Evaluation of $\displaystyle \int^{2}_{1}\frac{1}{x}dx$ using limit as a sum

Try: Using The formula $$\int^{b}_{a}f(x)dx = \lim_{h\rightarrow 0}h\times \sum^{n-1}_{r=1}f(a+rh)$$

where $nh=b-a$

above $a=1,b=2$ and $\displaystyle f(x)=\frac{1}{x}$ and $nh=1$

$$\int^{2}_{1}\frac{1}{x}dx = \lim_{h\rightarrow 0} \sum^{n-1}_{r=1}f(a+rh)=\lim_{h\rightarrow 0}h\cdot \sum^{n-1}_{r=0}f(1+rh) $$

$$\lim_{h\rightarrow 0}h\cdot \sum^{n-1}_{r=0}\frac{1}{1+rh}=\lim_{h\rightarrow 0}\bigg[\frac{h}{1+h}+\frac{h}{1+2h}+\cdots \cdots +\frac{h}{1+(r-1)h}\bigg]$$ i did not know how i proceed, struck here

could some help me. thanks

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    $\begingroup$ Well the answer is $\log 2$ and you need to know something about $\log 2$ before you can actually deal with the problem. $\endgroup$ – Paramanand Singh Oct 12 '18 at 6:48
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    $\begingroup$ If you accept that $\log 2=\lim_{n\to\infty}n(2^{1/n}-1)$ then you need to use a partition of $[1,2]$ via points $x_k=2^{k/n}$. If you accept that $\log 2=\sum_{i=1}^{\infty} (-1)^{i+1}(1/i)$ then your approach can be used to show that the integral is $\log 2$. $\endgroup$ – Paramanand Singh Oct 12 '18 at 6:51
  • $\begingroup$ In particular see method 3 of this answer math.stackexchange.com/a/155248/72031 which is helpful for your approach. $\endgroup$ – Paramanand Singh Oct 12 '18 at 7:03
  • $\begingroup$ I did not understand the line $x_{k}=2^{\frac{k}{n}}$ for $[1,2].$ thanks parmanand singh, Can you please explain me in detail. thanks $\endgroup$ – DXT Oct 12 '18 at 7:08

Consider the points $x=c^k$ with $c^n=2$.

$$\int_1^2\frac{dx}{x}\approx\sum_{k=0}^{n+1} \frac{\Delta c^k}{c^k}=\sum_{k=1}^n \frac{ c^{k+1}-c^k}{c^k}=n(c-1)=n\left(\sqrt[n]2-1\right).$$



Note that this is in fact a discrete version of an exponential change of variable, $x=e^t$, giving


The latter integral can be trivially computed as a sum.


$$\int_1^2\frac{dx}{x}\approx\frac1n\sum_{k=n+1}^{2n}\left(\dfrac kn\right)^{-1}=\sum_{k={n+1}}^{2n}\frac1{k}=H_{2n}-H_{n}$$ where $H_n$ denotes an harmonic number.


$$\lim_{n\to\infty}(H_{2n}-H_n)=\lim_{n\to\infty}(\log2n+\gamma+o(1)-\log n-\gamma+o(1))=\log2.$$

It is questionable whether this approach makes sense, as deriving the asymptotic expression of the harmonic numbers is much harder than the initial problem.

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    $\begingroup$ +1 especially for the remark at the end. $\endgroup$ – Paramanand Singh Oct 12 '18 at 8:28

Let's assume the following equation as given $$\log 2=\lim_{n\to\infty} n(2^{1/n}-1)\tag{1}$$ By definition of Riemann integral if $f:[a, b] \to\mathbb {R} $ is Riemann integrable on $[a, b] $ with integral $I$ then for every $\epsilon >0$ there is a $\delta>0$ such that for all partitions $P$ of $[a, b] $ with norm less than $\delta$ we have $|S(f, P) - I|<\epsilon $ and we write $\lim_{||P||\to 0}S(f,P)=I$.

To explain notation and terms we say that a set $P=\{x_0,x_1,\dots, x_n\} $ is a partition of interval $[a, b] $ if $$a=x_0<x_1<\dots<x_n=b$$ The norm of $P$ (denoted by $||P||$) is defined as $\max_{i=1}^{n}(x_i-x_{i-1})$. A Riemann sum for $f$ over a partition $P$ of $[a, b] $ (denoted by $S(f, P) $) is a sum of the form $$\sum_{i=1}^{n}f(t_i)(x_i-x_{i-1})$$ where points $t_i$ called tags are arbitrary points of the intervals $[x_{i-1},x_i]$ respectively.

Now it is well known that function $f$ defined by $f(x) =1/x$ is Riemann integrable on any interval $[a, b] $ if $0\notin[a,b]$ (because of continuity of $f$ in that interval) and in particular the Riemann integral $$I=\int_{1}^{2}\frac{dx}{x}$$ exists.

Let's choose partition $P$ of $[1,2]$ using points $x_k=2^{k/n}$. Then norm of $P$ is $\max_{k=1}^{n}(2^{k/n}-2^{(k-1)/n})$ and this tends to $0$ if $n\to \infty $. Also let's choose tags $t_k$ as $t_k=x_{k-1}=2^{(k-1)/n}$. Then the Riemann sum $$S(f, P) =\sum_{k=1}^{n}f(t_k)(x_k-x_{k-1})=\sum_{k=1}^{n}\frac{2^{k/n}-2^{(k-1)/n}}{2^{(k-1)/n}}=n(2^{1/n}-1)$$ and by our starting equation this tends to $\log 2$ as $n\to\infty$. Hence the value of the integral is $\log 2$.

Note: This is an expansion of my comment to the question and another answer based on same idea has been given by another user. I have tried here to give some more detail about Riemann integration and in particular highlighted the arbitrary nature of points of a partition. Typical introductory calculus texts almost exclusively use partition points in arithmetic progression and this answer in contrast uses points in geometric progression for creating a partition. The same technique can be used to evaluate the integral of $f(x) =x^p$ for any $p$.

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    $\begingroup$ Expanded version of mine, it seems ;-) $\endgroup$ – Yves Daoust Oct 12 '18 at 8:16
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    $\begingroup$ @YvesDaoust: see updated note at the end. $\endgroup$ – Paramanand Singh Oct 12 '18 at 8:20

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