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I am trying to evaluate $$I = \int_0^1 \frac{x \ln(1-x) \ln(x) \ln(1+x)}{1+x} \, dx$$

By expanding $\ln(1-x)$ and $\ln(1+x)$, I arrived at the double sum:

$$I = -\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{(-1)^{m+1}}{m n} \int_0^1 \frac{x^{m+n+1} \ln(x)}{1+x} \, dx.$$

I could compute the inside integral, and thus obtained:

$$\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{(-1)^{1+m}}{m} \frac{-1}{4n} \left(-\psi^{(1)}\left(\frac{2+m+n}{2}\right) + \psi^{(1)}\left(\frac{3+m+n}{2}\right)\right) $$

where $\psi$ denotes the polylogarithm.

Could not proceed from here. Any help is much appreciated.

mattTheMathLearner
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    Do you mean polygamma function for $\psi$? – David H Jan 01 '25 at 21:37
  • As David H says, $\psi^{(1)}(z)$ denotes a polygamma function, specifically the trigamma function (the derivative of the digamma function, or the second derivative of $\log{\Gamma(z)}$). Polylogarithms are typically denoted and defined by $\operatorname{Li}s(z):=\sum{n=1}^{\infty}\frac{z^n}{n^s}$, with $s\in\mathbb{C}$ and $|z|<1$ ($|z|$=1 requires $\operatorname{Re}(s)>1$) and analytic continuation allowing evaluation at other values. Your expression aligns with the trigamma function. – teadawg1337 Jan 01 '25 at 23:41
  • I'm able to reproduce your third equation from the second using $4.251.5$ and $4.251.6$ from Gradshteyn and Ryzhik (denoting even and odd exponents of the integral in the second equation, respectively), and it is indeed the trigamma function. However, you've made an error between the first and second steps: $\left(\sum_{m=1}^{\infty}a_mx^m\right)\cdot\left(\sum_{n=1}^{\infty}b_nx^n\right)\neq\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}a_mb_nx^{m+n}$. Multiplication of infinite series is more complicated; see here. – teadawg1337 Jan 02 '25 at 01:31

2 Answers2

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\begin{align} I = &\int_0^1 \frac{x \ln(1-x) \ln x \ln(1+x)}{1+x} \, dx\\ =& \int_0^1\underset{=I_1}{\ln(1-x) \ln x \ln(1+x) } dx - \frac12 \int_0^1 \ln(x) \ln(1+x)\overset{ibp}d[\ln^2(1+x)]\\ =&\ I_1 +\frac12\int_0^1 \underset{=I_2}{\frac{\ln(1-x) \ln^2(1+x)}{x} } dx -\frac12\int_0^1 \underset{=I_3}{\frac{\ln x \ln^2(1+x)}{1-x} } dx\\ \end{align} where $I_2= -\frac{\pi^4}{240}$ $$I_1= -6 + 4 \ln 2 - \ln^2 2 + \frac{5}{2} \zeta(2) - 3\zeta(2) \ln 2 + \frac{21}{8} \zeta(3) $$

To evaluate $I_3$, note that \begin{align}\int_0^1 &\frac{\ln^3\frac{2x}{1+x}}{1-x}{ dx} \overset{t= \frac{2x}{1+x}}=\int_0^1 \frac{\ln^3t}{1-t}dt -\int_0^1 \frac{\frac12\ln^3t}{1-\frac12t}dt =-\frac{\pi^4}{15}+6\text{Li}_4(\frac12)\tag1\\ \int_0^1 &\frac{\ln^3\frac{2x}{1+x}}{1-x}{ dx}=\int_0^1 \frac{\ln^3x +3\ln^2x\ln\frac2{1+x}+ 3\ln x\ln^2\frac2{1+x}+ \overset{t=\frac{1+x}2}{\ln^3\frac{2}{1+x} }}{1-x} dx\\ &= \int^{1/2}_0 \frac{\ln^3t}{1-t}dt +3\ln2\int_0^1 \frac{\ln^2x}{1-x}dx -3\int_0^1 \frac{\ln^2x\ln(1+x)}{1-x}dx\\ & \ +3\ln^2 2 \int_0^1\frac{\ln x}{1-x}dx -6\ln2 \int_0^1 \frac{\ln x\ln(1+x)}{1-x}dx+3I_3\tag2 \end{align} The resulting integrals above are relatively easy to compute, except

\begin{align} \int_0^1 &\frac{\ln^2x \ln (1+x)}{1-x}dx =\int_0^1 \frac{\ln^2x}{1-x}\left(\int_0^1 \frac {x}{1+x t}dt\right) dx\\ =&\int_0^1 \frac{1}{1+t}\left(2Li_3(1)-\int_0^1 \frac {\ln^2x}{1+tx}dx \right) dt\\ =&\>2 \int_0^1\frac{Li_3(-t)}{t}dt - 2 \int_0^1\frac{Li_3(-t)-Li_3(1)}{1+t}\>\overset{ibp}{dt}\\ = &\>2Li_4(-1)-2\ln2 Li_3(-1) - Li_2^2(-1)+2\ln2Li_3(1)\\= &\>\frac72\ln2\zeta(3)-\frac{19\pi^4}{720}\\ \end{align} As a result, (1) and (2) leads to $$I_3= 4\text{Li}_4(\frac12)+\frac{21}4\ln2\zeta(3)-\frac{7\pi^4}{144}-\frac{5\pi^2}{12}\ln^22+\frac16\ln^42 $$ Substitute $I_1$, $I_2$ and $I_3$ into $I$ to obtain $$I= -2\text{Li}_4(\frac12)-\frac{21}8\zeta(3)(1-\ln2)+\frac{\pi^2}2\left( \frac56-\ln2 +\frac5{12}\ln^22\right)\\ \>\>\>\>\>\>\>\>\> +\frac{\pi^4}{45} +4\ln2 -\ln^22-\frac1{12}\ln^42-6 $$

Quanto
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    Great answer! Your work is all correct, but the final answer's second term should be $+\frac{21}8\zeta(3)(1-\ln2)$ instead of $-\frac{21}8\zeta(3)(1-\ln2)$. When it's negative, it's off from the true value of $I$ by about $1.94$. – teadawg1337 Jan 02 '25 at 06:07
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Not a full answer, but this is what it is if we go for general method. $$f(a,b,c)=\int_{0}^{1}t^{a-1}(1-t)^{b-1}(1+t)^{c-1}dt=\int_{0}^{1}t^{a-1}(1-t)^{b-1}\sum_{k=0}^{\infty}\frac{\Gamma(c)}{\Gamma(k+1)\Gamma(c-k)}t^kdt$$$$=\sum_{k=0}^{\infty}\frac{\Gamma(c)}{\Gamma(k+1)\Gamma(c-k)}\int_{0}^{1}t^{a+k-1}(1-t)^{b-1}dt=\sum_{k=0}^{\infty}\frac{\Gamma(a+k)\Gamma(b)\Gamma(c)}{\Gamma(a+b+k)\Gamma(k+1)\Gamma(c-k)}$$
By algebra manipulation: $$\int_{0}^{1}\frac{x\ln(x)\ln(1-x)\ln(1+x)}{1+x}dt=\frac{{\partial}^3}{\partial a\ \partial b\ \partial c}f(a,b,c)\Big\vert_{(a,b,c)=(2,1,0)}\tag{1}$$ $$=\sum_{k=0}^{\infty}\left(\frac{{\partial}^2}{\partial a\ \partial b}\left(\frac{\Gamma(a+k)\Gamma(b)}{\Gamma(a+b+k)}\right)\Big\vert_{(a,b)=(2,1)}\right)\frac{\partial}{\partial c}\left(\frac{\Gamma(c)}{\Gamma(k+1)\Gamma(c-k)}\right)\Big\vert_{c=0}\tag{2}$$ $$\sum_{k=1}^{\infty}\left(\frac{\gamma+\psi^{(0)}(k+3)}{k+2}-\psi^{(1)}(k+3)\right)\frac{(-1)^{k+1}H_{k}}{k+2}\tag{3}$$ $$\sum_{k=1}^{\infty}\sum_{n=0}^{\infty}\left(\frac{1}{k+2}\left(\frac{1}{n+1}-\frac{1}{n+k+3}\right)-\frac{1}{(n+k+3)^2}\right)\frac{(-1)^{k+1}H_{k}}{k+2}\tag{4}$$ $$=\sum_{n=0}^{\infty}\frac{1}{n+1}\sum_{k=1}^{\infty}\frac{(-1)^{k+1}H_{k}}{(n+k+3)^2}\tag{5}$$ I stuck at this point because I don't know the closed form of $\sum_{k=1}^{\infty}\frac{(-1)^{k+1}H_{k}}{(n+k+3)^2}$, but whatever it is, it will lead to the answer given by Quanto.

Quý Nhân
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