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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.

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  • $\begingroup$ math.stackexchange.com/q/1839100/219995 $\endgroup$ Commented Mar 9, 2023 at 6:57
  • $\begingroup$ So it is some kind of duplicate that I didn't catch by eye or Approach0... bummer. But since the two current answer (excepting mine) demonstrate different "approaches" than the ones in your linked question, how should we proceed with this? Close it? Delete it? Just let it stay open? $\endgroup$
    – TheOutZ
    Commented Mar 9, 2023 at 7:58

3 Answers 3

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Define

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


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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  • $\begingroup$ Thank you for your answer. This almost looks so straightforward that I'm scratching my head at how I could have missed it, if it weren't for the ingenious trick with the derivative of the exponent introducing the log! That is a truly nice closed form. $\endgroup$
    – TheOutZ
    Commented Mar 9, 2023 at 7:53
  • $\begingroup$ I am a bit torn on whether I should accept your or my answer: My answer is the one that helped me the most, but your answer is probably the one that others could work better with, considering its only sums, binomials and factorials... What do you think? $\endgroup$
    – TheOutZ
    Commented Mar 9, 2023 at 8:07
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    $\begingroup$ @TheOutZ You should accept the answer YOU find most satisfying, and there's nothing wrong with accepting an answer you wrote yourself. If other users prefer my answer, well that's what upvotes are for. Cheers ;) $\endgroup$
    – David H
    Commented Mar 9, 2023 at 10:39
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    $\begingroup$ Well, that's a pretty convincing argument :D. Then I will accept mine but upvote yours and keep it in mind for future endeavours! Have a nice day :). $\endgroup$
    – TheOutZ
    Commented Mar 9, 2023 at 11:21
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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

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  • $\begingroup$ Yes, I was also able to obtain the "closed forms" of $I_r$ in the cases $r = 1,2$ in terms of harmonic numbers, where $I_1$ is exactly the same as yours and $(n+1)I_2 = -2\sum_{k = 1}^{n+1} \frac{H_k}{k}$. I wonder if $I_{\geq 3}$ admits a similar expression? $\endgroup$
    – TheOutZ
    Commented Mar 7, 2023 at 10:42
  • $\begingroup$ It seems that $I_3$ is described as A027462 in the OEIS. From there we also have a representation in terms of generalized harmonic numbers. Neat. $\endgroup$
    – TheOutZ
    Commented Mar 7, 2023 at 10:45
  • $\begingroup$ I was able to recognize the expression of $I_r$ in terms of generalized harmonic numbers and (complete) Bell polynomials. I will write up an answer shortly. $\endgroup$
    – TheOutZ
    Commented Mar 7, 2023 at 11:04
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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.

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