# Thread: Primitives of the natural logarithm

1. ## Primitives of the natural logarithm

All right, the idea of what I'm trying to do is find an easy way to generate the primitives of the natural logarithm without having to perform iteration after iteration of integration:
$\displaystyle P_{k+1}[f(x)]=\int P_k dx$
$\displaystyle P_0[ln{x}]=ln{x}$
$\displaystyle P_n[ln{x}]=?$
After performing a few integrations, I seem to have found a nice solution:
$\displaystyle P_n[ln{x}]=\frac{x^n}{n!} (ln(x)-H(n))$
Where H(n) is the Harmonic Series:
$\displaystyle H(n)=\sum_{k=0}^{n}{\frac{1}{k}}$
I'd appreciate it if anyone could verify or disprove my result.
If anyone would like to see my work, I'd gladly send them a .pdf of it.

2. If you find a general term that you think it is correct , it is time for you to prove your hypothesis by Mathematical Induction .

but at the same time , i have another method to do so

Consider

$\displaystyle \ln(x) = \lim_{a\to 0} \frac{ x^a - 1}{a}$

If we integrate it $\displaystyle n$ times ,

we can get

$\displaystyle \lim_{a\to 0} \frac{1}{a} \cdot \left( \frac{x^{a + n}}{ ( a+1)(a+2)...(a+n)} - \frac{x^n}{ n! } \right )$

then evaluate the limit by L'hospital rule

here the numerator is $\displaystyle \left( \frac{x^{a + n}}{ ( a+1)(a+2)...(a+n)} - \frac{x^n}{ n! } \right )$

and the denominator , obviously , is $\displaystyle a$

so $\displaystyle P_n = D[ \left( \frac{x^{a + n}}{ ( a+1)(a+2)...(a+n)} - \frac{x^n}{ n! } \right ) ]_{a\to 0}$

$\displaystyle = D[ \frac{x^{a + n}}{ ( a+1)(a+2)...(a+n)} ]$

( $\displaystyle D$ is the differential operator , wrt $\displaystyle a$ )

Then recall the logarithmic dervative

$\displaystyle P_n = \left [ \frac{x^{a + n}}{ ( a+1)(a+2)...(a+n)} \cdot [ \ln(x) - \frac{1}{a + 1} - \frac{1}{a+2} - ... - \frac{1}{a + n} ] \right ]_{a\to 0 }$

$\displaystyle =\frac{x^n}{n!} (ln(x)-H(n))$