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

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The discussion focuses on deriving the integral of $\frac{\ln^{n}(1-x)}{x}$ using integration by parts, leading to a formula involving the polylogarithm function $\text{Li}_{n}(x)$. The result shows that for $n > 1$, the integral can be expressed as a sum of terms involving $\ln x$ and polylogarithms, with a constant of integration. Participants also explored alternative methods, including substitutions and series expansions, to approach the integral. The conversation reflects a lack of engagement, possibly due to the perceived complexity or monotony of the topic. Ultimately, the integration by parts technique effectively reveals the relationship between logarithmic functions and polylogarithms in this context.
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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 $n$.
 
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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.
 
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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 .
 
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)
 
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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

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

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