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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 $n > 0$.

Help.

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.

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

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  • $\begingroup$ 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

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