8

To integrate $x^\alpha$ when $\alpha\neq1$ we subdivide the interval $[a,b]$ by the point of geometric progression: $$a, aq, aq^2, \ldots, aq^{n-1}, aq^n=b$$ 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 $i$-th 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 $n\rightarrow\infty$ 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 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 $n\rightarrow\infty$ we find:$$\int_a^b \frac 1x=\lim\limits_{n \to \infty}n\left(\sqrt[n]{b/a}-1\right)$$

How to evaluate this limit to get $\ln|a|-\ln|b|$?

Lucenaposition
  • 2,416
Stenio Ramos
  • 83
  • 1
  • 4
  • $\lim_{}n((b/a)^{\frac{1}{n}}-1)=\ln(b/a)\times \lim_{n\rightarrow \infty}(b/a)^{\frac{1}{n}}$ –  Jun 12 '15 at 02:56
  • Note $\int_{-\infty}^{+\infty} \exp(-x^4),dx = \frac{\pi}{\sqrt{2}\Gamma(3/4)}$. – GEdgar Jul 13 '24 at 14:16

2 Answers2

6

Suppose that $0\lt a\lt b$. We are looking for $$\lim_{n\to\infty}\frac{(b/a)^{1/n}-1}{1/n}$$ as $n\to\infty$. Equivalently, we will find the limit of $$\frac{(b/a)^t-1}{t}\tag{1}$$ as $t$ approaches $0$ from the right.

Note that $(b/a)^t=\exp(t\ln(b/a))$ and $$\lim_{t\to 0} \frac{\exp(t\ln(b/a))-1}{t}$$ is the value of the derivative of $\exp(t\ln(b/a))$ at $t=0$. That value is $\ln(b/a)$.

André Nicolas
  • 514,336
1

$$\lim_{t\to 0}\frac{a^t - 1}{t} = \ln{a}$$ so $$\lim_{n\to \infty}\frac{(b/a)^{1/n} - 1}{1/n} = \ln{b/a}$$

user23492240
  • 11