\begingroup$ Thanks a lot. I'll accept it several minutes later. ( It shows that I can't accept an answer in 6 minutes... ) $\endgroup$ – Richard Jun 14 '14 at 22:22

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