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I need to simplify $n$th derivative of $$f(x)=\frac{\log^3 x}{1-x}$$ where $0<x<1$

I tried writing $f(x)=u_1u_2u_3u_4$ where $u_1=u_2=u_3=\log x$ and $u_4=\frac{1}{1-x}$.

Using Leibniz rule for $n$th derivative we get $$f^{(n)}(x)=\sum_{k_1+k_2+k_3+k_4=n} \binom{n}{k_{1},k_{2},k_{3},k_{4}}u_1^{(k_{1})}u_2^{(k_{2})}u_3^{(k_{3})}u_4^{(k_{4})}$$
Now we have using $u_1^{(k_1)}=\frac{(-1)^{k_1+1} (k_1-1)!}{x^{k_1}}$ and $u_4^{(k_4)}=\frac{k_4!}{(1-x)^{1+k_4}}$ $$f^{(n)}(x)=\sum_{k_1+k_2+k_3+k_4=n} \left(\frac{n!}{k_1!k_2!k_3!k_4!}\right)\frac{(-1)^{k_1+1} (k_1-1)!(-1)^{k_2+1} (k_2-1)!(-1)^{k_3+1} (k_3-1)!k_4!}{x^{k_1+k_2+k_3} (1-x)^{1+k_4}} $$ $$f^{(n)}(x)=n!\sum_{k_1+k_2+k_3+k_4=n} \frac{(-1)^{k_1+k_2+k_3+1}}{k_1k_2k_3 x^{k_!+k_2+k_3}(1-x)^{1+k_4}} $$ $$f^{(n)}(x)=\frac{n!(-1)^{n+1}}{x^n}\sum_{k_1+k_2+k_3+k_4=n} \frac{(-1)^{k_4}}{k_1k_2k_3}\frac{x^{k_4}}{(1-x)^{1+k_4}} $$ Any help would be highly appreciated. Thank you.

Max
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1 Answers1

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Apply the Leibniz rule: $$\frac{d^n}{dx^n}\frac{\ln^3(x)}{1-x}=\sum_{m=0}^n\binom nm\frac{d^m}{dx^m}\ln^3(x)\frac{d^{n-m}}{dx^{n-m}}\frac1{1-x}$$

Now use the Stirling number of the first kind $\frac{d^n}{dx^n}\ln^a(x)$ formula and simplify the resulting factorials to get:

$$\frac{d^n}{dx^n}\frac{\ln^3 (x)}{1-x}=6n!\sum_{k=0}^4\sum_{m=0}^n\frac{S_m^{(k)}\ln^{3-k}(x)(1-x)^{m-n-1}}{\Gamma(4-k)\ln(x)^{k-3}x^mm!}$$

Expanding the outer sum and converting $S_m^{(k)}$ into (generalized) Harmonic numbers finally gives:

$$\boxed{\frac{d^n}{dx^n}\frac{\ln^3 (x)}{1-x}=\frac{n!\ln^3(x)}{(1-x)^{n+1}}+3(-1)^nn!\sum_{m=1}^n\frac{(x-1)^{m-n-1}}{mx^m}\left(\left((H_{m-1}-\ln(x)\right)^2-H_{m-1}^{(2)}\right)}$$

which is true. You can expand and sum over $\sum\limits_{m=1}^n\frac{(x-1)^{m-n-1}}{mx^m}$ and get a Lerch transcendent, but the sums over the harmonic numbers do not seem to have a special function form

Тyma Gaidash
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