# Evaluation of Integration using limit as a sum

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

• Well the answer is $\log 2$ and you need to know something about $\log 2$ before you can actually deal with the problem. – Paramanand Singh Oct 12 '18 at 6:48
• 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 $\log 2$. – Paramanand Singh Oct 12 '18 at 6:51
• In particular see method 3 of this answer math.stackexchange.com/a/155248/72031 which is helpful for your approach. – Paramanand Singh Oct 12 '18 at 7:03
• 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 – DXT Oct 12 '18 at 7:08

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

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

Then,

$$\lim_{n\to\infty}n\left(\sqrt[n]2-1\right)=\lim_{h\to0}\frac{2^h-1}h=\left.(2^h)'\right|_{h=0}=\log2.$$

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

$$\int_1^2\frac{dx}x=\int_{\log1}^{\log2}\frac{e^t\,dt}{e^t}=\int_0^{\log2}dt.$$

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.

Now,

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

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

• +1 especially for the remark at the end. – 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 $$a=x_0 The norm of $$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 $$p$$.

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