The $n$ th derivative of $\ln(xy)/(1-xy)$ with respect to $x$.

Question

I'm looking for a closed form of

\begin{align} \frac{d^n}{dx^n}\left( \frac{\ln(xy)}{1-xy} \right) \end{align}

I tried using the Taylor series: \begin{align} \frac{\ln(xy)}{1-xy} = \frac{\sum_{k=1}^{\infty} \frac{(-1)^{k}}{k} (1-xy)^{k}} {1-xy} = \sum_{k=1}^{\infty} \frac{(-1)^{k}}{k} (1-xy)^{k-1} \end{align}

I think there is a wrong sign above. I'm not sure how to proceed and if this is useful for a closed form.

Answer 1

Note that $y$ is treated as a constant parameter here. Let $z=xy$. Then, the required expression is $$E=y^n\frac{d^n}{dz^n}\frac{\ln z }{1-z}.$$ We have by the general Leibniz rule that $$\frac{d^n}{dz^n}\frac{\ln z }{1-z}=\sum_{k=0}^n\binom{n}{k}\left(\frac{d^k}{dz^k}\ln z\right)\left(\frac{d^{n-k}}{dz^{n-k}}\frac{1}{1-z}\right).$$ Thus, $$E=y^n\Biggl(\frac{n!\ln z}{(1-z)^{n+1}}+\sum_{k=1}^n\binom{n}{k}\frac{(-1)^{k-1}(k-1)!}{z^k}\left(\frac{(n-k)!}{(1-z)^{n-k+1}}\right)\Biggr).$$ Plugging in $z=xy$, we get $$E=n!y^n\left(\frac{\ln(xy)}{(1-xy)^{n+1}}+\sum_{k=1}^n\frac{(-1)^{k-1}}{k(xy)^k(1-xy)^{n-k+1}}\right).$$ You can also write $$E=\frac{n!}{1-xy}\left(\frac{y^n\ln(xy)}{(1-xy)^{n}}+\sum_{k=1}^n\frac{(-1)^{k-1}y^{n-k}}{kx^k(1-xy)^{n-k}}\right).$$

Answer 2

It's not a trully an answer.With CAS:

$$\frac{\partial ^n}{\partial x^n}\frac{\ln (x y)}{1-x y}=\\-\frac{(-x)^{-n} n! \, _2F_1\left(1,1+n;2+n;1-\frac{1}{x y}\right)}{(1+n) x y}+(-x)^{-n} (-x y)^n (1-x y)^{-1-n} n! \ln \left(\frac{1}{x y}\right)+y^n (1-x y)^{-1-n} n! \ln (x y)$$

Mathematica code:

 D[Log[x y]/(1 - x y), {x, n}] ==
 -(((-x)^-n n! Hypergeometric2F1[1, 1 + n, 2 + n, 1 - 1/(x y)])/((1 + n) x y)) + (-x)^-n (-x y)^n (1 - x y)^(-1 - n) n! Log[1/(x y)] + y^n (1 - x y)^(-1 - n) n! Log[x y]]