Evaluation of Integration using limit as a sum

Question

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

Answer 1

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.

Answer 2

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

Answer 3

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 $p$.