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I have a follow up question to my question here: Finding $\int_0^1 \frac{1}{x}\; dx$ using definition

In this question I asked about $\int_{0}^1\frac{1}{x}\; dx$ and I think I understand how this integral is $\infty$.

I think the different integral $$ \int_1^2 \frac{1}{x}\;dx $$ is actually equal to a number. My question is how one can find this integral using the limit definition.

After working on this a bit, I get $$ \int_1^2 \frac{1}{x}\;dx = \lim_{n\to \infty} \sum_{i=1}^{n} \frac{1}{1+i/n}\frac{1}{n}\\ =\lim_{n\to \infty}\frac{1}{n} \sum_{i=1}^{n} \frac{1}{1+i/n}\\ =\lim_{n\to \infty}\sum_{i=1}^{n} \frac{1}{n+i} $$ So I need to find $$ \sum_{i=1}^{n} \frac{1}{n+i} $$ but I am not sure how I can do this. I get how you can find sums like $\sum i$, $\sum i^2$, and $\sum i^3$.

robjohn
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John Doe
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2 Answers2

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The better subdivision in terms of simplifying the Riemann sum is with a geometric series, set $x_k=q^k$ with $q^N=2$ then the Riemann sum with the left interval points is $$ \sum_{k=0}^{N-1}f(x_k)(x_{k+1}-x_k)=\sum_{k=0}^{N-1}\frac{q^k(q-1)}{q^k}=N(q-1)=N(\sqrt[N]2-1) $$ and it is a classical result that that limit is $\ln(2)$. You could consider it a difference quotient $$ \frac{e^{\ln(2)\cdot(1/N)}-e^0}{(1/N)-0} $$ with the obvious derivative as limit.

Lutz Lehmann
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The Riemann Sum in the question is not quite right (though it gives the same result). The Riemann Sum really should be $$ \begin{align} \int_1^2\color{#C00}{\frac1x}\,\color{#090}{\mathrm{d}x} &=\lim_{n\to\infty}\sum_{k=n}^{2n}\color{#C00}{\frac1{k/n}}\color{#090}{\frac1n}\\ &=\lim_{n\to\infty}\sum_{k=n}^{2n}\frac1k\\ &=\lim_{n\to\infty}\sum_{k=n}^{2n}\left[\log\left(1+\frac1k\right)+O\!\left(\frac1{k^2}\right)\right]\\[6pt] &=\lim_{n\to\infty}\left[\log\left(\frac{2n+1}n\right)+O\!\left(\frac1{n}\right)\right]\\[12pt] &=\bbox[5px,border:2px solid #C0A000]{\log(2)} \end{align} $$ where we have used $x=\log(1+x)+O\!\left(x^2\right)$ and a telescoping sum employing $\log\left(1+\frac1k\right)=\log(k+1)-\log(k)$.

Approach Using $\boldsymbol{\int x^\alpha\,\mathrm{d}x=\frac1{\alpha+1}x^{\alpha+1}}+C$ $$ \begin{align} \int_1^2\frac1x\,\mathrm{d}x &=\lim_{n\to\infty}\int_1^2x^{-1+1/n}\,\mathrm{d}x\\[9pt] &=\lim_{n\to\infty}n\left(2^{1/n}-1^{1/n}\right)\\[3pt] &=\log(2)\lim_{n\to\infty}\frac{e^{\log(2)/n}-1}{\log(2)/n-0}\\[3pt] &=\log(2)\,\left.\frac{\mathrm{d}}{\mathrm{d}x}e^x\,\right|_{x=0}\\[9pt] &=\bbox[5px,border:2px solid #C0A000]{\log(2)} \end{align} $$

robjohn
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  • One should say that $x^{-1 + 1/n}$ converges uniformly to $x^{-1}$ on $[1,2]$ because $$ \sup_{x \in [1,2]} \left\lvert \frac{1}{x} - \frac{1}{x^{1 - 1/n}} \right\rvert = \sup_{x \in [1,2]} \left\lvert \frac{x^{-1/n} - 1}{x^{1 - 1/n}} \right\rvert \le 1 - 2^{-1/n}$$ since the numerator is biggest at $x = 2$ and the denominator is smallest at $x = 1$. And $1 - 2^{-1/n} \to 0$ as $n \to \infty$. – Trevor Gunn Nov 01 '17 at 13:08