# A tricky integral to evaluate

I came across this integral in some work. So, I would like to ask:

QUESTION. Can you evaluate this integral with proofs? $$\int_0^1\frac{\log x\cdot\log(x+2)}{x+1}\,dx.$$

• I think it is bad form silently to edit your question to ask for proofs, after @CarloBeenakker already gave an answer to the original version that did not request proofs. (That doesn't mean you can't ask for proofs, just that I think it's polite to point out that that is a new request.) – LSpice Mar 1 at 15:11
• @Carlo Beenakker and all: I apologize for the edit. I was always under the impression that when you evaluate something, you also give a proof. But, that was not happening earlier, so to clarify I added "with proof". – T. Amdeberhan Mar 1 at 15:13
• @LSpice: I do not think that answers without proofs for such questions belong on MathOverflow in the first place, so editing the question to indicate this implicitly understood convention is not bad form. – Dmitri Pavlov Mar 1 at 17:13
• I'd split the difference between LSpice and Dmitri by saying that answers which are just numbers you got from computer software are okay, in so far as they can certainly be helpful for getting a complete solution, but that asking for a better answer with a true proof is totally reasonable too. – Sam Hopkins Mar 1 at 17:47
• This appears to be a duplicate of this MSE question. – Timothy Budd Mar 2 at 7:42

For $$t\in(0,1]$$, let $$$$I(t):=\int_0^1\frac{\log(x)\,\log(1+t(x+1))}{x+1}\,dx$$$$ (so that the integral in question is $$I(1)$$),
\begin{aligned} J(t)&:=-\text{Li}_3(2+1/t)+\text{Li}_3(-2 t-1)+\text{Li}_3(t+1) \\ &-\text{Li}_3(2t+1) +\text{Li}_2(2+1/t) (\log (2 t+1)+i \pi ) \\ &+\text{Li}_2(-2 t-1)(-\log (2 t+1)-i \pi ) \\ &+\text{Li}_2(t+1) (-\log (t+1)-i \pi )+\text{Li}_2(2 t+1) (\log (2 t+1)+i \pi ) \\ &+\frac{1}{6} (\log ^3(t)+(-3 \log ^2(t+1)-6 i \pi \log (t+1)+4 \pi ^2) \log (t) \\ &+3 \pi (-i \log ^2(t+1)+2 i \log ^2(2 t+1)+2 \pi \log (t+1)-4 \pi \log (2 t+1))) \\ &+\frac{3 \zeta (3)}{4}-\frac{5 i \pi ^3}{12}, \end{aligned} \begin{aligned} I_1(t)&:=6tI'(t)=6t\int_0^1\frac{\log(x)}{1+t(x+1)}\,dx \\ & =6 \text{Li}_2\left(\frac{t+1}{2 t+1}\right)-3 \log ^2(t+1)+3 \log ^2(2 t+1) \\ &+6 \log (t) \log \left(\frac{t+1}{2 t+1}\right)-\pi ^2, \end{aligned} \begin{aligned} J_1(t)&:=6tJ'(t)=6 \text{Li}_2\left(2+1/t\right)+3 \log ^2(t)-3 \log ^2(t+1) \\ &-6 (\log (2 t+1)+i \pi ) \log (t)+6 \log (t+1) \log (2 t+1) \\ &+6 i \pi \log (2 t+1)-2 \pi ^2. \end{aligned} Then $$I'_1=J'_1$$ and $$I_1(0+)=J_1(0+)$$, so that $$I_1=J_1$$, and hence $$I'=J'$$. Also, $$I(0)=I(0+)=0=J(0+)$$, so that $$I=J$$, and the integral in question is \begin{aligned} I(1)&=J(1)=\text{Li}_3(-3)-2 \text{Li}_3(3)+i \pi \left(-\text{Li}_2(-3)+2 \text{Li}_2(3)+\log ^2(3)\right) \\ &-\text{Li}_2(-3) \log (3)+\text{Li}_2(3) \log (9)+\frac{13 \zeta (3)}{8}-\frac{2 i \pi ^3}{3}-\pi ^2 \log (9) \\ &=-0.651114\dots. \end{aligned}

• Educated guessing vs. "backward pioneering"... – Wolfgang Mar 2 at 9:39

$$\int_0^1\frac{\ln x\cdot\ln(x+2)}{x+1}\,dx=$$ $$=\text{Li}_3\left(-\tfrac{1}{3}\right)-2 \,\text{Li}_3\left(\tfrac{1}{3}\right)+\tfrac{1}{2} \ln 3\left[ \text{Li}_2\left(\tfrac{1}{9}\right)-6\, \text{Li}_2\left(\tfrac{1}{3}\right) -\tfrac{2}{3} \ln ^2 3\right]+\tfrac{13}{8} \zeta (3).$$ I checked that this combination of polylog's evaluates to $$-0.651114$$, equal to a numerical evaluation of the integral.

Update: As Timothy Budd pointed out, that this combination of polylog's simplifies to $$-\frac{13}{24}\zeta(3)$$ is proven by Przemo at MSE.
The identities that enable this simplification are $$\text{Li}_3\left(-\tfrac{1}{3}\right)-2 \,\text{Li}_3\left(\tfrac{1}{3}\right) = -\tfrac{1}{6} \ln^3 3 + \tfrac{1}{6}\pi^2 \ln 3 - \tfrac{13}{6} \zeta(3),$$ $$\text{Li}_2(\tfrac{1}{9})=2\,\text{Li}_2(-\tfrac{1}{3})+2\,\text{Li}_2(\tfrac{1}{3}),$$ $$2\text{Li}_2\left(-\tfrac{1}{3}\right)-4 \,\text{Li}_2\left(\tfrac{1}{3}\right) = \ln^2 3 -\tfrac{1}{3}\pi^2 .$$

• Shouldn't you be able to eliminate the imaginary parts? – user174996 Mar 1 at 14:40
• @CarloBeenakker: is there an even simpler answer and proof? – T. Amdeberhan Mar 1 at 14:50
• I computed numerically that the value is $-\frac{13\zeta(3)}{24}$, so it seems. Hope this shades some light into the problem. – T. Amdeberhan Mar 1 at 19:14
• @T.Amdeberhan: On 18 April 2017 you knew already (with proof) that the integral equals $-\frac{13\zeta(3)}{24}$. See your own response at mathoverflow.net/questions/267485/… So it is not clear why you say that "I computed numerically etc." Of course it is legitimate and useful to ask for simpler proofs. – GH from MO Mar 2 at 10:54
• @GHfromMO: Thank you, this is helpful. I completely forgot about this encounter. – T. Amdeberhan Mar 2 at 19:54