Integration of 1/x as a limit of a sum

Question

This is from R.Courant book Example "Introduction to Calculus and Analysis vol.1 "

To integrate $x^\alpha$ when $\alpha\neq1$

we subdivide the interval [a,b] by the point of geometric progression: $$a, aq, aq^2, .., aq^{n-1}, aq^n=b$$ where $q=\sqrt[n]{b/a}$. We then only need to evaluate the sum of geometric series. Given the points of division $x_i=aq^i$ the length of the ith cell is given by: $$\Delta x_i=aq^i-aq^{i-1}=aq^i(q-1)/q$$ The largest $x_i$ is the last: $$x_n=b(q-1)/q$$ For $n\quad\rightarrow\quad\infty$ the number q tends to 1 and hence the length o $\Delta x_n$ of the largest cell and then also the lengths of all cells tend to zero. For the intermediate points $\xi_i$ we choose the right hand-hand endpoints $x_i$ of each cell. The sum $$F_n=\sum_{i=1}^n (\xi_i)^\alpha \Delta x_i=\sum_{i=1}^n (aq^i)^\alpha aq^i\frac{q-1}{q}=a^{\alpha+1} \frac{q-1}{q} \sum_{i=1}^n (q^{1+\alpha})^i$$

Now for $\alpha=-1$ we get $F_n=n(q-1)/q$ . Observing that $q=\sqrt[n]{b/a}$ tends to 1 as $n\quad\rightarrow\quad\infty$ we find:$$\int_a^b \frac 1x=\lim\limits_{n \to \infty}n(\sqrt[n]{b/a}-1)$$

The Question: if $F_n=n(q-1)/q$ shouldn't it be $$\int_a^b \frac 1x=\lim\limits_{n \to \infty}n \frac{(\sqrt[n]{b/a}-1)}{\sqrt[n]{b/a})}?$$ Where did the number q from denominator go?

Answer 1

After some time of thinking I came up with a simple idea:

if $q=\sqrt[n]{b/a}$ then it tends to 1 as n tends to infinity.

We have $F_n=n(q-1)/q$ and $$\lim\limits_{n \to \infty}n \frac{(q-1)}{q} = \lim\limits_{n \to \infty}n \frac{(\sqrt[n]{b/a}-1)}{\sqrt[n]{b/a})} = \int_a^b \frac 1x$$

We can write the limit as: $\lim\limits_{n \to \infty}n (\sqrt[n]{b/a}-1)*\lim\limits_{n \to \infty}(\frac 1{\sqrt[n]{b/a}})$ where the first limit is indeterminate of the form $\infty*0$ and the second limit just equels one. Then we have the result: $$\int_a^b \frac 1x=\lim\limits_{n \to \infty}n(\sqrt[n]{b/a}-1)$$

Answer 2

$\int_a^b \frac{1}{x} dx = \sum_{i=1}^n \frac{1}{\xi_i} \Delta x_i$. The Points $\xi_i$ on which the function is evaluated is given by $\xi_i = aq^i$. Moreover it holds: $\Delta x_i = aq^{i+1}-aq^i = aq^{i}(q-1)$. Note that the "forward" difference $\Delta x_i = x_{i+1}-x_i$ was used. Hence: $\sum_{i=1}^n \frac{1}{\xi_i} \Delta x_i = \sum_{i=1}^n \frac{aq^{i}(q-1)}{aq^{i}} = n(q-1)$. There is no $q$ in the denominator because it was used the difference $\Delta x_i = x_{i+1}-x_i$ instead of $\Delta x_i = x_{i}-x_{i-1}$ (also called "backward" difference).