# How to evaluate $\int _0^1\frac{\ln ^2\left(1-x\right)\ln ^5\left(1+x\right)}{1+x}\:dx$

I've been trying to evaluate $$\int _0^1\frac{\ln ^2\left(1-x\right)\ln ^5\left(1+x\right)}{1+x}\:dx$$ But I can't find a way to simplify it. Integration by parts is not valid since we face convergence issues. Subbing the $$1-x$$ term is also not quite helpful here. I also tried to use the sub $$\frac{1}{1+x}$$ but this is not useful either. Using algebraic identities isn't that useful either, since we run into similar difficulty integrals. Since these attempts do not lead to anything, how can I approach it?

• I would start with $$\int_0^1\frac{\log^2(1-x)\log^5(1+x)}{1+x}\,dx=\left.\left(\frac{\partial^2}{\partial \alpha^2}\frac{\partial^5}{\partial \beta^5}\int_0^1 (1-x)^\alpha(1+x)^{\beta-1}\,dx\right)\right|_{(\alpha,\beta)=(0,0)}$$Or you could begin by IBP with $u=\log^2(1-x)$ and $v=\frac15\log^4(1+x)$ and proceed similarly. – Mark Viola Aug 20 '20 at 16:21
• @MarkViola that 1st idea is pretty clever, but im not sure how to proceed with it, also doesnt the u*v product diverges? for the IBP – user809806 Aug 20 '20 at 16:29
• It would diverge, so you need to write it as a limit: $$\lim_{\varepsilon\to 0^+}\left(\frac15 \log(\varepsilon) \log(2-\varepsilon) +\frac25\int_{0}^{1-\varepsilon} \frac{\log(1-x)\log^4(1+x)}{1-x}\,dx)\right)$$ – Mark Viola Aug 20 '20 at 17:00
• Set 1+x=y then expand $\ln^2(y-2)$ in series . – Ali Shadhar Aug 20 '20 at 17:23
• Could you elaborate on the expansion? i dont know it, maybe even post a solution? thanks – user809806 Aug 20 '20 at 21:04

$$\int_0^1\frac{\ln^2(1-x)\ln^5(1+x)}{1+x}dx\overset{1+x\to x}{=}\int_1^2\frac{\ln^2(2-x)\ln^5x}{x}dx$$

$$=\ln^2(2)\int_1^2\frac{\ln^5x}{x}dx+2\ln(2)\int_1^2\frac{\ln(1-x/2)\ln^5x}{x}dx+\int_1^2\frac{\ln^2(1-x/2)\ln^5x}{x}dx$$ write $$\ln(1-x/2)=-\sum_{n=1}^\infty\frac{x^n}{n2^n}$$ for the first integral and write $$\ln^2(1-x/2)=2\sum_{n=1}^\infty(\frac{H_n}{n2^n}-\frac{1}{n^22^n})x^n$$ for the third integral

$$=\frac16\ln^8(2)+\sum_{n=1}^\infty\left(\frac{2H_n}{n}-\frac{2}{n^2}-\frac{2\ln(2)}{n}\right)\int_1^2 \frac{x^{n-1}\ln^5x}{2^n}dx$$

$$=\frac16\ln^8(2)+\sum_{n=1}^\infty\left(\frac{2H_n}{n}-\frac{2}{n^2}-\frac{2\ln(2)}{n}\right)$$ $$\left(\frac{\ln^5(2)}{n}-\frac{5\ln^4(2)}{n^2}+\frac{20\ln^3(2)}{n^3}-\frac{60\ln^2(2)}{n^4}+\frac{120\ln(2)}{n^5}-\frac{120}{n^6}+\frac{120}{n^62^n}\right)$$

$$=\frac16\ln^8(2)-2\ln^6(2)\zeta(2)+12\ln^5(2)\zeta(3)-\frac{85}{2}\ln^4(2)\zeta(4)+40\ln^3(2)[5\zeta(5)-\zeta(2)\zeta(3)]$$ $$-30\ln^2(2)[11\zeta(6)-2\zeta^2(3)]-240\ln(2)\left[\zeta(4)\zeta(3)+\zeta(2)\zeta(5)-4\zeta(7)+\text{Li}_7\left(\frac12\right)\right]$$ $$-300\zeta(8)+240\zeta(3)\zeta(5)-240\text{Li}_8\left(\frac12\right)+240\sum_{n=1}^\infty\frac{H_n}{n^72^n}\approx 0.113272.$$

There is no closed form for your integral as $$\sum_{n=1}^\infty\frac{H_n}{n^72^n}$$ has no closed form.

Remark:

The closed form of $$\int_0^1\frac{\ln^2(1-x)\ln^a(1+x)}{1+x}dx$$ can be expressed in terms of $$\ln, \pi, \zeta$$ and $$\text{Li}_r$$ plus $$\sum_{n=1}^\infty\frac{H_n}{n^{a+2}2^n}$$. What I know of is that there is no closed form for such series with $$a>2$$ and the case $$a=2$$ is calculated here.

• Bravo, Ali! You showed only elementary manipulations and simple series are necessary to get the desired result! – user97357329 Aug 21 '20 at 22:49
• Thank you @user97357329 . I enjoy using simple techniques for solving hard problems. – Ali Shadhar Aug 21 '20 at 22:52
• very very nice Ali. – user178256 Aug 22 '20 at 8:34
• @user178256 Thank you for the kind words Sir. – Ali Shadhar Aug 22 '20 at 11:27
• @Dennis Orton yes Dennis i see that based on the algebraic identity you provided we may find another integral because the rest are trivial. You can post that here as a bonus. – Ali Shadhar Aug 22 '20 at 11:30

For a proof of the following, see here. This is also worth reading.

• $$I=-\frac{20}{3} \pi ^2 \zeta(\bar5,1)-160 \zeta(\bar7,1)-40\zeta(\bar5,1,\bar1,1)-160 \text{Li}_5\left(\frac{1}{2}\right) \zeta (3)-\frac{10}{9} \pi ^4 \text{Li}_4\left(\frac{1}{2}\right)+960 \text{Li}_8\left(\frac{1}{2}\right)-\frac{5 \pi ^2 \zeta (3)^2}{2}+\frac{7125 \zeta (3) \zeta (5)}{16}+\frac{34}{3} \zeta (3) \log ^5(2)-\frac{50}{9} \pi ^2 \zeta (3) \log ^3(2)+\frac{2075}{12} \zeta (5) \log ^3(2)+60 \zeta (3)^2 \log ^2(2)-\frac{23}{9} \pi ^4 \zeta (3) \log (2)-\frac{325}{24} \pi ^2 \zeta (5) \log (2)+\frac{2075}{2} \zeta (7) \log (2)-\frac{79 \pi ^8}{648}+\frac{25 \log ^8(2)}{168}-\frac{5}{18} \pi ^2 \log ^6(2)-\frac{109}{216} \pi ^4 \log ^4(2)-\frac{73}{189} \pi ^6 \log ^2(2)$$

For those enjoy evaluating integrals via elementary ways, the following might be edifying. Let $$\text{NL}(a,b,c)=\int_0^1 \log ^a(1-x) \log ^b(x) \log ^c(x+1) \, dx$$, then

• $$\small \frac{7}{216} \text{NL}(0,3,4)+\frac{1}{24} \text{NL}(1,4,2)+\frac{11}{180} \text{NL}(1,5,1)+\frac{1}{12} \text{NL}(2,3,2)-\frac{1}{60} \text{NL}(5,0,2)=8 \text{Li}_4\left(\frac{1}{2}\right) \zeta (3)+\frac{5}{3} \pi ^2 \text{Li}_4\left(\frac{1}{2}\right)-\frac{124 \text{Li}_4\left(\frac{1}{2}\right)}{3}+4 \pi ^2 \text{Li}_5\left(\frac{1}{2}\right)-\frac{80 \text{Li}_5\left(\frac{1}{2}\right)}{3}-\frac{88 \text{Li}_6\left(\frac{1}{2}\right)}{3}-24 \text{Li}_7\left(\frac{1}{2}\right)-\frac{8}{3} \text{Li}_4\left(\frac{1}{2}\right) \log ^2(2)+2 \pi ^2 \text{Li}_4\left(\frac{1}{2}\right) \log (2)+\frac{16}{3} \text{Li}_4\left(\frac{1}{2}\right) \log (2)-\frac{16}{3} \text{Li}_5\left(\frac{1}{2}\right) \log (2)-\frac{153 \pi ^2 \zeta (3)}{16}+\frac{2809 \zeta (3)}{12}-\frac{3209 \pi ^2 \zeta (5)}{288}+\frac{3269 \zeta (5)}{16}+\frac{30545 \zeta (7)}{192}-\frac{2617 \zeta (3)^2}{96}-\frac{7433 \pi ^4 \zeta (3)}{17280}-\frac{16}{9} \zeta (3) \log ^3(2)-\frac{7}{8} \pi ^2 \zeta (3) \log ^2(2)+\frac{79}{3} \zeta (3) \log ^2(2)+\frac{77}{4} \zeta (5) \log ^2(2)+\frac{147}{8} \zeta (3)^2 \log (2)+\frac{31}{6} \pi ^2 \zeta (3) \log (2)-\frac{436}{3} \zeta (3) \log (2)-\frac{317}{3} \zeta (5) \log (2)+\frac{3407 \pi ^6}{36288}+\frac{277 \pi ^4}{144}+\frac{737 \pi ^2}{18}-\frac{3052}{3}-\frac{1}{210} \log ^7(2)-\frac{11 \log ^6(2)}{270}+\frac{11}{180} \pi ^2 \log ^5(2)+\frac{2 \log ^5(2)}{45}+\frac{5}{72} \pi ^2 \log ^4(2)-\frac{11 \log ^4(2)}{18}-\frac{29}{540} \pi ^4 \log ^3(2)-\frac{10}{27} \pi ^2 \log ^3(2)+\frac{68 \log ^3(2)}{9}+\frac{9}{80} \pi ^4 \log ^2(2)+\frac{35}{9} \pi ^2 \log ^2(2)-110 \log ^2(2)-\frac{5}{96} \pi ^6 \log (2)-\frac{19}{18} \pi ^4 \log (2)-\frac{58}{3} \pi ^2 \log (2)+\frac{1760 \log (2)}{3}$$
• User 628759: << This integral is beyond reach of, as far as I'm concerned, ability of all but one or two users on MSE.>> You know the skill of ALL people reading MSE? For sure, it's a long work to write such solution. – FDP Aug 21 '20 at 11:40
• @ User 628759: For me , the problem is the first three terms of the close form. That probably means this integral is the sum of a doable integral using elementary techniques and another one, a monstruous one. – FDP Aug 21 '20 at 13:01

Generalization:

$$I_a=\int_0^1\frac{\ln^2(1-x)\ln^a(1+x)}{1+x}dx\overset{1+x\to x}{=}\int_1^2\frac{\ln^2(2-x)\ln^ax}{x}dx$$

$$=\ln^2(2)\int_1^2\frac{\ln^ax}{x}dx+2\ln(2)\int_1^2\frac{\ln(1-x/2)\ln^ax}{x}dx+\int_1^2\frac{\ln^2(1-x/2)\ln^ax}{x}dx$$ write $$\ln(1-x/2)=-\sum_{n=1}^\infty\frac{x^n}{n2^n}$$ for the first integral and write $$\ln^2(1-x/2)=2\sum_{n=1}^\infty(\frac{H_n}{n2^n}-\frac{1}{n^22^n})x^n$$ for the third integral

$$=\frac{\ln^{a+3}(2)}{a+1}+\sum_{n=1}^\infty\left(\frac{2H_n}{n}-\frac{2}{n^2}-\frac{2\ln(2)}{n}\right)\int_1^2 \frac{x^{n-1}\ln^ax}{2^n}dx$$

$$=\frac{\ln^{a+3}(2)}{a+1}+\sum_{n=1}^\infty\left(\frac{2H_n}{n}-\frac{2}{n^2}-\frac{2\ln(2)}{n}\right)\left(\frac{(-1)^{a-1}a!}{n^{a+1}2^n}+a!\sum_{k=1}^{a+1}\frac{(-1)^{k-1}\ln^{a-k+1}(2)}{n^k(a-k+1)!}\right)$$

$$=\frac{\ln^{a+3}(2)}{a+1}-2(-1)^aa!\left[\sum_{n=1}^\infty\frac{H_n}{n^{a+2}2^n}-\text{Li}_{a+3}\left(\frac12\right)-\ln(2)\text{Li}_{a+2}\left(\frac12\right)\right]$$

$$+2a!\sum_{k=1}^{a+1}\frac{(-1)^{k-1}\ln^{a-k+1}(2)}{(a-k+1)!}\left[\sum_{n=1}^\infty\frac{H_n}{n^{k+1}}-\zeta(k+2)-\ln(2)\zeta(k+1)\right]$$

$$\because \quad\sum_{n=1}^\infty\frac{H_n}{n^r}=\frac{r+2}{2}\zeta(r+1)-\frac12\sum_{j=1}^{r-2}\zeta(j+1)\zeta(r-j)$$

$$\therefore\quad I_a=\frac{\ln^{a+3}(2)}{a+1}-2(-1)^aa!\left[\sum_{n=1}^\infty\frac{H_n}{n^{a+2}2^n}-\text{Li}_{a+3}\left(\frac12\right)-\ln(2)\text{Li}_{a+2}\left(\frac12\right)\right]$$ $$+2a!\sum_{k=1}^{a+1}\frac{(-1)^{k-1}\ln^{a-k+1}(2)}{(a-k+1)!}\left[\frac{k+1}{2}\zeta(k+2)-\ln(2)\zeta(k+1)-\frac12\sum_{j=1}^{k-1}\zeta(j+1)\zeta(k-j+1)\right]$$

Some cases:

$$I_3=12 \mathcal{H}_5-12 \text{Li}_6\left(\frac{1}{2}\right)-12 \text{Li}_5\left(\frac{1}{2}\right) \ln (2)+6 \zeta^2 (3)+8 \zeta (3) \ln ^3(2)-12 \zeta(2) \zeta (3) \ln (2)+36 \zeta (5) \ln (2)-9\zeta(6)+\frac{1}{4}\ln ^6(2)-2\zeta(2) \ln ^4(2)-\frac{27}{2} \zeta(4) \ln ^2(2)$$

$$I_4=-48 \mathcal{H}_6+48 \text{Li}_7\left(\frac{1}{2}\right)+48 \text{Li}_6\left(\frac{1}{2}\right) \ln (2)-48\zeta(4) \zeta (3)-48 \zeta(2)\zeta (5)+144 \zeta (7)+10 \zeta (3) \ln ^4(2)-24 \zeta(2) \zeta (3) \ln ^2(2)+96 \zeta (5) \ln ^2(2)+24 \zeta^2 (3) \ln (2)+\frac{1}{5}\ln ^7(2)$$ $$-2\zeta(2) \ln ^5(2)-26\zeta(4) \ln ^3(2)-84\zeta(6) \ln (2)$$

$$I_5=240 \mathcal{H}_7-240 \text{Li}_8\left(\frac{1}{2}\right)-240 \text{Li}_7\left(\frac{1}{2}\right) \ln (2)+240 \zeta (3) \zeta (5)+12 \zeta (3) \ln ^5(2)-40\zeta(2) \zeta (3) \ln ^3(2)+200 \zeta (5) \ln ^3(2)+60 \zeta^2 (3) \ln ^2(2)-240\zeta(4) \zeta (3) \ln (2)$$ $$-240\zeta(2) \zeta (5) \ln (2)+960 \zeta (7) \ln (2)-300\zeta(8)+\frac{1}{6}\ln ^8(2)-2\zeta(2) \ln ^6(2)$$ $$-\frac{85}{2} \zeta(4) \ln ^4(2)-330\zeta(6) \ln ^2(2)$$

$$I_6=-1440 \mathcal{H}_8+1440 \text{Li}_9\left(\frac{1}{2}\right)+1440 \text{Li}_8\left(\frac{1}{2}\right) \ln (2)-1440\zeta(6) \zeta (3)-1440\zeta(4)\zeta (5)-1440 \zeta(2) \zeta (7)+5760 \zeta (9)+14 \zeta (3) \ln ^6(2)-60\zeta(2) \zeta (3) \ln ^4(2)+360 \zeta (5) \ln ^4(2)+120 \zeta^2 (3) \ln ^3(2)-720\zeta(4) \zeta (3) \ln ^2(2)-720\zeta(2) \zeta (5) \ln ^2(2)+3600 \zeta (7) \ln ^2(2)+1440 \zeta (3) \zeta (5) \ln (2)+\frac{1}{7}\ln ^9(2)-2\zeta(2) \ln ^7(2)-63\zeta(4) \ln ^5(2)-900\zeta(6) \ln ^3(2)-3240\zeta(8) \ln (2)$$

Where $$\displaystyle\mathcal{H}_r=\sum_{n=1}^\infty\frac{H_n}{n^r2^n}$$

• (+1) Very well, Ali Shather. Impressively simple and elegant. – user97357329 Aug 22 '20 at 8:28
• @user97357329 Thank you. Unfortunately the closed form involves that sum that has no closed form for the case $a>2$. I hope its useful. – Ali Shadhar Aug 22 '20 at 11:47

Bonus: $$\int _0^1\frac{\ln ^4\left(1-x\right)\ln ^3\left(1+x\right)}{1+x}\:dx$$ Let's make use of the algebraic identity

$$a^4b^3=\frac{1}{70}\left(a+b\right)^7-\frac{1}{70}\left(a-b\right)^7-\frac{1}{5}a^6b-\frac{3}{5}a^2b^5-\frac{1}{35}b^7$$.

$$=\int _0^1\frac{1}{70}\frac{\ln ^7\left(1-x^2\right)}{1+x}-\frac{1}{70}\frac{\ln ^7\left(\frac{1-x}{1+x}\right)}{1+x}-\frac{1}{5}\frac{\ln ^6\left(1-x\right)\ln \left(1+x\right)}{1+x}$$ $$-\frac{3}{5}\frac{\ln ^2\left(1-x\right)\ln ^5\left(1+x\right)}{1+x}-\frac{1}{35}\frac{\ln ^7\left(1+x\right)}{1+x}\:dx$$

Here we can use the same strategy found in the book (Almost) Impossible integrals, Sums, and Series page $$80$$. $$\frac{1}{70}\int _0^1\frac{\ln ^7\left(1-x^2\right)}{1+x}\:dx=\frac{1}{70}\underbrace{\int _0^1\frac{\ln ^7\left(1-x^2\right)}{1-x^2}\left(1-x\right)\:dx}_{t=x^2}$$ $$=\frac{1}{140}\int _0^1\frac{\ln ^7\left(1-t\right)}{1-t}\frac{\left(1-\sqrt{t}\right)}{\sqrt{t}}\:dt=-\frac{1}{2240}\int _0^1\:\frac{\ln ^8\left(1-t\right)}{t^{\frac{3}{2}}}\:dt$$ $$=-\frac{1}{2240}\lim_{\alpha\rightarrow -1/2\\\beta\rightarrow 0}\frac{\partial ^8}{\partial \beta ^8}\operatorname{B}\left(\alpha ,\beta \right)$$ $$=-\frac{7017}{16}\zeta \left(8\right)+1296\ln \left(2\right)\zeta \left(7\right)-\frac{711}{2}\ln ^2\left(2\right)\zeta \left(6\right)-72\zeta \left(2\right)\zeta ^2\left(3\right)+432\zeta \left(3\right)\zeta \left(5\right)$$ $$+\frac{96}{5}\ln ^5\left(2\right)\zeta \left(3\right)-96\ln ^3\left(2\right)\zeta \left(2\right)\zeta \left(3\right)+288\ln ^3\left(2\right)\zeta \left(5\right)+144\ln ^2\left(2\right)\zeta ^2\left(3\right)$$ $$-324\ln \left(2\right)\zeta \left(3\right)\zeta \left(4\right)-432\ln \left(2\right)\zeta \left(2\right)\zeta \left(5\right)+\frac{8}{35}\ln ^8\left(2\right)-\frac{16}{5}\ln ^6\left(2\right)\zeta \left(2\right)-54\ln ^4\left(2\right)\zeta \left(4\right)$$

$$-\frac{1}{70}\underbrace{\int _0^1\frac{\ln ^7\left(\frac{1-x}{1+x}\right)}{1+x}\:dx}_{t=\frac{1-x}{1+x}}=-\frac{1}{70}\int _0^1\frac{\ln ^7\left(t\right)}{1+t}\:dt=72\sum _{k=1}^{\infty }\frac{\left(-1\right)^{k+1}}{k^8}$$ $$=\frac{1143}{16}\zeta \left(8\right)$$

$$-\frac{1}{5}\underbrace{\int _0^1\frac{\ln ^6\left(1-x\right)\ln \left(1+x\right)}{1+x}\:dx}_{t=1-x}$$ $$=-\frac{1}{10}\int _0^1\frac{\ln ^6\left(t\right)\ln \left(1-\frac{t}{2}\right)}{1-\frac{t}{2}}\:dt-\frac{1}{10}\ln \left(2\right)\int _0^1\frac{\ln ^6\left(t\right)}{1-\frac{t}{2}}\:dt$$ $$=144\sum _{k=1}^{\infty }\frac{H_k}{k^7\:2^k}-144\sum _{k=1}^{\infty }\frac{1}{k^8\:2^k}-144\ln \left(2\right)\sum _{k=1}^{\infty }\frac{1}{k^7\:2^k}$$ $$=144\sum _{k=1}^{\infty }\frac{H_k}{k^7\:2^k}-144\operatorname{Li}_8\left(\frac{1}{2}\right)-144\ln \left(2\right)\operatorname{Li}_7\left(\frac{1}{2}\right)$$

Thanks to Ali Shather we may use the value of this integral which he evaluated nicely in this same thread. $$-\frac{3}{5}\int _0^1\frac{\ln ^2\left(1-x\right)\ln ^5\left(1+x\right)}{1+x}\:dx$$ $$=180\zeta \left(8\right)-144\zeta \left(3\right)\zeta \left(5\right)-576\ln \left(2\right)\zeta \left(7\right)+144\ln \left(2\right)\zeta \left(2\right)\zeta \left(5\right)+144\ln \left(2\right)\zeta \left(3\right)\zeta \left(4\right)$$ $$+198\ln ^2\left(2\right)\zeta \left(6\right)-36\ln ^2\left(2\right)\zeta ^2\left(3\right)-120\ln ^3\left(2\right)\zeta \left(5\right)+24\ln ^3\left(2\right)\zeta \left(2\right)\zeta \left(3\right)+\frac{51}{2}\ln ^4\left(2\right)\zeta \left(4\right)$$ $$-\frac{36}{5}\ln ^5\left(2\right)\zeta \left(3\right)-144\sum _{k=1}^{\infty }\frac{H_k}{k^7\:2^k}+144\operatorname{Li}_8\left(\frac{1}{2}\right)+\frac{6}{5}\ln ^6\left(2\right)\zeta \left(2\right)$$ $$+144\ln \left(2\right)\operatorname{Li}_7\left(\frac{1}{2}\right)-\frac{1}{10}\ln ^8\left(2\right)$$

$$-\frac{1}{35}\underbrace{\int _0^1\frac{\ln ^7\left(1+x\right)}{1+x}\:dx}_{t=\ln\left(1+x\right)}=-\frac{1}{280}\ln ^8\left(2\right)$$

Collecting the results we find $$\int _0^1\frac{\ln ^4\left(1-x\right)\ln ^3\left(1+x\right)}{1+x}\:dx=-\frac{1497}{8}\zeta \left(8\right)+288\zeta \left(3\right)\zeta \left(5\right)+720\ln \left(2\right)\zeta \left(7\right)$$

$$-288\ln \left(2\right)\zeta \left(2\right)\zeta \left(5\right)-180\ln \left(2\right)\zeta \left(3\right)\zeta \left(4\right)-72\zeta \left(2\right)\zeta ^2\left(3\right)-\frac{315}{2}\ln ^2\left(2\right)\zeta \left(6\right)$$ $$+108\ln ^2\left(2\right)\zeta ^2\left(3\right)+168\ln ^3\left(2\right)\zeta \left(5\right)-72\ln ^3\left(2\right)\zeta \left(2\right)\zeta \left(3\right)-\frac{57}{2}\ln ^4\left(2\right)\zeta \left(4\right)$$ $$+12\ln ^5\left(2\right)\zeta \left(3\right)-2\ln ^6\left(2\right)\zeta \left(2\right)+\frac{1}{8}\ln ^8\left(2\right)$$ Notice how the harmonic series canceled out giving us an actual closed form.

• Nice manipulations. Just to point out that your final integral is the $I_4$ here. – pisco Aug 23 '20 at 3:46