Integrals and Miscellaneous 7

$$I=\int_0^1\ln^3t\ln(1-t)dt$$4th derivative of the beta function:\begin{eqnarray}\frac{\partial^4 B}{\partial x^3\partial y}&&=B\{\psi(y)-\psi(x+y)\}\Bigl[\{\psi(x)-\psi(x+y)\}^3+3\{\psi(x)-\psi(x+y)\}\{\psi'(x)-\psi'(x+y)\}\\&&\quad+\{\psi''(x)-\psi''(x+y)\}\Bigr]\\&&\quad-B\Bigl[3\{\psi(x)-\psi(x+y)\}^2\psi'(x+y)+3\{\psi'(x)-\psi'(x+y)\}\psi'(x+y)\\&&\quad+3\{\psi(x)-\psi(x+y)\}\psi''(x+y)+\psi'''(x+y)\Bigr]\end{eqnarray}and we find$$I=\left.\frac{\partial^4 B}{\partial x^3\partial y}\right|_{x=y=1}$$Using $\psi(z+1)=\psi(z)+1/z$, $\psi'(1)=\zeta(2)$, $\psi''(1)=-2\zeta(3)$, $\psi'''(1)=6\zeta(4)$, we get$$\int_0^1\ln^3t\ln(1-t)dt=24-\pi^2-6\zeta(3)-\frac{\pi^4}{15}$$

$$I=\int_0^1\ln^2t\ln^2(1-t)dt$$We can see$$I=\left.\frac{\partial^4 B}{\partial x^2\partial^2 y}\right|_{x=y=1}$$4th derivative of the beta function can be written:\begin{eqnarray}\frac{\partial^4 B}{\partial x^2\partial y^2}&&=B\Bigl[\{\psi(x)-\psi(x+y)\}^2\{\psi(y)-\psi(x+y)\}^2+\{\psi'(x)-\psi'(x+y)\}\{\psi(y)-\psi(x+y)\}^2\\&&\quad+\{\psi(x)-\psi(x+y)\}^2\{\psi'(y)-\psi'(x+y)\}+\{\psi'(x)-\psi'(x+y)\}^2\{\psi'(y)-\psi'(x+y)\}\\&&\quad-2\psi''(x+y)\{\psi(x)+\psi(y)-2\psi(x+y)\}\\&&\quad-4\psi'(x+y)\{\psi(x)-\psi(x+y)\}\{\psi(y)-\psi(x+y)\}\\&&\quad+2\psi'(x+y)^2-\psi'''(x+y)\Bigr]\end{eqnarray}Using $\psi(z+1)=\psi(z)+1/z$, $\psi'(1)=\zeta(2)$, $\psi''(1)=-2\zeta(3)$, $\psi'''(1)=6\zeta(4)$, we get$$\int_0^1\ln^2t\ln^2(1-t)dt=24-\zeta(4)-8\zeta(3)-8\zeta(2)$$

\begin{eqnarray*}\sum_{n=1}^\infty \frac{H_{n}^2}{2^n}&&=\frac{\pi^2}{6}+\ln^22\\\sum_{n=1}^\infty \frac{H_{n}^2}{\phi^{2n}}&&=\frac{\pi^2}{15}\phi\\\sum_{n=1}^\infty (-1)^n\frac{H_{n}^2}{2^n}&&=\frac{1}{3}\left(\Li_2\left(\frac{1}{4}\right)-\frac{\pi^2}{6}+2\ln^2\frac{3}{2}+\ln^22\right)\\\sum_{n=1}^\infty (-1)^n\frac{H_{n}^2}{n}&&=-\frac{3}{4}\zeta(3)+\frac{\pi^2}{12}\ln2-\frac{\ln^3 2}{3}\\\sum_{n=1}^\infty \frac{H_{n}^2}{2^n n}&&=\frac{7}{8}\zeta(3)\\\sum_{n=1}^\infty \frac{H_{n}^2}{n\phi^{2n}}&&=\frac{4}{5}\zeta(3)-\frac{\pi^2}{15}\ln\phi\\\sum_{n=1}^\infty (-1)^n\frac{H_{n}^2}{2^nn}&&=\frac{1}{4}\Li_3\left(\frac{1}{4}\right)+\frac{\ln\frac{2}{3}}{2}\Li_2\left(\frac{1}{4}\right)-\frac{7}{8}\zeta(3)\\&&\quad+\frac{\pi^2}{12}\ln3-\frac{3}{2}\ln^22\ln3+\ln^23\ln2\\&&\quad-\frac{\ln^33}{3}+\frac{2}{3}\ln^32\end{eqnarray*}

\begin{eqnarray*}\sum_{n=1}^\infty\frac{H_n^2}{n^2}x^n&&=\Li_4(x)-2\Li_4(1-x)+2\ln(1-x)\:\Li_3(1-x)\\&&\quad+\frac{1}{2}\Li_2(x)^2-\ln^2(1-x)\:\Li_2(1-x)-\frac{1}{3}\ln x\ln^3(1-x)+\frac{\pi^4}{45}\end{eqnarray*}

Substituting some values to $x$, we obtain\begin{eqnarray*}\sum_{n=1}^\infty\frac{H_n^2}{n^2}&&=\frac{17}{360}\pi^4\\\sum_{n=1}^\infty\frac{H_{n-1}^2}{n^2}&&=\frac{11}{360}\pi^4\\\sum_{n=1}^\infty\frac{H_n^2}{2^nn^2}&&=-\Li_4\left(\frac{1}{2}\right)+\frac{37\pi^4}{1440}-\frac{7}{4}\zeta(3)\ln2+\frac{\pi^2\ln^22}{24}-\frac{\ln^42}{24}\\\sum_{n=1}^\infty(-1)^n\frac{H_n^2}{n^2}&&=2\Li_4\left(\frac{1}{2}\right)-\frac{41\pi^4}{1440}+\frac{7}{4}\zeta(3)\ln2-\frac{\pi^2}{12}\ln^22+\frac{\ln^42}{12}\end{eqnarray*}

proof here:

調和数を含んだ級数(Euler-sum)とゼータ関数 part4

$$O_n:=1+\frac{1}{3}+\cdots+\frac{1}{2n-1}\quad,\quad O_0=0$$We can get such generating functions as\begin{eqnarray*}\sum_{n=1}^\infty O_n x^{2n-1}&&=\frac{1}{2(1-x^2)}\ln\frac{1+x}{1-x}\\\sum_{n=1}^\infty \frac{O_n}{n} x^{2n}&&=\frac{1}{4}\ln^2\frac{1-x}{1+x}\\\sum_{n=1}^\infty \frac{O_n}{n^2} x^{2n}&&=\frac{7}{4}\zeta(3)+\Li_3\left(\frac{x-1}{x+1}\right)-\Li_3\left(\frac{1-x}{1+x}\right)\\&&\quad +\ln\frac{1-x}{1+x}\left[\Li_2\left(\frac{1-x}{1+x}\right)-\Li_2\left(\frac{x-1}{x+1}\right)\right]\\&&\quad+\frac{1}{2}\ln x\ln^2\frac{1-x}{1+x}\end{eqnarray*}Substituting some values to $x$, we obtain\begin{eqnarray*}\sum_{n=1}^\infty\frac{O_n}{4^n}&&=\frac{\ln3}{3}\\\sum_{n=1}^\infty\frac{O_n}{2^n}&&=\sqrt{2}\ln(\sqrt{2}+1)\\\sum_{n=1}^\infty(-1)^n\frac{O_n}{2^n}&&=-\frac{\sqrt{2}}{3}\arctan\frac{1}{\sqrt{2}}\\\sum_{n=1}^\infty\frac{O_n}{n4^n}&&=\frac{\ln^2 3}{4}\\\sum_{n=1}^\infty\frac{O_n}{n2^n}&&=\ln^2(\sqrt{2}-1)\\\sum_{n=1}^\infty(-1)^n\frac{O_n}{n}&&=-\frac{\pi^2}{16}\\\sum_{n=1}^\infty(-1)^n\frac{O_n}{n3^n}&&=-\frac{\pi^2}{36}\\\sum_{n=1}^\infty\frac{O_n}{n^2}&&=\frac{7}{4}\zeta(3)\\\sum_{n=1}^\infty\frac{O_n}{9^nn^2}&&=\frac{1}{4}\Li_3\left(\frac{1}{4}\right)+\frac{\ln2}{2}\Li_2\left(\frac{1}{4}\right)\\&&\quad+\frac{2}{3}\ln^32-\frac{1}{2}\ln^22\ln3\\\sum_{n=1}^\infty(-1)^n\frac{O_n}{n^2}&&=\frac{7}{4}\zeta(3)-\pi G\\\sum_{n=1}^\infty(-1)^n\frac{O_n}{3^n n^2}&&=\frac{35}{36}\zeta(3)+\frac{\pi^2}{36}\ln3+\frac{\sqrt{3}}{27}\pi^3-\frac{\pi\sqrt{3}}{54}\psi'\left(\frac{1}{6}\right)\end{eqnarray*}

Proof here:

調和数を含んだ級数(Euler-sum)とゼータ関数 part5

$$\Li_3(1+i)+\Li_3(1-i)=\frac{35}{32}\zeta(3)+\frac{\pi^2}{16}\ln2$$

PROOF.\begin{eqnarray*}\Li_3(i)&&=\sum_{n=1}^\infty\frac{i^n}{n^3}\\&&=\sum_{n=1}^\infty\frac{(-1)^n}{(2n)^3}+i\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^3}\\&&=\frac{1}{8}\Li_3(-1)+i\beta(3)\\&&=-\frac{3}{32}\zeta(3)+\frac{\pi^3i}{32}\end{eqnarray*}Recall the trilogarithm identity(*)\begin{eqnarray}\Li_3(x)+&&\Li_3(1-x)+\Li_3\left(1-\frac{1}{x}\right)\\&&=\zeta(3)+\frac{\pi^2}{6}\ln x-\frac{\ln^2x\ln(1-x)}{2}+\frac{\ln^3x}{6}\end{eqnarray}Setting $x=i$, we obtain$$\Li_3(1+i)+\Li_3(1-i)=\frac{35}{32}\zeta(3)+\frac{\pi^2}{16}\ln2$$

$$\Li_2(1\pm i)=\frac{\pi^2}{16}\pm i\left(\frac{\pi}{4}\ln2+G\right)$$PROOF.

In the same way as 2022/7/5B, we get$$\Li_2(\pm i)=-\frac{\pi^2}{48}\pm iG$$Recall the reflection formula$$\Li_2(x)+\Li_2(1-x)=\frac{\pi^2}{6}-\ln x\ln(1-x)$$Setting $x=\pm i$, we find$$\Li_2(1\pm i)=\frac{\pi^2}{16}\pm i\left(\frac{\pi}{4}\ln2+G\right)$$

$$\Li_3(e^{-\frac{\pi}{3}i})-\Li_3(-e^{-\frac{\pi}{3}i})=\frac{7}{9}\zeta(3)-\frac{\pi^3}{18}i$$PROOF.

From the series representation $\Li_3(z)=\sum_{n=1}^\infty n^{-3}z^n$, we see\begin{eqnarray*}\Li_3(e^{-\frac{\pi}{3}i})-\Li_3(-e^{-\frac{\pi}{3}i})&&=2\sum_{n=0}^\infty\frac{e^{-\frac{\pi}{3}i(2n+1)}}{(2n+1)^3}\\&&=2e^{-\frac{\pi}{3}i}\sum_{n=0}^\infty\frac{1}{(6n+1)^3}-2\sum_{n=0}^\infty\frac{1}{27(2n+1)^3}+2e^{-\frac{5\pi}{3}i}\sum_{n=0}^\infty\frac{1}{(6n+5)^3}\end{eqnarray*}Recall two equations$$\sum_{n=0}^\infty\frac{1}{(6n+1)^3}=\frac{91}{216}\zeta(3)+\frac{\sqrt{3}\pi^3}{108}$$$$\sum_{n=0}^\infty\frac{1}{(6n+5)^3}=\frac{91}{216}\zeta(3)-\frac{\sqrt{3}\pi^3}{108}$$,which we obtained in the previous article "2022/5/3 C" in

Integrals and Miscellaneous 4

Hence,$$\Li_3(e^{-\frac{\pi}{3}i})-\Li_3(-e^{-\frac{\pi}{3}i})=\frac{7}{9}\zeta(3)-\frac{\pi^3}{18}i$$

\begin{eqnarray*}\int_0^x\frac{\ln^2(1-t^2)}{t}dt&&=-\Li_3(1-x^2)+\ln (1-x^2)\Li_2(1-x^2)\\&&\quad+\ln x\ln^2(1-x^2)+\zeta(3)\end{eqnarray*}PROOF.

Recall the equation\begin{eqnarray*}\int_0^x\frac{\ln^2(1-t)}{t}dt&&=-2\Li_3(1-x)+2\ln (1-x)\Li_2(1-x)\\&&\quad+\ln x\ln^2(1-x)+2\zeta(3)\end{eqnarray*}, which we obtained in "2022/6/18 B" here:

Integrals and Miscellaneous 6

Setting $x\to x^2$, we have\begin{eqnarray*}\int_0^{x^2}\frac{\ln^2(1-t)}{t}dt&&=-2\Li_3(1-x^2)+2\ln (1-x^2)\Li_2(1-x^2)\\&&\quad+2\ln x\ln^2(1-x^2)+2\zeta(3)\end{eqnarray*}Substituting $t=u^2$ in the integral,$$\int_0^{x^2}\frac{\ln^2(1-t)}{t}dt=2\int_0^{x}\frac{\ln^2(1-u^2)}{u}du$$Hence,\begin{eqnarray*}\int_0^x\frac{\ln^2(1-t^2)}{t}dt&&=-\Li_3(1-x^2)+\ln (1-x^2)\Li_2(1-x^2)\\&&\quad+\ln x\ln^2(1-x^2)+\zeta(3)\end{eqnarray*}

$0\le x\le 1$,\begin{eqnarray*}\int_0^x\frac{\ln(1+t)\ln(1-t)}{t}dt&&=-\frac{1}{2}\Li_3(1-x^2)+\Li_3(1-x)+\Li_3\left(\frac{1}{1+x}\right)\\&&+\frac{1}{2}\ln(1-x^2)\Li_2(1-x^2)-\ln(1-x)\Li_2(1-x)+\ln(1+x)\Li_2\left(\frac{1}{1+x}\right)\\&&+\ln x\ln(1-x)\ln(1+x)+\frac{1}{3}\ln^3(1+x)-\frac{3}{2}\zeta(3)\end{eqnarray*}PROOF.

\begin{equation}\int_0^x\frac{\ln^2(1-t^2)}{t}dt=\int_0^x\frac{\ln^2(1+t)}{t}dt+\int_0^x\frac{\ln^2(1-t)}{t}dt+2\int_0^x\frac{\ln(1+t)\ln(1-t)}{t}dt\tag{1}\end{equation}We have already obtained the value of LHS in "2022/7/10A", namely\begin{eqnarray}\int_0^x\frac{\ln^2(1-t^2)}{t}dt&&=-\Li_3(1-x^2)+\ln (1-x^2)\Li_2(1-x^2)\\&&\quad+\ln x\ln^2(1-x^2)+\zeta(3)\tag{2}\end{eqnarray}Besides, recall two integral values:\begin{eqnarray}\int_0^x\frac{\ln^2(1-t)}{t}dt&&=-2\Li_3(1-x)+2\ln (1-x)\Li_2(1-x)\\&&\quad+\ln x\ln^2(1-x)+2\zeta(3)\tag{3}\end{eqnarray}\begin{eqnarray}\int_0^x\frac{\ln^2(1+t)}{t}dt&&=-2\Li_3\left(\frac{1}{1+x}\right)-2\ln(1+x)\Li_2\left(\frac{1}{1+x}\right)\\&&\quad+\ln x\ln^2(1+x)-\frac{2}{3}\ln^3(1+x)+2\zeta(3)\tag{4}\end{eqnarray}, which we got in "2022/6/18B" and "2022/6/18C" here:

Integrals and Miscellaneous 6

Therefore, with the help of (1)(2)(3)(4),\begin{eqnarray*}\int_0^x\frac{\ln(1+t)\ln(1-t)}{t}dt&&=-\frac{1}{2}\Li_3(1-x^2)+\Li_3(1-x)+\Li_3\left(\frac{1}{1+x}\right)\\&&+\frac{1}{2}\ln(1-x^2)\Li_2(1-x^2)-\ln(1-x)\Li_2(1-x)+\ln(1+x)\Li_2\left(\frac{1}{1+x}\right)\\&&+\ln x\ln(1-x)\ln(1+x)+\frac{1}{3}\ln^3(1+x)-\frac{3}{2}\zeta(3)\end{eqnarray*}

$$\int_0^1\frac{\ln(1+t)\ln(1-t)}{t}dt=-\frac{5}{8}\zeta(3)$$PROOF.

See the equation in "2022/7/10B" above and set $x=1$. You can use the identities of $\Li_3(1/2)$ , $\Li_2(1/2)$ here:

多重対数関数（ポリログ）の関係式一覧・証明付き

From @infseriesbot

$$I:=\int_0^1\frac{\ln(1+x)\ln(1+x^2)}{x}dx=\frac{\pi}{2}G-\frac{33}{32}\zeta(3)$$PROOF.
Expand $\ln(1+x^2)$ to get\begin{eqnarray*}I = -\sum_{n=1}^\infty\frac{(-1)^n}{n}\int_0^1 x^{2n-1}\ln(1+x)dx \end{eqnarray*}Integrating by parts gives\begin{eqnarray*}&=&-\frac{\ln2}{2}\sum_{n=1}^\infty\frac{(-1)^n}{n^2}+\frac{1}{2}\sum_{n=1}^\infty\frac{(-1)^n}{n^2}\int_0^1\frac{x^{2n}}{1+x}dx\\&=&-\frac{\ln2}{2}\sum_{n=1}^\infty\frac{(-1)^n}{n^2}+\frac{1}{2}\sum_{n=1}^\infty\frac{(-1)^n}{n^2}\int_0^1\left(x^{2n-1}-x^{2n-2}+\cdots +x-1+\frac{1}{x+1}\right)dx \\&=&-\frac{\ln2}{2}\sum_{n=1}^\infty\frac{(-1)^n}{n^2}+\frac{1}{2}\sum_{n=1}^\infty\frac{(-1)^n}{n^2}\left(\ln2+\frac{H_n}{2}-O_n\right)\\&=& \frac{1}{4}\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^2}-\frac{1}{2}\sum_{n=1}^\infty\frac{(-1)^nO_n}{n^2}\end{eqnarray*}where $O_n$ denotes odd harmonic number $1+\frac{1}{3}\cdots+\frac{1}{2n-1}$. We obtained these two series in closed-form here and here, namely$$\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^2}=-\frac{5}{8}\zeta(3)\;,\;\sum_{n=1}^\infty(-1)^n\frac{O_n}{n^2}=\frac{7}{4}\zeta(3)-\pi G$$Hence,$$\int_0^1\frac{\ln(1+x)\ln(1+x^2)}{x}dx=\frac{\pi}{2}G-\frac{33}{32}\zeta(3)$$

$$\int \ln^2\frac{1-x}{1+x}\frac{dx}{x}=?$$

Substituting $t=\frac{1-x}{1+x}$, \begin{eqnarray*}LHS&&=-2\int\frac{\ln^2 t}{(1-t)(1+t)}dt\\&&=-\int\frac{\ln^2t}{1-t}dt-\int\frac{\ln^2 t}{1+t}dt\end{eqnarray*}Integrating $\Li_3(x)$ and $\Li_3(-x)$ by parts$$\Li_3(x)=\ln x\Li_2(x)+\frac{1}{2}\ln^2 x\ln(1-x)+\frac{1}{2}\int_0^x\frac{\ln^2 t}{1-t}dt$$$$\Li_3(-x)=\ln x\Li_2(-x)+\frac{1}{2}\ln^2 x\ln(1+x)-\frac{1}{2}\int_0^x\frac{\ln^2 t}{1+t}dt$$Therefore,\begin{eqnarray*}\int \ln^2\frac{1-x}{1+x}\frac{dx}{x}&&=-2\left[\Li_3(t)-\Li_3(-t)\right]+2\ln t\left[\Li_2(t)-\Li_2(-t)\right]+\ln^2t\ln\frac{1-t}{1+t}\end{eqnarray*}Hence,\begin{eqnarray*}\int \ln^2\frac{1-x}{1+x}\frac{dx}{x}&&=-2\left[\Li_3\left(\frac{1-x}{1+x}\right)-\Li_3\left(\frac{x-1}{x+1}\right)\right]\\&&\quad+2\ln \frac{1-x}{1+x}\left[\Li_2\left(\frac{1-x}{1+x}\right)-\Li_2\left(\frac{x-1}{x+1}\right)\right]+\ln^2\frac{1-x}{1+x}\ln x\\&&\quad +\mathrm{const.}\end{eqnarray*}

$$\int \ln^3\frac{1-x}{1+x}\frac{dx}{x}=?$$

Substituting $t=\frac{1-x}{1+x}$, \begin{eqnarray*}LHS&&=-2\int\frac{\ln^3 t}{(1-t)(1+t)}dt\\&&=-\int\frac{\ln^3t}{1-t}dt-\int\frac{\ln^3 t}{1+t}dt\end{eqnarray*}Integrating $\Li_4(x)$ and $\Li_4(-x)$ by parts gives (See also "2022/7/13A" above),\begin{eqnarray*}\int\frac{\ln^3x}{1-x}dx&&=-6\Li_4(x)+6\ln x\Li_3(x)+3\ln^2x\Li_2(x)+\ln^3x\ln(1-x)\\\int\frac{\ln^3x}{1+x}dx&&=6\Li_4(-x)-6\ln x\Li_3(-x)+3\ln^2x\Li_2(-x)+\ln^3x\ln(1+x)\end{eqnarray*}Hence,\begin{eqnarray*}\int \ln^3\frac{1-x}{1+x}\frac{dx}{x}&&=6\left[\Li_4\left(\frac{1-x}{1+x}\right)-\Li_4\left(\frac{x-1}{x+1}\right)\right]\\&&\quad-6\ln\frac{1-x}{1+x}\left[\Li_3\left(\frac{1-x}{1+x}\right)-\Li_3\left(\frac{x-1}{x+1}\right)\right]\\&&\quad+3\ln^2\frac{1-x}{1+x}\left[\Li_2\left(\frac{1-x}{1+x}\right)-\Li_2\left(\frac{x-1}{x+1}\right)\right]\\&&\quad+\ln^3\frac{1-x}{1+x}\ln x+\mathrm{const.}\end{eqnarray*}

$$O_n:=1+\frac{1}{3}+\cdots+\frac{1}{2n-1}\quad,\quad O_0=0$$\begin{eqnarray*}\sum_{n=1}^\infty \frac{O_n^2}{\phi^{2n}} &&=-\frac{3}{2}\ln^2\phi+3\ln2\ln\phi-\Li_2(-\phi^{-3})\\\sum_{n=1}^\infty \frac{O_n^2}{4^nn}&&=\frac{1}{4}\Li_3\left(\frac{1}{9}\right)-\frac{1}{4}\zeta(3)-\frac{\ln 3}{4}\Li_2\left(\frac{1}{4}\right)\\&&\quad+\ln3\Li_2\left(\frac{1}{3}\right)+\frac{1}{2}\ln2\ln3\ln\frac{3}{2}+\frac{\ln^33}{12}\end{eqnarray*}Proof here:

調和数を含んだ級数(Euler-sum)とゼータ関数 part6