I know that the function, $$f(x) = \frac 1x$$tends to infinity when $x$ approaches $0$, but I'm unable to figure out how to calculate this integral - $$\int_0^n\frac 1x dx$$where $n > 0$.


Edit: I wanted to put the limits from $0$ to $n$, but mistakenly I did the opposite initially, my bad. I've reversed it now though. So, answer would be multiplication by $(−1)$ to the answers based on $n$ to $0$ as limits.


You may observe that, for $n>0$, $$ \int_n^0\frac 1x dx=-\lim_{b \to 0^+}\int_b^n\frac 1x dx=-\lim_{b \to 0^+}[\log x]_b^n=-\log n+\lim_{b \to 0^+}\log x=-\infty. $$

  • $\begingroup$ What if $n$ is $\infty$? $\endgroup$ – Aditya Agarwal Aug 14 '15 at 15:03
  • $\begingroup$ @Olivier: Thanks for the answer. $\endgroup$ – PalashV Aug 14 '15 at 15:06
  • $\begingroup$ @PalashV You are welcome! $\endgroup$ – Olivier Oloa Aug 14 '15 at 15:07
  • $\begingroup$ @AdityaAgarwal When one writes $n$ it can't be $\infty$. But one can let $ n \to \infty$. $\endgroup$ – Olivier Oloa Aug 14 '15 at 15:08
  • $\begingroup$ Yes, that is what I meant. $\endgroup$ – Aditya Agarwal Aug 14 '15 at 15:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.