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I believe I understand how to find definite integrals of polynomials using the limit definition. So, for example, to find $\int_0^1 x^2$ one ends up needing $\sum_{i=1}^n i^2$. My question is how one would do an integral like $$ \int_0^1 \frac{1}{x}\;dx $$ using the definition. From my work I end up having to find $$ \sum_{i=1}^n \frac{1}{i} $$ and I don't know this sum.

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  • $\begingroup$ en.wikipedia.org/wiki/Harmonic_number $\endgroup$
    – Zubzub
    Oct 31, 2017 at 19:04
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    $\begingroup$ It diverges.... $\endgroup$
    – user223391
    Oct 31, 2017 at 19:05
  • $\begingroup$ You can't integrate from $0$. $\endgroup$
    – user65203
    Oct 31, 2017 at 19:11

1 Answer 1

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I think you're talking about using the Riemann sum to calculate integrals. For example, $$\int_0^1x^2dx=\lim_{n\rightarrow\infty}\sum_{i=1}^n\left(\frac{i}{n}\right)^2\frac{1}{n}=\lim_{n\rightarrow\infty}\frac{1}{n^3}\sum_{i=1}^ni^2=\lim_{n\rightarrow\infty}\frac{n(n+1)(2n+1)}{6n^3}=\frac{1}{3}.$$

But the integral $$\int_0^1\frac{dx}{x}=\lim_{n\rightarrow\infty}\sum_{i=1}^n\frac{n}{i}\frac{1}{n}=\sum_{i=1}^\infty\frac{1}{i}=\infty$$

diverges because the harmonic series diverges.

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