Integration of $ \frac{1}{x}$ from First Principles 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. 
Integration of 1/x
 A: 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
$$\lim_{x\to0}{b^x-1\over x}=\log b\ .$$
A: 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
$$
\lim_{n\to\infty}n(e^{y/n}-1)=\lim_{u\to0^+}y\frac{e^{yu}-1}{yu}=y=\ln x
$$
