MHB Integration by Parts: Showing $\ln^n(1-x)$

[polygamma]

Integration by parts

By repeatedly integrating by parts show that for $ n >1 $,

$$ \int \frac{\ln^{n}(1-x)}{x} \ dx = \ln x \ln^{n}(1-x) + \sum_{k=1}^{n} (-1)^{k-1} \frac{n!}{(n-k)!} \text{Li}_{k+1}(1-x) \ln^{n-k} (1-x) + C$$

where $\text{Li}_{n}(x)$ is the polylogarithm function of order $n$.

 

[polygamma]

This problem didn't' generate much interest. Perhaps it was too boring.$ \displaystyle \int \frac{\ln^{n}(1-x)}{x} \ dx $Let $ \displaystyle u = \ln^{n}(1-x)$ and $ \displaystyle dv = \frac{dx}{x}$.$ = \displaystyle - \ln x \ln^{n}(1-x) + n \int \frac{\ln x \ln^{n-1} (1-x)}{1-x} \ dx $Let $\displaystyle u = \ln^{n-1}(1-x)$ and $\displaystyle dv = \frac{\ln x}{1-x} \ dx$.$ \displaystyle = \ln x \ln^{n}(1-x) + \text{Li}_{2}(1-x) \ln^{n-1}(1-x) + n (n-1) \int \frac{\text{Li}_{2}(1-x) \ln^{n-2}(1-x)}{1-x} \ dx $Let $\displaystyle u = \ln^{n-2}(1-x)$ and $\displaystyle dv = \frac{\text{Li}_{2}(1-x)}{1-x} \ dx $.$ = \displaystyle \ln x \ln^{n}(1-x) + n \text{Li}_{2}(1-x) \ln^{n-1}(1-x) - n(n-1)\text{Li}_{3}(1-x) \ln^{n-2}(1-x)$

$ \displaystyle - n(n-1)(n-2) \int \frac{\text{Li}_{3} \ln^{n-3} (1-x)}{1-x} \ dx $$ = \displaystyle \ln x \ln^{n}(1-x) + n \text{Li}_{2}(1-x) \ln^{n-1}(1-x) - n(n-1)\text{Li}_{3}(1-x) \ln^{n-2}(1-x) $

$ \displaystyle + n(n-1)(n-2) \text{Li}_{4}(1-x) \ln^{n-3}(1-x) + n(n-1)(n-2)(n-3) \int \frac{\text{Li}_{4}(1-x)\ln^{n-4}(1-x)}{1-x} \ dx $$ \displaystyle = \ldots = \ln x \ln^{n}(1-x) + n \text{Li}_{2}(1-x) \ln^{n-1}(1-x) - n(n-1)\text{Li}_{3}(1-x) \ln^{n-2}(1-x)$

$\displaystyle + \ldots + (-1)^{n} (n-1)! \text{Li}_{n} \ln (1-x) \ dx + (-1)^{n} n! \int \frac{\text{Li}_{n}(1-x)}{1-x} \ dx $$ \displaystyle = \ln x \ln^{n}(1-x) + n \text{Li}_{2}(1-x) \ln^{n-1}(1-x) - n(n-1)\text{Li}_{3}(1-x) \ln^{n-2}(1-x)$

$ \displaystyle + \ldots + (-1)^{n} (n-1)! \text{Li}_{n} \ln (1-x) \ dx + (-1)^{n-1} n! \text{Li}_{n+1}(1-x) + C $$\displaystyle = \ln x \ln^{n}(1-x) + \sum_{k=1}^{n} (-1)^{k-1} \frac{n!}{(n-k)!} \text{Li}_{k+1}(1-x) \ln^{n-k} (1-x) + C $But actually you can express it more succinctly since $ \displaystyle \ln x = - \text{Li}_{1}(1-x)$.So $ \displaystyle \int \frac{\ln^{n}(1-x)}{x} \ dx = \sum_{k=0}^{n} (-1)^{k-1} \frac{n!}{(n-k)!} \text{Li}_{k+1}(1-x) \ln^{n-k} (1-x) + C $And it's valid for $n=1$ as well.

 

[alyafey22]

I proved a similar forumla a while back

$$\int^a_0 \frac{\log^n(x)}{1-x}\, dx = \log^n(a) \sum_{k=0}^{n}(-1)^k\frac{ n!}{(n-k)!}\,\frac{\text{Li}_{k+1}(a)}{\log^k(a)} $$

I used the binomial expansion .

 

[DreamWeaver]

I'd be inclined to skin this particular transcendental cat slightly differently... Firstly, apply the reflection substitution $$x \to 1-x$$, to obtain

$$\int \frac{\log^n(1-x)}{x}\,dx=\int \frac{(\log x)^n}{(1-x)}\, dx$$Then expand the denominator as an infinite series$$\frac{1}{1-x}=\sum_{k=0}^{\infty}x^k$$

Insert this into the indefinite integral, and then integrate term-by-term...$$\int \frac{\log^n(1-x)}{x}\,dx=\sum_{k=0}^{\infty} \int x^k(\log x)^n\, dx$$
EDIT: You beat me to it, Zaid... I must confess, I never bother with integrals like these in indefinite form, so like you, I did the parametric case... (Cool)

 

[alyafey22]

Here is how I did it Using a substitution we obtain

$$a\int^{1}_0 \frac{\log^n(ax)}{1-ax}\, dx $$

which can be written as

$$a\int^{1}_0 \frac{\left(\log(a)+\log(x) \right)^n}{1-ax}\, dx $$

Now using that binomial expansion we have

$$a\int^{1}_0 \sum_{k=0}^{n} {n \choose k} \log^{n-k}(a) \log^k(x) \frac{ dx}{1-ax}\, $$

arranging to obtain

$$a \sum_{k=0}^{n}{n \choose k}\log^{n-k}(a)\int^{1}_0 \frac{\log^k(x) }{1-ax}\,dx $$

$$\sum_{k=0}^{n}(-1)^ k {n \choose k}\log^{n-k}(a) \Gamma(k+1) \text{Li}_{k+1}(a) $$