Primitives of the natural logarithm

**m.walunga** · December 28th 2009, 01:39 PM

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:

$P_{k+1}[f(x)]=\int P_k dx$
$P_0[ln{x}]=ln{x}$
$P_n[ln{x}]=?$

After performing a few integrations, I seem to have found a nice solution:

$P_n[ln{x}]=\frac{x^n}{n!} (ln(x)-H(n))$

Where H(n) is the Harmonic Series:

$H(n)=\sum_{k=0}^{n}{\frac{1}{k}}$

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.

**simplependulum** · December 28th 2009, 10:26 PM

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

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

If we integrate it $n$ times ,

we can get

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

then evaluate the limit by L'hospital rule

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

and the denominator , obviously , is $a$

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

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

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

Then recall the logarithmic dervative

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

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

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