I'm interested in finding the area under $f(x)=\frac{1}{x}$ without resorting to the first part of the Fundamental Theorem of Calculus. This has been my attempt so far, not sure how to continue as the harmonic series diverges.
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The mistake is that you should use the points $x_i = a + \frac{(b-a)i}{N}$ instread of $x_i = \frac{(b-a)i}{N}$ when computing $f(x_i)$ that goes into the sum (and the sum should start at $i=0$ instead of $i=1$). As you have written it you are evaluating the integral $\int_0^{b-a}\frac{dx}{x}$ and this integral does not exist. – Winther Dec 25 '16 at 17:15
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Ah I see! Thank you very much. That does make sense. :) – Chung Ren Khoo Dec 25 '16 at 17:17
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If you allow the FTOC, but no prior knowledge of $\int1/x dx$, this can be done with limits and such. – Simply Beautiful Art Dec 25 '16 at 17:45
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I've made the necessary adjustments but I'm not sure if I've defined the limits correctly as I just end up with zero. http://puu.sh/t0Kzd/58cd882fd0.png – Chung Ren Khoo Dec 25 '16 at 17:53
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@Winther How did you break it up into two different sums? (Thanks again btw. Been cracking my head over this for days now and I need to prove this rigorously before I can tackle integration of sine and cosine.) – Chung Ren Khoo Dec 25 '16 at 18:40
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We have $\sum_{k=m}^n f(k) = \color{blue}{f(m) + f(m+1) + \ldots + f(n)} = \color{red}{f(1) + f(2) + \ldots + f(m-1)} + \color{blue}{f(m) + f(m+1) + \ldots +f(n)} - \color{red}{[f(1) + f(2) + \ldots + f(m-1)]} = \sum_{k=1}^n f(k) - \sum_{k=1}^{m-1} f(k)$ (adding and subtracting the same term to be able to write it this way) – Winther Dec 25 '16 at 18:43
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@Winther Apologies for reviving this old thread, but after toiling over it til 330 in the morning I had to take a short break ... I still don't quite get how you can write it in the form of $\sum_{i=1}^{Nb} -\sum_{i=1}^{Na-1}\frac{1}{i}$. I think I get where Na-1 and Nb comes from, but how is it that the general form of the summation is just $\frac{1}{i}$ ? Specifically, from (1) to (2) http://puu.sh/t15pj/d79196feec.png . Thanks again. – Chung Ren Khoo Dec 26 '16 at 01:19
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(Sorry there's a typo, it should be $\sum_{i=1}^{Nb} \frac{1}{i}- \sum_{i=1}^{Na-1} \frac{1}{i}$. Missed the $\frac{1}{i}$. Can't seem to edit my comments, probably because I'm new to MSE...) – Chung Ren Khoo Dec 26 '16 at 01:21
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That comment was reffering to the last line in your link http://puu.sh/t0Kzd/58cd882fd0.png I now see that I misread the last line so it's good thing that you spotted that :) If you want to solve it this way then notice that it is one the form $\sum_{i=0}^N\frac{1}{A + i} = \sum_{i=0}^{A+N} \frac{1}{i} - \sum_{i=0}^{A-1} \frac{1}{i}$. A better way to solve this problem is to go for the geometrical partioning used in the answers below. It's simpler and it avoids using $H_n \sim \log(n) + \gamma$ (which is often proved using that the integral of $1/x$ is $\log(x)$ so could be a bit circular). – Winther Dec 26 '16 at 01:33
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Above $A = \frac{aN}{b-a}$ – Winther Dec 26 '16 at 01:34
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Hmm. True. The proof that I'm aware of for the Euler-Mascheroni constant does involve the integral of $1/x$ being $\ln |x|$ .... Argh! I don't suppose there is a way to prove $\gamma$ without $\int \frac{1}{x} = \ln |x|$ ? – Chung Ren Khoo Dec 26 '16 at 02:14
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@Winther For $\sum_{i=0}^{A+N} \frac{1}{i} - \sum_{i=0}^{A-1}\frac{1}{i}$, wouldn't the first term yield an undefined result? When $i=0$, $\frac{1}{i} =\frac{1}{0}$. – Chung Ren Khoo Dec 26 '16 at 02:27
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Yes. The sum should start at $i=1$. Also see some of the answers in this question for showing that $H_n - \log(n)$ converges (which is all you need) without integrals: http://math.stackexchange.com/questions/607039/how-to-show-gamma-aka-eulers-constant-is-convergent – Winther Dec 26 '16 at 02:28
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@Winther Since $A=\frac{aN}{b-a}$, it might not necessarily be an integer so A+N might also not be an integer. Doesn't the summation notation mean $\sum_{i=a}^b f(i) = f(a)+f(a+1)+...+f(b-1)+f(b)$ ? If so, then $\sum_{i=1}^{A+N} \frac{1}{i} = \frac{1}{1}+\frac{1}{2}+...+\frac{1}{A+N-1}+\frac{1}{A+N}$ but if A is not an integer then it seems impossible to get A+N by adding 1's. Or .. could it?! :o – Chung Ren Khoo Dec 26 '16 at 02:51
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Yes these are the hairy details needed to be filled in if you want to do it this way. One way is to use $\frac{1}{\lfloor A\rfloor + i + 1} \leq \frac{1}{A + i} \leq \frac{1}{\lfloor A\rfloor + i}$ where $\lfloor \cdot \rfloor$ is the floor-function. You should be able to squeeze your limit between two sums you are able to evaluate. – Winther Dec 26 '16 at 02:54
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@Winther Thanks for all the help so far, I really do appreciate it. I'm guessing this is the squeeze theorem? Similar to the method used to prove that $\lim_{n\rightarrow\infty} \frac{\sin\theta}{\theta}=1$? Not exactly sure how to use that to prove this though ... Apologies for taking up so much of your time. – Chung Ren Khoo Dec 26 '16 at 03:49
2 Answers
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In order to compute the integral $$\int_1^b{1\over x}\>dx\>,\qquad b>1,$$ using Riemann sums choose an $N\gg 1$, and put $\rho:=b^{1/N}$. Use the partition $$1=\rho^0<\rho^1<\rho^2<\ldots<\rho^N=b\ ,$$ i.e., $x_k:=\rho^k$ $(0\leq k\leq N)$, of the interval $[1,b]$, and consider the Riemann sum $$\sum_{k=1}^N{1\over x_{k-1}}(x_k-x_{k-1})=\sum_{k=1}^N {1\over \rho^{k-1}}\bigl(\rho^k-\rho^{k-1}\bigr)=N(\rho-1)\ .$$ It follows that $$\int_1^b{1\over x}\>dx=\lim_{N\to\infty}{b^{1/N}-1\over1/N}=\log b\ ,$$ whereby we have made use of the standard limit $$\lim_{x\to0}{b^x-1\over x}=\log b\ .$$
Christian Blatter
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Thank you! This is a rather Interesting way of doing it ... I don't quite get the last line though. How is $\lim_{N\rightarrow\infty} \frac{b^{\frac{1}{N}}-1}{\frac{1}{N}} = \log b$ ? – Chung Ren Khoo Dec 26 '16 at 01:12
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1Nice proof (+1). Re question in the comment above, let expression on LHS be equal to say $a$. Rearranging gives $b=\lim_{N\to \infty} (1+\tfrac aN)^{\tfrac 1N}= e^a$. Hence $a=\ln b$. – Hypergeometricx Aug 19 '19 at 17:12
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You can compute, for $x>1$, $$ \int_1^x\frac{1}{t}\,dt $$ using a subdivision of the interval $[1,x]$ into $n$ parts in geometric progression. Thus the points are $$ 1,\quad q=\sqrt[n]{x},\quad q^2=\sqrt[n]{x^2},\quad\dots,\quad q^n=\sqrt[n]{x^n}=x $$ If you choose the right endpoint, you get an approximation by defect: $$ \sum_{k=1}^n\frac{1}{q^k}(q^k-q^{k-1})= \sum_{k=1}^n\left(1-\frac{1}{q}\right)=n\left(1-\frac{1}{\sqrt[n]{x}}\right) $$ Choosing the left endpoint, $$ \sum_{k=1}^n\frac{1}{q^{k-1}}(q^k-q^{k-1})= n(\sqrt[n]{x}-1) $$ For $0<x<1$ you get the same thing.
Thus $$ \int_{1}^x\frac{1}{t}\,dt=\lim_{n\to\infty}n(\sqrt[n]{x}-1) $$ Now, if $x=e^y$, the limit becomes $$ \lim_{n\to\infty}n(e^{y/n}-1)=\lim_{u\to0^+}y\frac{e^{yu}-1}{yu}=y=\ln x $$
Paramanand Singh
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egreg
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What does approximation by defect mean? And subdivision of $[1,x]$? How did you get $\sum_{k=1}^n \frac{1}{q^k}(q^k -q^{k-1})$? – Chung Ren Khoo Dec 26 '16 at 13:57
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@ChungRenKhoo From below, if you prefer. The summation is the Riemann sum corresponding to taking the rightmost point in the subinterval. – egreg Dec 26 '16 at 14:00