Table of Integrals, Series, and Products, 8th Edition

1.5 The Logarithm

1.51 Series representation

1.511¹²

$\ln \left(1+x\right)=x-\frac{1}{2}{x}^2+\frac{1}{3}{x}^3-\frac{1}{4}{x}^4+\mathrm{\dots }={\displaystyle \sum _{k=1}^{\infty }{\left(-1\right)}^{k+1}\frac{x^k}{k}=xF\left(1,1;\text{?}2;-x\right)}$

$\left[-1<x\le 1\right]$

1.512

1. 

$\ln x=(x-1)-\frac{1}{2}{\left(x-1\right)}^2+\frac{1}{3}{\left(x-1\right)}^3-\cdots ={\displaystyle \sum _{k=1}^{\infty }{\left(-1\right)}^{k+1}\frac{{\left(x-1\right)}^k}{k}}$

$\left[0<x\le 2\right]$

2. 

$\ln x=2\left[\frac{x-1}{x+1}+\frac{1}{3}{\left(\frac{x-1}{x+1}\right)}^3+\frac{1}{5}{\left(\frac{x-1}{x+1}\right)}^5+\cdots \right]=2{\displaystyle \sum _{k=1}^{\infty }\frac{1}{2k-1}}{\left(\frac{x-1}{x+1}\right)}^{2k-1}$

$\left[0<x\right]$

3. 

$\ln x=\frac{x-1}{x}+\frac{1}{2}{\left(\frac{x-1}{x}\right)}^2+\frac{1}{3}{\left(\frac{x-1}{x}\right)}^3+\cdots ={\displaystyle \sum _{k=1}^{\infty }\frac{1}{k}}{\left(\frac{x-1}{x}\right)}^k$

  AD (644.6)

$\left[x\ge \frac{1}{2}\right]$

4.* 

$\ln$