How to evaluate $\int_0^1\frac{\ln x\ln(1+x^2)}{1-x^2}dx$ in an elegant way?

How to prove that:

$$\int_0^1\frac{\ln x\ln(1+x^2)}{1-x^2}dx=\frac74\zeta(3)-\frac34\ln2 \zeta(2)-\frac{\pi}{2}G$$

where $$\zeta$$ is the Riemann zeta function and $$G$$ is Catalan constant.

I came across this integral while working on evaluating some harmonic series.

I am tagging "harmonic series" as its pretty related to logarithmic integrals.

• Per the title: is there an inelegant way to prove the identity that you specifically want to avoid? Or are you satisfied by any proof of the identity? – Clayton Oct 16 '19 at 22:40
• of course any proof is fine but " elegant way" is always better. – Ali Shadhar Oct 16 '19 at 23:01
• It might be useful in the future to have the following. Consider: $$X=\int_0^1 \frac{\ln x\ln(1+x^2)}{1+x}dx;\quad Y=\int_0^1 \frac{\ln x\ln(1+x^2)}{1-x}dx$$ $$X+Y=2\int_0^1 \frac{\ln x\ln(1+x^2)}{1-x^2}dx=\frac72 \zeta(3)-\frac32 \zeta(2)\ln 2 -\pi G$$ $$X-Y=-2\int_0^1 \frac{x\ln x (1+x^2)}{1-x^2}dx\overset{x^2=t}=-\frac12 \int_0^1 \frac{\ln x\ln(1+x)}{1-x}dx=\frac34 \zeta(3)\ln 2 -\frac{\zeta(3)}{2}$$ See: tapatalk.com/groups/integralsandseries/… (copy paste the Latex here to read). And we can extract the values of $X,Y$. – Zacky Oct 16 '19 at 23:44
• very nice idea. – Ali Shadhar Oct 16 '19 at 23:52
• @LeBlanc by the way the last integral can be done as follows $$\int_0^1\frac{\ln x\ln(1+x)}{1-x}\ dx=\int_0^1\frac{\ln(1-x)\ln(2-x)}{x}\ dx\\=\ln2\int_0^1\frac{\ln(1-x)}{x}\ dx+\int_0^1\frac{\ln(1-x)\ln(1-x/2)}{x}\ dx\\=-\ln2\zeta(2)-\sum_{n=1}^\infty\frac1{2^n}\int_0^1x^{n-1}\ln(1-x)\ dx\\=-\ln2\zeta(2)+\sum_{n=1}^\infty\frac{H_n}{n^22^n}\\=-\ln2\zeta(2)+\zeta(3)-\frac12\ln2\zeta(2)\\=\zeta(3)-\frac32\ln2\zeta(2)$$ – Ali Shadhar Oct 17 '19 at 1:53

We will start by using the following substitution: $$\frac{1-x}{1+x}=t\Rightarrow x=\frac{1-t}{1+t}\Rightarrow dx=\frac{2}{(1+t)^2}dt$$ $$\Rightarrow I=\int_0^1 \frac{\ln x\ln(1+x^2)}{1-x^2}dx=\frac12\int_0^1 \frac{[\ln(1-t)-\ln(1+t)][\ln2+\ln(1+t^2)-2\ln(1+t)]}{t}dt$$ Now we are going to use the following result to evaluate a part from above: $$\small \int_0^1 \frac{[m\ln(1+x)+n\ln(1-x)][q\ln(1+x)+p\ln(1-x)]}{x}dx=\left(\frac{mq}{4}-\frac{5}{8}(mp+nq)+2np\right)\zeta(3)$$ $$\Rightarrow I=\frac{7}{8}\zeta(3)-\frac34\zeta(2)\ln 2+\frac12{\int_0^1 \frac{[\ln(1-t)-\ln(1+t)]\ln(1+t^2)}{t}dt}$$ The last integral is $$I-J=\frac74\zeta(3)-\pi G$$ which appears in the following post. $$\Rightarrow I =\frac{7}{8}\zeta(3)-\frac34\zeta(2)\ln 2+\frac78\zeta(3)-\frac{\pi}{2}G=\frac74\zeta(3)-\frac34\zeta(2)\ln 2-\frac{\pi}{2}G$$

• nice .. I have a good idea :) – Ali Shadhar Oct 16 '19 at 23:12

\begin{align} J&=\int_0^1 \frac{\ln(1+x^2)\ln x}{1-x^2}\,dx \end{align} On $$[0;1]$$, define the function $$R$$ by, \begin{align} R(x)&=\int_0^x \frac{\ln t}{1-t^2}\,dt\\ &=\int_0^1 \frac{x\ln(tx)}{1-t^2x^2}\,dt \end{align} Observe that, \begin{align}R(0)&=0\\ R(1)&=\int_0^1 \frac{\ln t}{1-t^2}\,dt\\ &=\int_0^1 \frac{\ln t}{1-t}\,dt-\int_0^1 \frac{t\ln t}{1-t^2}\,dt\\ &=\int_0^1 \frac{\ln t}{1-t}\,dt-\frac{1}{4}\int_0^1 \frac{\ln t}{1-t}\,dt\\ &=\frac{3}{4}\int_0^1 \frac{\ln t}{1-t}\,dt\\ &=\frac{3}{4}\times -\frac{\pi^2}{6}\\ &=-\frac{\pi^2}{8}\\ J&=\Big[R(x)\ln(1+x^2)\Big]_0^1-\int_0^1 \frac{2xR(x)}{1+x^2}\,dx\\ &=-\frac{\pi^2}{8}\ln 2-\int_0^1\int_0^1 \frac{2x^2\ln(tx)}{(1+x^2)(1-t^2x^2)}\,dt\,dx\\ &=-\frac{\pi^2}{8}\ln 2-\int_0^1\int_0^1 \frac{2x^2\ln t}{(1+x^2)(1-t^2x^2)}\,dt\,dx-\int_0^1\int_0^1 \frac{2x^2\ln t}{(1+x^2)(1-t^2x^2)}\,dt\,dx\\ &=-\frac{\pi^2}{8}\ln 2-\int_0^1 \left(\frac{x\ln x\ln(1+x)}{1+x^2}-\frac{x\ln x\ln(1-x)}{1+x^2}\right)\,dx-\\ &\int_0^1 \left(\frac{\ln t\ln(1+t)}{t}-\frac{\ln t\ln(1-t)}{t}-\frac{t\ln t\ln(1+t)}{1+t^2}+\frac{t\ln t\ln(1-t)}{1+t^2}-\frac{\frac{\pi}{2}\ln t}{1+t^2}\right)\,dt\\ &=-\frac{\pi^2}{8}\ln 2-\int_0^1 \frac{\ln t\ln(1+t)}{t}\,dt+\int_0^1 \frac{\ln t\ln(1-t)}{t}\,dt+\frac{\pi}{2}\int_0^1 \frac{\ln t}{1+t^2}\,dt\\ &=-\frac{\pi^2}{8}\ln 2-\frac{\pi}{2}\text{G}-\frac{1}{2}\Big[\ln^2 t\ln(1+t)\Big]_0^1+\frac{1}{2}\int_0^1 \frac{\ln^2 t}{1+t}\,dt+\\ &\frac{1}{2}\Big[\ln^2 t\ln(1-t)\Big]_0^1+\frac{1}{2}\int_0^1 \frac{\ln^2 t}{1-t}\,dt\\ &=-\frac{\pi^2}{8}\ln 2-\frac{\pi}{2}\text{G}+\int_0^1 \frac{\ln^2 t}{1-t^2}\,dt\\ &=-\frac{\pi^2}{8}\ln 2-\frac{\pi}{2}\text{G}+\int_0^1 \frac{\ln^2 t}{1-t}\,dt-\int_0^1 \frac{t\ln^2 t}{1-t^2}\,dt\\ &=-\frac{\pi^2}{8}\ln 2-\frac{\pi}{2}\text{G}+\int_0^1 \frac{\ln^2 t}{1-t}\,dt-\frac{1}{8}\int_0^1 \frac{\ln^2 t}{1-t}\,dt\\ &=-\frac{\pi^2}{8}\ln 2-\frac{\pi}{2}\text{G}+\frac{7}{8}\int_0^1 \frac{\ln^2 t}{1-t}\,dt\\ &=-\frac{\pi^2}{8}\ln 2-\frac{\pi}{2}\text{G}+\frac{7}{8}\times 2\zeta(3)\\ &=\boxed{-\frac{\pi^2}{8}\ln 2-\frac{\pi}{2}\text{G}+\frac{7}{4}\zeta(3)} \end{align} NB: I assume, \begin{align} \int_0^1 \frac{\ln x}{1-x}\,dx&=-\zeta(2)\\ &=-\frac{\pi^2}{6}\\ \int_0^1 \frac{\ln^2 x}{1-x}\,dx&=2\zeta(3) \end{align}

• Very nice approach. Sorry just saw it (+1) – Ali Shadhar Aug 20 '20 at 1:14

@LeBlanc proved here

$$I=\Im\int_0^1 \frac{\operatorname{Li}_2(ix)}{1+x^2}dx=\frac{7}{8}\zeta(3)-\frac{\pi}{4}G\tag1$$

on the other hand and by using $$\operatorname{Li}_2(y)=-\int_0^1\frac{y\ln u}{1-yu}du$$, we can write

\begin{align} I&=-\Im\int_0^1\frac{1}{1+x^2}\left(\int_0^1\frac{ix\ln u}{1-ixu}\ du\right)\ dx\\ &=-\Im\int_0^1\ln u\left(\int_0^1\frac{ix}{(1+x^2)(1-ixu)}\ dx\right)\ du\\ &=-\Im\int_0^1\ln u\left(\frac{i\ln2}{2}\frac{1}{1-u^2}-\frac{i\ln(1-iu)}{1-u^2}+\frac{\pi}{4}\frac{u}{1-u^2}\right)\ du\\ &=\frac12\int_0^1\frac{\ln u\ln(1+u^2)}{1-u^2}du-\frac{\ln2}{2}\underbrace{\int_0^1\frac{\ln u}{1-u^2}du}_{-\frac34\zeta(2)}\tag2 \end{align}

From (1) and (2) we get the closed form of our integral.

Note that what I did in the second last line is I ignored the last term $$\frac{\pi}{4}\frac{u}{1-u^2}$$ as we are interested in only the imaginary parts and I used $$\Re \ln(1-iu)=\ln\sqrt{1+u^2}$$.