Limits

lim x → 0 + log a ⁡ ( x ) = − ∞ if  a > 1 {\displaystyle \lim _{x\to 0^{+}}\log _{a}(x)=-\infty \quad {\mbox{if }}a>1} lim x → 0 + log a ⁡ ( x ) = ∞ if  a < 1 {\displaystyle \lim _{x\to 0^{+}}\log _{a}(x)=\infty \quad {\mbox{if }}a<1} lim x → ∞ log a ⁡ ( x ) = ∞ if  a > 1 {\displaystyle \lim _{x\to \infty }\log _{a}(x)=\infty \quad {\mbox{if }}a>1} lim x → ∞ log a ⁡ ( x ) = − ∞ if  a < 1 {\displaystyle \lim _{x\to \infty }\log _{a}(x)=-\infty \quad {\mbox{if }}a<1} lim x → 0 + x b log a ⁡ ( x ) = 0 if  b > 0 {\displaystyle \lim _{x\to 0^{+}}x^{b}\log _{a}(x)=0\quad {\mbox{if }}b>0} lim x → ∞ log a ⁡ ( x ) x b = 0 if  b > 0 {\displaystyle \lim _{x\to \infty }{\frac {\log _{a}(x)}{x^{b}}}=0\quad {\mbox{if }}b>0}

The last limit is often summarized as "logarithms grow more slowly than any power or root of x".

Derivatives of logarithmic functions

d d x ln ⁡ x = 1 x , {\displaystyle {d \over dx}\ln x={1 \over x},} d d x log b ⁡ x = 1 x ln ⁡ b , {\displaystyle {d \over dx}\log _{b}x={1 \over x\ln b},}

Where x > 0 {\displaystyle x>0} , b > 0 {\displaystyle b>0} , and b ≠ 1 {\displaystyle b\neq 1} .

Integral definition

ln ⁡ x = ∫ 1 x 1 t d t {\displaystyle \ln x=\int _{1}^{x}{\frac {1}{t}}dt}

Integrals of logarithmic functions

∫ log a ⁡ x d x = x ( log a ⁡ x − log a ⁡ e ) + C {\displaystyle \int \log _{a}x\,dx=x(\log _{a}x-\log _{a}e)+C}

To remember higher integrals, it's convenient to define:

x [ n ] = x n ( log ⁡ ( x ) − H n ) {\displaystyle x^{\left[n\right]}=x^{n}(\log(x)-H_{n})}

Where H n {\displaystyle H_{n}} is the nth Harmonic number.

x [ 0 ] = log ⁡ x {\displaystyle x^{\left[0\right]}=\log x} x [ 1 ] = x log ⁡ ( x ) − x {\displaystyle x^{\left[1\right]}=x\log(x)-x} x [ 2 ] = x 2 log ⁡ ( x ) − 3 2 x 2 {\displaystyle x^{\left[2\right]}=x^{2}\log(x)-{\begin{matrix}{\frac {3}{2}}\end{matrix}}\,x^{2}} x [ 3 ] = x 3 log ⁡ ( x ) − 11 6 x 3 {\displaystyle x^{\left[3\right]}=x^{3}\log(x)-{\begin{matrix}{\frac {11}{6}}\end{matrix}}\,x^{3}}

Then,

d d x x [ n ] = n x [ n − 1 ] {\displaystyle {\frac {d}{dx}}\,x^{\left[n\right]}=n\,x^{\left[n-1\right]}} ∫ x [ n ] d x = x [ n + 1 ] n + 1 + C {\displaystyle \int x^{\left[n\right]}\,dx={\frac {x^{\left[n+1\right]}}{n+1}}+C}