Closed form of the recursion formula for $\int_{0}^{1} \ln^r(1-x)x^n \, dx$

Question

It is easy to see that we have the following recursive formula for the indefinite integral of log powers: $$J_r := \int \ln^r(1-x) \, dx \stackrel{\text{sub.}+IBP}{=} -(1-x)\ln^r(1-x) - r\int \ln^{r-1}(1-x) \, dx = -(1-x)\ln^r(1-x) - rJ_{r-1}$$

with initial condition $J_0 = x$. It is also fairly easy to obtain the explicit formula for this integral: $$J_r = (-1)^r (r!) x + (x-1)\sum_{j = 0}^{r-1} (-1)^j (r)_j \ln^{r-j}(1-x) \space\space\space(+C)$$ where $(r)_j = \prod_{k = 0}^{j-1} (r-k)$ is the falling factorial. We can use this to find a recursive relationship for the integral $$I_{r,n} := \int_{0}^{1} \ln^r(1-x)x^n \, dx$$ mentioned in the title. Skipping the finicky calculation part (applying IBP once using $J_r$ and moving $I_{r,n}$ to the LHS), we can calculate that

$$I_{r,n} = \frac{(-1)^r r!}{(n+1)^2} - \frac{n}{n+1}\left(\sum\limits_{j = 1}^{r-1} (-1)^j (r)_j I_{r-j,n} - \sum\limits_{j = 0}^{r-1}(-1)^{j} (r)_j I_{r-j,n-1}\right)$$

with initial conditions $$I_{0,k} = \frac{1}{k+1} \text{ and } I_{r,0} = (-1)^r r!$$ The first initial condition is trivial, the second one can be calculated in the same way as the integral $J_r$ above, imposing the boundaries straight from the beginning.

My question is:

Does the relationship for $I_{r,n}$ entail a closed form (i.e. explicit) expression in terms of $r$ and $n$? If not, would it at least be possible to evaluate $I_{r,n}$ for some fixed $r \in \mathbb{N}$?

Some background: This integral surprise-attacked me while trying to find the value of the integral $$-\int_{0}^{1} \frac{\ln^r(t)\ln(1-t)}{1-t} \, dt$$ by series expansion. I will add additional details if requested.

Answer 1

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Define

$$\mathcal{I}_{n,p}:=\int_{0}^{1}\mathrm{d}x\,x^{n}\ln^{p}{\left(1-x\right)};~~~\small{n\in\mathbb{Z}_{\ge0}\land p\in\mathbb{N}}.$$

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Given fixed but arbitrary $n\in\mathbb{Z}_{\ge0}\land p\in\mathbb{N}$, we find

$$\begin{align} \mathcal{I}_{n,p} &=\int_{0}^{1}\mathrm{d}x\,x^{n}\ln^{p}{\left(1-x\right)}\\ &=\int_{0}^{1}\mathrm{d}y\,\left(1-y\right)^{n}\ln^{p}{\left(y\right)};~~~\small{\left[x=1-y\right]}\\ &=\int_{0}^{1}\mathrm{d}y\,\ln^{p}{\left(y\right)}\sum_{k=0}^{n}\binom{n}{k}\left(-y\right)^{k}\\ &=\sum_{k=0}^{n}\left(-1\right)^{k}\binom{n}{k}\int_{0}^{1}\mathrm{d}y\,y^{k}\ln^{p}{\left(y\right)}\\ &=\sum_{k=0}^{n}\left(-1\right)^{k}\binom{n}{k}\int_{0}^{1}\mathrm{d}y\,\frac{\partial^{p}}{\partial k^{p}}\left(y^{k}\right)\\ &=\sum_{k=0}^{n}\left(-1\right)^{k}\binom{n}{k}\frac{d^{p}}{dk^{p}}\int_{0}^{1}\mathrm{d}y\,y^{k}\\ &=\sum_{k=0}^{n}\left(-1\right)^{k}\binom{n}{k}\frac{d^{p}}{dk^{p}}\left(\frac{1}{k+1}\right)\\ &=\sum_{k=0}^{n}\left(-1\right)^{k}\binom{n}{k}\frac{(-1)^{p}p!}{\left(k+1\right)^{p+1}}\\ &=(-1)^{p}p!\sum_{k=0}^{n}\frac{\left(-1\right)^{k}\binom{n}{k}}{\left(k+1\right)^{p+1}}.\\ \end{align}$$

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Answer 2

Quite tedious calculations even for a CAS !

Let $$I_r=(n+1)\int \log ^r(1-x)\,\, x^n\,dx$$ $$I_1=-H_{n+1}$$ $$I_2=\left(H_{n+1}\right){}^2-\psi ^{(1)}(n+2)+\frac{\pi ^2}{6}$$ $$I_3=-\frac{1}{2} H_{n+1} \left(2 \left(H_{n+1}\right){}^2-6 \psi ^{(1)}(n+2)+\pi ^2\right)-\psi ^{(2)}(n+2)-2 \zeta (3)$$

The next are really too long

Answer 3

Thanks to @Claude Leibovici and some helpful lookups in the OEIS, I was able to recognize the "closed forms" of some $I_{r,n}$ for fixed $r$. After some backtracing through my notes, it looks very promising that $$I_{r,n} = (-1)^r \frac{Y_r(0!H_{n+1}^{(1)},\ldots,(r-1)!H_{n+1}^{(r)})}{n+1}$$ is a closed forms desirable for my purposes, where $Y_r$ is the complete Bell polynomial of order $r$ evaluated at multiples of the generalized harmonic numbers.

The only thing missing is to check that this closed form actually obeys the recursion. Plugging in the initial values we get (ignoring the notational issues with $Y_0$ at the moment) $$I_{0,n} = (-1)^0 \frac{Y_0(?)}{n+1} = \frac{1}{n+1} \text{ and } I_{r,0} = (-1)^r \frac{Y_r(0!,\ldots,(r-1)!)}{1} = (-1)^r r!$$ and looking at the recursion we have \begin{align*} &\frac{(-1)^r r!}{(n+1)^2} - \frac{n}{n+1}\left(\sum\limits_{j = 1}^{r-1} (-1)^j (r)_j I_{r-j,n} - \sum\limits_{j = 0}^{r-1}(-1)^{j} (r)_j I_{r-j,n-1}\right) \\ =&\frac{(-1)^r r!}{(n+1)^2} - \frac{n}{n+1}\left(\sum_{j = 1}^{r-1} (-1)^j (r)_j \frac{(-1)^{r-j}}{n+1} Y_{r-j}(0!H_{n+1},\ldots,(r-j-1)!H_{n+1}^{(r-j)})-\sum_{j = 0}^{r-1} (-1)^j (r)_j \frac{(-1)^{r-j}}{n} Y_{r-j}(0!H_{n},\ldots,(r-j-1)!H_{n}^{(r-j)})\right) \\ =&(-1)^r\left(\frac{r!}{(n+1)^2} - \frac{n}{n+1}\left(\sum_{j = 1}^{r-1} (r)_j \frac{1}{n+1} Y_{r-j}(0!H_{n+1},\ldots,(r-j-1)!H_{n+1}^{(r-j)})-\sum_{j = 0}^{r-1} (r)_j \frac{1}{n} Y_{r-j}(0!H_{n},\ldots,(r-j-1)!H_{n}^{(r-j)})\right)\right) \end{align*}

Well... this is where I am currently stuck. How can we proceed to show that the expression above equals $(-1)^r \frac{Y_r(0!H_{n+1},\ldots,(r-1)!H_{n+1}^{(r)})}{n+1}$? It also seems very likely that we can absorb the extra term outside of the sums...

I have asked for an alternative proof of this identity in another thread.