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Let $$I_n = \int_0^1 \frac{\log^n (1-x) \log^{n-1} (1+x)}{1+x} dx$$ In a recently published article, $I_n$ are evaluated for $n\leq 6$: $$\begin{aligned}I_1 &= \frac{\log ^2(2)}{2}-\frac{\pi ^2}{12} \\ I_2 &= 2 \zeta (3) \log (2)-\frac{\pi ^4}{360}+\frac{\log ^4(2)}{4}-\frac{1}{6} \pi ^2 \log ^2(2) \\ I_3 &= \small 6 \zeta (3)^2+6 \zeta (3) \log ^3(2)-2 \pi ^2 \zeta (3) \log (2)+24 \zeta (5) \log (2)-\frac{23 \pi ^6}{2520}+\frac{\log ^6(2)}{6}-\frac{1}{4} \pi ^2 \log ^4(2)-\frac{1}{12} \pi ^4 \log ^2(2) \\ I_4 &= \small{-12 \pi ^2 \zeta (3)^2+288 \zeta (3) \zeta (5)+12 \zeta (3) \log ^5(2)-12 \pi ^2 \zeta (3) \log ^3(2)+168 \zeta (5) \log ^3(2)+108 \zeta (3)^2 \log ^2(2)-2 \pi ^4 \zeta (3) \log (2)-48 \pi ^2 \zeta (5) \log (2)+720 \zeta (7) \log (2)-\frac{499 \pi ^8}{25200}+\frac{\log ^8(2)}{8}-\frac{1}{3} \pi ^2 \log ^6(2)-\frac{19}{60} \pi ^4 \log ^4(2)-\frac{1}{6} \pi ^6 \log ^2(2)} \end{aligned}$$ Based on these evidences, the author (me) made the conjecture that

For positive integer $n$, $I_n$ is in the algebra over $\mathbb{Q}$ generated by $\log(2)$ and $\{\zeta(m) | m\in \mathbb{Z}, m\geq 3\}$.

The closed-form of $I_5, I_6$ also satisfy this conjecture. $I_5$ is:

-20\pi^4\zeta(3)^2+7200\zeta(5)^2-960\pi^2\zeta(3)\zeta(5)+14400\zeta(3)\zeta(7)+20\zeta(3)\log^7(2)-40\pi^2\zeta(3)\log^5(2)+600\zeta(5)\log^5(2)+600\zeta(3)^2\log^4(2)-\frac{76}{3}\pi^4\zeta(3)\log^3(2)-560\pi^2\zeta(5)\log^3(2)+8640\zeta(7)\log^3(2)-360\pi^2\zeta(3)^2\log^2(2)+10080\zeta(3)\zeta(5)\log^2(2)+1440\zeta(3)^3\log(2)-\frac{20}{3}\pi^6\zeta(3)\log(2)-112\pi^4\zeta(5)\log(2)-2400\pi^2\zeta(7)\log(2)+40320\zeta(9)\log(2)-\frac{149\pi^{10}}{1320}+\frac{\log^{10}(2)}{10}-\frac{5}{12}\pi^2\log^8(2)-\frac{7}{9}\pi^4\log^6(2)-\frac{19}{18}\pi^6\log^4(2)-\frac{47}{60}\pi^8\log^2(2)

$I_6$ is:

10800\zeta(3)^4-100\pi^6\zeta(3)^2-36000\pi^2\zeta(5)^2-3360\pi^4\zeta(3)\zeta(5)-72000\pi^2\zeta(3)\zeta(7)+1123200\zeta(5)\zeta(7)+1209600\zeta(3)\zeta(9)+30\zeta(3)\log^9(2)-100\pi^2\zeta(3)\log^7(2)+1560\zeta(5)\log^7(2)+2100\zeta(3)^2\log^6(2)-140\pi^4\zeta(3)\log^5(2)-3000\pi^2\zeta(5)\log^5(2)+47520\zeta(7)\log^5(2)-3000\pi^2\zeta(3)^2\log^4(2)+90000\zeta(3)\zeta(5)\log^4(2)+24000\zeta(3)^3\log^3(2)-\frac{380}{3}\pi^6\zeta(3)\log^3(2)-2040\pi^4\zeta(5)\log^3(2)-43200\pi^2\zeta(7)\log^3(2)+739200\zeta(9)\log^3(2)-1140\pi^4\zeta(3)^2\log^2(2)+388800\zeta(5)^2\log^2(2)-50400\pi^2\zeta(3)\zeta(5)\log^2(2)+777600\zeta(3)\zeta(7)\log^2(2)-7200\pi^2\zeta(3)^3\log(2)-47\pi^8\zeta(3)\log(2)-560\pi^6\zeta(5)\log(2)+302400\zeta(3)^2\zeta(5)\log(2)-8880\pi^4\zeta(7)\log(2)-201600\pi^2\zeta(9)\log(2)+3628800\zeta(11)\log(2)-\frac{4714153\pi^{12}}{5045040}+\frac{\log^{12}(2)}{12}-\frac{1}{2}\pi^2\log^{10}(2)-\frac{37}{24}\pi^4\log^8(2)-\frac{253}{63}\pi^6\log^6(2)-\frac{527}{72}\pi^8\log^4(2)-\frac{223}{36}\pi^{10}\log^2(2)

Question: How to prove the conjecture for general $n$?

Any suggestion is appreciated.


Some remarks:

  1. Even $I_3,I_4,I_5,I_6$ are extremely challenging, someone brave enough might want to embark on finding them independently.

  2. $I_n$ is not related to beta function in an obvious way, so the well-known differentiation trick does not work here.

  3. For any $I_n$, the algorithm outlined in the article should produce closed-form of $I_n$ in a finite amount of time if the conjecture is true. However, the algorithm is a bit mechanical, so benefits little toward a proof for general $n$.

  4. Perhaps I am missing something, this conjecture is elementary to state, so it might have an easy proof and I was being negligent.
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    $\begingroup$ Are you interested to see $I_{3-6}$ evaluated independently here in this question, or you're only after the conjecture? $\endgroup$
    – Zacky
    Oct 29, 2019 at 11:34
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    $\begingroup$ I would try making the change of variable $(1+x)/2=y$ and then connect the result with the fact that $$ \int_0^{1/2} \frac{\log^p(1-x)\log^{p+1}(x)}{1-x}\textrm{d}x=-\frac{1}{2(1+p)}\log^{2(p+1)}(2)+\frac{1}{2}\lim_{\substack{x\to0 \\ y \to 1}}\frac{\partial^{2p+1}}{\partial x^p \partial y^{p+1}}\operatorname{B}(x,y),$$ which appears in the book (Almost) Impossible Integrals, Sums, and Series, on page $9$. $\endgroup$ Oct 29, 2019 at 13:59
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    $\begingroup$ $$I_3=\int_0^{1/2} \left(\frac{\log ^5(2)}{1-t}+\frac{2 \log ^4(2) \log (1-t)}{1-t}+\frac{3 \log ^4(2) \log (t)}{1-t}+\frac{6 \log ^3(2) \log (1-t) \log (t)}{1-t}+\frac{2 \log (2) \log (1-t) \log ^3(t)}{1-t}+\frac{6 \log ^2(2) \log (1-t) \log ^2(t)}{1-t}+\frac{3 \log ^2(2) \log ^2(1-t) \log (t)}{1-t}+\frac{3 \log (2) \log ^2(1-t) \log ^2(t)}{1-t}+\frac{\log ^3(2) \log ^2(1-t)}{1-t}+\frac{3 \log ^3(2) \log ^2(t)}{1-t}+\frac{\log ^2(2) \log ^3(t)}{1-t}+\frac{\log ^2(1-t) \log ^3(t)}{1-t}\right) \textrm{d}t.$$ $\endgroup$ Oct 29, 2019 at 15:20
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    $\begingroup$ @pisco for higher weights like $I_4$ we may also need to deal with the integrals of the type, $\displaystyle \int_0^{1/2} \frac{\log ^n(1-x) \log ^n(x)}{1-x} \textrm{d}x$, where we can use again a result inspired from (Almost) Impossible Integrals, Sums, and Series, that is $$\int_0^{1/2} \frac{\log ^n(2 (1-x)) \log ^n(2 x)}{1-x}\textrm{d}x=\int_0^1 \frac{\log ^n(1+x) \log ^n(1-x)}{1+x} \textrm{d}x.$$ Combined with the algebraic identities we should get rid of all troublesome series. $\endgroup$ Oct 29, 2019 at 15:44
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    $\begingroup$ @AliShather Well, it's a tricky integral. Then one can do it without using Beta function as Cornel did here facebook.com/… $\endgroup$ Oct 30, 2019 at 16:15

4 Answers 4

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Denote $f(k,j)=\int_0^{\frac{1}{2}} \frac{\log ^j(1-y) \log ^k(y)}{1-y} \, dy$. Then for $j, k>1$ (RHS denotes Beta derivatives)

$$U(k,j):=jf(k,j-1)+kf(j,k-1)=-(-\log(2))^{j+k}+ k \left( \partial_a^{k-1} \partial_b^j B\right) (0,1)$$

Which is direct by IBP, separation, Beta derivatives and reflection $y\to 1-y$: $$\small jf(k,j-1)= -(-\log (2))^{j+k}+ k \int_0^{\frac{1}{2}} \frac{\log ^j(1-y) \log ^{k-1}(y)}{y} \, dy$$ $$\small =-(-\log (2))^{j+k}+ k \left(\int_0^{1}-\int_{\frac{1}{2}}^1 \right) \frac{\log ^j(1-y) \log ^{k-1}(y)}{y} \, dy$$ $$\small =-(-\log(2))^{j+k}+ k \left( \partial_a^{k-1} \partial_b^j B\right) (0,1)-kf(j,k-1)$$ Thus taking $\frac{\binom{n-1}{j-1} \binom{n}{k}}{\binom{n}{j} \binom{n-1}{k-1}}=\frac{j}{k}$ into account yields the important $\color{blue}{formula}$

$$\small \binom{n}{k} \binom{n-1}{m-k} f(k,m-k)+\binom{n}{m+1-k} \binom{n-1}{k-1} f(m+1-k,k-1)=\frac{\binom{n}{k}\binom{n-1}{m-k} }{-k+m+1}U(k,m+1-k)$$

Now let $y\to\frac{1-x}{2}$ $$I_n=\int_0^{\frac{1}{2}} \frac{\log ^n(2 y) \log ^{n-1}(2 (1-y))}{1-y} \, dy$$ Apply Binomial thm twice, extract $k=0$ $$I_n=\sum _{k=1}^n \sum _{j=0}^{n-1} \binom{n}{k} \binom{n-1}{j} f(k,j) \log ^{2n-j-k-1}(2)+\int_0^{\frac{1}{2}} \frac{\log ^n(2) \log ^{n-1}(2 (1-y))}{1-y} \, dy$$ Take Cauchy product $$I_n=\sum _{m=1}^{2n-1} \sum _{k+j=m}\binom{n}{k} \binom{n-1}{j} f(k,j) \log ^{2n-m-1}(2)+\frac{\log ^{2 n}(2)}{n}$$ Take care of range of $j,k$ $$\scriptsize I_n=\sum _{m=1}^n \sum _{k=1}^m \binom{n}{k} \binom{n-1}{m-k} f(k,m-k) \log ^{2n-m-1}(2)+ \sum _{m=n+1}^{2 n-1} \sum _{k=m-n+1}^n \binom{n}{k} \binom{n-1}{m-k} f(k,m-k) \log ^{2n-m-1}(2)+\frac{\log ^{2 n}(2)}{n}$$ Let $k\to m+1-k$, take average $$\scriptsize I_n=\frac{1}{2} \sum _{m=1}^n \sum _{k=1}^m \left(\binom{n}{k} \binom{n-1}{m-k} f(k,m-k)+\binom{n}{m+1-k} \binom{n-1}{k-1} f(m+1-k,k-1)\right) \log ^{2n-m-1}(2)+\frac{1}{2} \sum _{m=n+1}^{2 n-1} \sum _{k=m-n+1}^n \left(\binom{n}{k} \binom{n-1}{m-k} f(k,m-k)+\binom{n}{m+1-k} \binom{n-1}{k-1} f(m+1-k,k-1)\right) \log ^{2n-m-1}(2)+\frac{\log ^{2 n}(2)}{n}$$ Use the $\color{blue}{formula}$ to simplify $$\scriptsize I_n=\frac{1}{2} \sum _{m=1}^n \sum _{k=1}^m \frac{\binom{n}{k}\binom{n-1}{m-k} \log ^{-m+2 n-1}(2) }{-k+m+1} U(k,m+1-k)+\frac{1}{2} \sum _{m=n+1}^{2 n-1} \sum _{k=m-n+1}^n \frac{\binom{n}{k}\binom{n-1}{m-k} \log ^{-m+2 n-1}(2) }{-k+m+1}U(k,m+1-k)+\frac{\log ^{2 n}(2)}{n}$$ Expand $U(k,m+1-k)$

$$ \scriptsize I_n=\frac{1}{2} \left(\sum _{m=1}^n \sum _{k=1}^m +\sum _{m=n+1}^{2 n-1} \sum _{k=m-n+1}^n\right) \frac{\binom{n}{k}\binom{n-1}{m-k} \log ^{-m+2 n-1}(2) }{-k+m+1}\left(k \underset{a\to 0}{\text{lim}}\underset{b\to 1}{\text{lim}}\frac{\partial ^{m}B(a,b)}{\partial a^{k-1}\, \partial b^{-k+m+1}}+(-1)^m \log ^{m+1}(2)\right)+\frac{\log ^{2 n}(2)}{n}$$

This is the final expression of $I_n$. According to Lemma $2.3$ in OP's article, all Beta derivatives in this expression lie in the algebra $\mathbb{Q}(\pi^2, \zeta(3), \zeta(5), \zeta(7), \cdots)$, whence after adding up $\log(2)$ terms, $I_n$ lies in the extended $\mathbb{Q}(\log(2), \pi^2, \zeta(3), \zeta(5), \zeta(7), \cdots)$. QED.

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  • $\begingroup$ Creative (+1). I'm sure this generalization doesnt exist in the literature. Congrats $\endgroup$ Jun 13, 2020 at 19:00
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Too long for a comment: By using the start I described in comments and then algebraic identities, I could reduce $\mathcal{I_4}$ to

$$\mathcal{I_4}=\log ^8(2)+\frac{31}{420} \log^2(2)\pi^6+4 \log (2) \underbrace{\int_0^1 \frac{\log ^3(1-t) \log ^3(t)}{t} \textrm{d}t}_{\text{Beta function}}+\log ^7(2)\int_0^{1/2} \frac{1}{1-t}\textrm{dt}\\+3 \log ^6(2)\int_0^{1/2}\frac{ \log (1-t)}{1-t}\textrm{d}t+4 \log ^6(2)\int_0^{1/2}\frac{ \log (t)}{1-t}\textrm{d}t+12 \log ^5(2) \int_0^{1/2}\frac{\log (1-t) \log (t)}{1-t}\textrm{d}t+12 \log ^3(2)\underbrace{\int_0^{1/2}\frac{ \log (1-t) \log ^3(t)}{1-t}\textrm{d}t}_{\text{Reducible to K}}+4 \log ^3(2)\int_0^{1/2} \frac{\log ^3(1-t) \log (t)}{1-t}\textrm{d}t\\+3 \log ^5(2)\int_0^{1/2}\frac{ \log ^2(1-t)}{1-t}\textrm{d}t+6 \log ^5(2)\int_0^{1/2}\frac{ \log ^2(t)}{1-t}\textrm{d}t+\frac{3}{5} \log ^2(2)\int_0^{1/2}\frac{ \log ^5(t)}{1-t}\textrm{d}t\\-\frac{3}{5} \log ^2(2)\int_0^{1/2}\frac{ \log ^5(1-t)}{1-t}\textrm{d}t+\log ^4(2) \int_0^{1/2} \frac{\log ^3(1-t)}{1-t}\textrm{d}t+4 \log ^4(2)\int_0^{1/2}\frac{ \log ^3(t)}{1-t}\textrm{d}t\\+\log ^3(2) \int_0^{1/2}\frac{\log ^4(t)}{1-t}\textrm{d}t+\underbrace{\int_0^{1/2}\frac{\log ^3(1-t) \log ^4(t)}{1-t}\textrm{d}t}_{\text{Reducible to $J_3$}}+18 \log ^4(2) \underbrace{\int_0^{1/2}\frac{ \log (1-t) \log ^2(t)}{1-t}\textrm{d}t}_{\textrm{Reducible to $J_1$}}+12 \log ^4(2)\int_0^{1/2}\frac{ \log ^2(1-t) \log (t)}{1-t}\textrm{d}t+3 \log ^2(2)\int_0^{1/2}\frac{ \log ^4(1-t) \log (t)}{1-t}\textrm{d}t\\+18 \log ^3(2)\underbrace{\int_0^{1/2}\frac{ \log ^2(1-t) \log ^2(t)}{1-t}\textrm{d}t}_{\text{Reducible to $K$}}+18 \log ^2(2) \underbrace{\int_0^{1/2}\frac{\log ^2(1-t) \log ^3(t)}{1-t} \textrm{d}t}_ {\text{Reducible to $J_2$}}.$$

I considered the auxiliary results

$$J_n=\int_0^{1/2} \frac{\log^n(1-x)\log^{n+1}(x)}{1-x}\textrm{d}x=-\frac{1}{2(1+n)}\log^{2(n+1)}(2)+\frac{1}{2}\lim_{\substack{x\to0 \\ y \to 1}}\frac{\partial^{2n+1}}{\partial x^n \partial y^{n+1}}\operatorname{B}(x,y)$$ and $$ K=\int_{0}^{1/2} \frac{\log^2(x)\log^2(1-x)}{x}\textrm{d}x$$ $$=\frac{1}{8}\zeta(5)-2\zeta(2)\zeta(3)-\frac{2}{3}\log^3(2)\zeta(2)+\frac{7}{4}\log^2(2)\zeta(3)-\frac{1}{15}\log^5(2) $$ $$+4\log(2)\operatorname{Li}_4\left(\frac{1}{2}\right)+4\operatorname{Li}_5\left(\frac{1}{2}\right),$$ which are both calculated in the book (Almost) Impossible Integrals, Sums, and Series.

A short note: For the generalization the key is to figure out which groups of integrals to take together for transformations after the first step I described in comments, where to further use algebraic identities to get those expected magical cancellations like in the case above. The rest is trivial. Also, I skipped giving references for the trivial integrals above.

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  • $\begingroup$ outstanding approach (+1). $\endgroup$ Oct 29, 2019 at 19:09
  • $\begingroup$ @AliShather Thank you! $\endgroup$ Oct 29, 2019 at 19:11
  • $\begingroup$ A nice solution of $I_4$, I wish your method could be generalized to arbitrary $n$ though. $\endgroup$
    – pisco
    Oct 30, 2019 at 3:31
  • $\begingroup$ @pisco Thank you. The way used above should work for every $n$, but to put all in a generalized form requires some tedious work. $\endgroup$ Oct 30, 2019 at 6:47
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    $\begingroup$ This is indeed an inspiring way. Actually using IBP, the calculation of nontrivial part $K$ is not needed. By generalizing your method I proved the general proposition. Much appreciation. $\endgroup$ Jun 2, 2020 at 15:12
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My approach to $I_3$:

Starting with the algebraic identity $20a^3b^2=(a+b)^5+(a-b)^5-2a^5-10ab^4$ we can write

$$20\int_0^1\frac{\ln^3(1-x)\ln^2(1+x)}{1+x}\ dx\\=\int_0^1\frac{\ln^5(1-x^2)}{1+x}+\int_0^1\frac{\ln^5\left(\frac{1-x}{1+x}\right)}{1+x}-2\int_0^1\frac{\ln^5(1-x)}{1+x}-10\int_0^1\frac{\ln(1-x)\ln^4(1+x)}{1+x}\ dx$$


The first integral can be calculated the same way Cornel did here

$$\int_0^1\frac{\ln^5(1-x^2)}{1+x}dx=\int_0^1(1-x)\frac{\ln^5(1-x^2)}{1-x^2}dx\overset{x^2=y}{=}\frac12\int_0^1\frac{1-\sqrt{y}}{\sqrt{y}}.\frac{\ln^5(1-y)}{1-y}dy$$ $$\overset{IBP}{=}-\frac1{24}\int_0^1\frac{\ln^6(1-y)}{y^{3/2}}dy=-\frac{1}{24}\lim_{x\mapsto-1/2\\y\mapsto1}\frac{\partial^6}{\partial y^6}\text{B}(x,y)$$

$$\boxed{=\frac{16}3\ln^62-40\ln^42\zeta(2)+160\ln^32\zeta(3)-270\ln^22\zeta(4)+720\ln2\zeta(5)\\-240\ln2\zeta(2)\zeta(3)-\frac{1185}{4}\zeta(6)+120\zeta^2(3)}$$


The second integral can be simplified via subbing $\frac{1-x}{1+x}=y$:

$$\int_0^1\frac{\ln^5\left(\frac{1-x}{1+x}\right)}{1+x}\ dx=\int_0^1\frac{\ln^5 y}{1+y}\ dy\\=-\sum_{n=1}^\infty (-1)^n\int_0^1 y^{n-1}\ln^5 y\ dy=5!\sum_{n=1}^\infty\frac{(-1)^n}{n^6}=\boxed{-\frac{465}{4}\zeta(6)}$$


and lets set $1-x=y$ for the third integral:

$$\int_0^1\frac{\ln^5(1-x)}{1+x}\ dx=\int_0^1\frac{\ln^5y}{2-y}\ dy\\=\sum_{n=1}^\infty\frac1{2^n}\int_0^1 y^{n-1}\ln^5 y\ dy=-5!\sum_{n=1}^\infty\frac{1}{2^n n^6}=\boxed{-120\operatorname{Li}_6(1/2)}$$


For the last integral, we set $1+x=y$

$$\int_0^1\frac{\ln(1-x)\ln^4(1+x)}{1+x}\ dx=\int_1^2\frac{\ln(2-y)\ln^4y}{y}\ dy\\=\ln2\int_1^2\frac{\ln^4y}{y}\ dx+\int_1^2\frac{\ln(1-y/2)\ln^4y}{y}\ dy\\=\frac15\ln^62-\sum_{n=1}^\infty\frac{1}{n2^n}\int_1^2 y^{n-1}\ln^4y\ dy\\=\frac15\ln^62-\sum_{n=1}^\infty\frac1{n2^n}\left(24\frac{2^n}{n^5}-24\frac{2^n\ln2}{n^4}+12\frac{2^n\ln^22}{n^3}-4\frac{2^n\ln^32}{n^2}+\frac{2^n\ln^42}{n}-\frac{24}{n^5}\right)\\\boxed{=\frac15\ln^62-24\zeta(6)+24\ln2\zeta(5)-12\ln^22\zeta(4)+4\ln^32\zeta(3)-\ln^42\zeta(2)+24\operatorname{Li}_6(1/2)}$$

By combining the boxed results, the closed form of $I_3$ follows:

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    $\begingroup$ (+1) for the nice approach. $\endgroup$ Oct 30, 2019 at 19:12
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    $\begingroup$ thank you for reminding me how to solve the tricky integral :) $\endgroup$ Oct 30, 2019 at 19:18
  • $\begingroup$ @ali shather I noticed an error in $$\int_0^1\frac{\ln^5(1-x^2)}{1+x}\ dx=-\frac{8295}{28}\zeta(6)-40\zeta(2)Log^4(2)+...=-30.382721029123626...$$ The other terms are correct. $\endgroup$
    – user178256
    Oct 30, 2019 at 23:17
  • $\begingroup$ @user178256 thank you.. i forgot the term $-40\ln^42\zeta(2)$.. edited now $\endgroup$ Oct 31, 2019 at 1:31
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Generalization using the Stirling number

Start with $1-x\to x$

$$I_n=\int_0^1\frac{\ln^n(1-x)\ln^{n-1}(1+x)}{1+x}dx=\int_0^1\frac{\ln^n(x)\ln^{n-1}(2-x)}{2-x}dx$$

$$=\int_0^1\frac{\ln^n(x)\left[\ln(2)+\ln(1-x/2)\right]^{n-1}}{2-x}dx$$

$$=\int_0^1\frac{\ln^n(x)\sum_{k=0}^{n-1}{n-1\choose k}\ln^{n-k-1}(2)\ln^k(1-x/2)}{2-x}dx$$

$$=\frac12\sum_{k=0}^{n-1}{n-1\choose k}\ln^{n-k-1}(2)\int_0^1\frac{\ln^n(x)\ln^k(1-x/2)}{1-x/2}dx$$

$$\overset{IBP}{=}n\sum_{k=0}^{n-1}{n-1\choose k}\ln^{n-k-1}(2)\int_0^1\frac{\ln^{n-1}(x)\ln^{k+1}(1-x/2)}{x}dx$$

Recall Stirling number of first kind

$$\frac{\ln^k(1+x)}{k!}=\sum_{j=k}^\infty(-1)^{j-k} \begin{bmatrix} j \\ k \end{bmatrix}\frac{x^j}{j!}$$

where if we replace $x$ by $-x/2$ and $k$ by $k+1$ we have

$$\frac{\ln^{k+1}(1-x/2)}{(k+1)!}=\sum_{j=k+1}^\infty(-1)^{j-k-1} \begin{bmatrix} j \\ k+1 \end{bmatrix}\frac{x^j}{j!}$$

Thus

$$I_n=\sum_{j=k+1}^\infty\sum_{k=0}^{n-1}(-1)^{j-k-1} \begin{bmatrix} j \\ k+1 \end{bmatrix}\frac{n}{j!}{n-1\choose k}\ln^{n-k-1}(2)\int_0^1\ln^{n-1}(x)x^{j-1}dx$$

$$=\sum_{j=k+1}^\infty\sum_{k=0}^{n-1}(-1)^{j-k-1} \begin{bmatrix} j \\ k+1 \end{bmatrix}\frac{n}{j!}{n-1\choose k}\ln^{n-k-1}(2)\left[\frac{(-1)^{n-1}(n-1)!}{j^n}\right]$$

$$=\sum_{j=k+1}^\infty\sum_{k=0}^{n-1}(-1)^{j-k+n} \frac{n!}{j!j^n}\begin{bmatrix} j \\ k+1 \end{bmatrix}{n-1\choose k}\ln^{n-k-1}(2)$$

Not sure if this double sum can be further simplified.

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