# The quadratic and cubic versions of a tough intregral

In this post, Proving that $\int_0^1 \frac{\log \left(\frac{1}{t}\right) \log (t+2)}{t+1} \, dt=\frac{13}{24} \zeta (3)$, it's proved that $$I_1=\int_0^1 \frac{\log \left(\frac{1}{t}\right) \log (t+2)}{t+1} \, dt=\frac{13}{24} \zeta (3)$$ but then some natural questions arise. Might we possibly hope to find nice closed forms for the following integrals too?

$$I_2=\int_0^1 \frac{\log \left(\frac{1}{t}\right) \log^2 (t+2)}{t+1} \, dt$$ $$I_3=\int_0^1 \frac{\log \left(\frac{1}{t}\right) \log^3 (t+2)}{t+1} \, dt$$

What tools, strategies would you like to propose?

UPDATE: According to David H's comment we have that

$$I_2=\int_0^1 \frac{\log \left(\frac{1}{t}\right) \log^2 (t+2)}{t+1} \, dt$$ $$=\operatorname{Li}_{4}{\left(\frac13\right)}-\frac12\,\operatorname{Li}_{4}{‌​\left(-\frac13\right)}+\frac{13}{12}\ln{(3)}\zeta{(3)}-\frac14\ln^{2}{(3)}\zeta{(‌​2)}+\frac{1}{48}\ln^{4}{(3)}-\frac78\,\zeta{(4)}$$ that is also confirmed numerically by Mathematica.

• For the quadratic version, I find $I_2=\operatorname{Li}_{4}{\left(\frac13\right)}-\frac12\,\operatorname{Li}_{4}{\left(-\frac13\right)}+\frac{13}{12}\ln{(3)}\zeta{(3)}-\frac14\ln^{2}{(3)}\zeta{(2)}+\frac{1}{48}\ln^{4}{(3)}-\frac78\,\zeta{(4)}$, though I suspect it can be simplified further. Jul 2, 2015 at 12:46
• @DavidH actually, yes, it works! Jul 2, 2015 at 13:04

Here is my evaluation of the quadratic case. The starting point for the derivation is the algebraic identity

$$6ab^2=\left(a+b\right)^3+\left(a-b\right)^3-2a^3.$$

Then,

\begin{align} \mathcal{I}_{2} &=\int_{0}^{1}\frac{\ln{\left(\frac{1}{t}\right)}\ln^{2}{\left(t+2\right)}}{t+1}\,\mathrm{d}t\\ &=\small{\int_{0}^{1}\frac{\frac16\left(\ln{\left(\frac{1}{t}\right)}+\ln{\left(t+2\right)}\right)^3+\frac16\left(\ln{\left(\frac{1}{t}\right)}-\ln{\left(t+2\right)}\right)^3-\frac13\ln^{3}{\left(\frac{1}{t}\right)}}{t+1}\,\mathrm{d}t}\\ &=\int_{0}^{1}\frac{\frac16\ln^{3}{\left(\frac{t+2}{t}\right)}+\frac16-\ln^{3}{\left(t(t+2)\right)}+\frac13\ln^{3}{\left(t\right)}}{t+1}\,\mathrm{d}t\\ &=\frac16\int_{0}^{1}\frac{\ln^{3}{\left(\frac{t+2}{t}\right)}}{t+1}\,\mathrm{d}t-\frac16\int_{0}^{1}\frac{\ln^{3}{\left(t(t+2)\right)}}{t+1}\,\mathrm{d}t\\ &~~~~~+\frac13\int_{0}^{1}\frac{\ln^{3}{\left(t\right)}}{t+1}\,\mathrm{d}t\\ &=\frac16\int_{0}^{1}\frac{\ln^{3}{\left(\frac{t+2}{t}\right)}}{t+1}\,\mathrm{d}t-\frac16\int_{0}^{1}\frac{\ln^{3}{\left((t+1)^2-1\right)}}{t+1}\,\mathrm{d}t\\ &~~~~~+\frac13\int_{0}^{1}\frac{\ln^{3}{\left(t\right)}}{t+1}\,\mathrm{d}t\\ &=\frac16\int_{0}^{1}\frac{\ln^{3}{\left(\frac{t+2}{t}\right)}}{t+1}\,\mathrm{d}t-\frac16\int_{0}^{1}\frac{\ln^{3}{\left((t+1)^2-1\right)}}{t+1}\,\mathrm{d}t\\ &~~~~~+\frac13\int_{0}^{1}\frac{\ln^{3}{\left(t\right)}}{t+1}\,\mathrm{d}t\\ &=\frac16\int_{\frac12}^{1}\frac{\ln^{3}{\left(\frac{1+y}{1-y}\right)}}{y}\,\mathrm{d}y-\frac16\int_{\frac12}^{1}\frac{\ln^{3}{\left(\frac{1}{y^2}-1\right)}}{y}\,\mathrm{d}y\\ &~~~~~+\frac13\int_{0}^{1}\frac{\ln^{3}{\left(t\right)}}{t+1}\,\mathrm{d}t;~~~\small{\left[\frac{1}{1+t}=y\right]}\\ &=-\frac16\int_{\frac12}^{1}\frac{\ln^{3}{\left(\frac{1-y}{1+y}\right)}}{y}\,\mathrm{d}y-\frac{1}{12}\int_{0}^{3}\frac{\ln^{3}{\left(x\right)}}{1+x}\,\mathrm{d}x\\ &~~~~~+\frac13\int_{0}^{1}\frac{\ln^{3}{\left(t\right)}}{t+1}\,\mathrm{d}t;~~~\small{\left[y=\frac{1}{\sqrt{1+x}}\right]}\\ &=-\frac13\int_{0}^{\frac13}\frac{\ln^{3}{\left(x\right)}}{1-x^2}\,\mathrm{d}x-\frac{1}{12}\int_{0}^{3}\frac{\ln^{3}{\left(x\right)}}{1+x}\,\mathrm{d}x\\ &~~~~~+\frac13\int_{0}^{1}\frac{\ln^{3}{\left(t\right)}}{t+1}\,\mathrm{d}t;~~~\small{\left[\frac{1-y}{1+y}=x\right]}\\ &=-\frac16\int_{0}^{\frac13}\frac{\ln^{3}{\left(x\right)}}{1-x}\,\mathrm{d}x-\frac16\int_{0}^{\frac13}\frac{\ln^{3}{\left(x\right)}}{1+x}\,\mathrm{d}x\\ &~~~~~\small{-\frac{1}{12}\int_{0}^{1}\frac{\ln^{3}{\left(x\right)}}{1+x}\,\mathrm{d}x-\frac{1}{12}\int_{1}^{3}\frac{\ln^{3}{\left(x\right)}}{1+x}\,\mathrm{d}x+\frac13\int_{0}^{1}\frac{\ln^{3}{\left(t\right)}}{t+1}\,\mathrm{d}t}\\ &=-\frac16\int_{0}^{\frac13}\frac{\ln^{3}{\left(x\right)}}{1-x}\,\mathrm{d}x-\frac16\int_{0}^{\frac13}\frac{\ln^{3}{\left(x\right)}}{1+x}\,\mathrm{d}x\\ &~~~~~-\frac{1}{12}\int_{1}^{3}\frac{\ln^{3}{\left(x\right)}}{1+x}\,\mathrm{d}x+\frac14\int_{0}^{1}\frac{\ln^{3}{\left(t\right)}}{t+1}\,\mathrm{d}t\\ &=-\frac16\int_{0}^{\frac13}\frac{\ln^{3}{\left(x\right)}}{1-x}\,\mathrm{d}x-\frac16\int_{0}^{\frac13}\frac{\ln^{3}{\left(x\right)}}{1+x}\,\mathrm{d}x\\ &~~~~~+\frac{1}{12}\int_{\frac13}^{1}\frac{\ln^{3}{\left(w\right)}}{w\left(1+w\right)}\,\mathrm{d}w+\frac14\int_{0}^{1}\frac{\ln^{3}{\left(t\right)}}{t+1}\,\mathrm{d}t;~~~\small{\left[\frac{1}{x}=w\right]}\\ &=\small{-\frac16\int_{0}^{\frac13}\frac{\ln^{3}{\left(x\right)}}{1-x}\,\mathrm{d}x-\frac{1}{12}\int_{0}^{\frac13}\frac{\ln^{3}{\left(x\right)}}{1+x}\,\mathrm{d}x-\frac{1}{12}\int_{0}^{\frac13}\frac{\ln^{3}{\left(x\right)}}{1+x}\,\mathrm{d}x}\\ &~~~~~\small{+\frac{1}{12}\int_{\frac13}^{1}\frac{\ln^{3}{\left(w\right)}}{w}\,\mathrm{d}w-\frac{1}{12}\int_{\frac13}^{1}\frac{\ln^{3}{\left(w\right)}}{1+w}\,\mathrm{d}w+\frac14\int_{0}^{1}\frac{\ln^{3}{\left(t\right)}}{t+1}\,\mathrm{d}t}\\ &=-\frac16\int_{0}^{\frac13}\frac{\ln^{3}{\left(x\right)}}{1-x}\,\mathrm{d}x-\frac{1}{12}\int_{0}^{\frac13}\frac{\ln^{3}{\left(x\right)}}{1+x}\,\mathrm{d}x\\ &~~~~~-\frac{1}{48}\ln^{4}{(3)}+\frac16\int_{0}^{1}\frac{\ln^{3}{\left(t\right)}}{t+1}\,\mathrm{d}t\\ &=-\frac{1}{18}\int_{0}^{1}\frac{\ln^{3}{\left(\frac{w}{3}\right)}}{1-\frac13w}\,\mathrm{d}w-\frac{1}{36}\int_{0}^{1}\frac{\ln^{3}{\left(\frac{w}{3}\right)}}{1+\frac13w}\,\mathrm{d}w\\ &~~~~~-\frac{1}{48}\ln^{4}{(3)}+\frac16\int_{0}^{1}\frac{\ln^{3}{\left(t\right)}}{t+1}\,\mathrm{d}t;~~~\small{\left[3x=w\right]}\\ &=\small{-\frac{1}{18}\int_{0}^{1}\frac{\ln^{3}{\left(w\right)}}{1-\frac13w}\,\mathrm{d}w+\frac{\ln{(3)}}{6}\int_{0}^{1}\frac{\ln^{2}{\left(w\right)}}{1-\frac13w}\,\mathrm{d}w-\frac{\ln^{2}{(3)}}{6}\int_{0}^{1}\frac{\ln{\left(w\right)}}{1-\frac13w}\,\mathrm{d}w}\\ &~~~~~\small{+\frac{\ln^{3}{(3)}}{18}\int_{0}^{1}\frac{\mathrm{d}w}{1-\frac13w}-\frac{1}{36}\int_{0}^{1}\frac{\ln^{3}{\left(w\right)}}{1+\frac13w}\,\mathrm{d}w+\frac{\ln{(3)}}{12}\int_{0}^{1}\frac{\ln^{2}{\left(w\right)}}{1+\frac13w}\,\mathrm{d}w}\\ &~~~~~\small{-\frac{\ln^{2}{(3)}}{12}\int_{0}^{1}\frac{\ln{\left(w\right)}}{1+\frac13w}\,\mathrm{d}w+\frac{\ln^{3}{(3)}}{36}\int_{0}^{1}\frac{\mathrm{d}w}{1+\frac13w}-\frac{1}{48}\ln^{4}{(3)}+\frac16\int_{0}^{1}\frac{\ln^{3}{\left(t\right)}}{t+1}\,\mathrm{d}t}.\\ \end{align}

Using the polylogarithmic integral representation,

$$\int_{0}^{1}\frac{\ln^{n}{\left(y\right)}}{1-zy}\,\mathrm{d}y=\frac{(-1)^{n}n!}{z}\,\operatorname{Li}_{n+1}{\left(z\right)};~~~\small{n\in\mathbb{Z}^{+}\land1\ge z\neq0}$$

we have,

\begin{align} \mathcal{I}_{2} &=\small{\operatorname{Li}_{4}{\left(\frac13\right)}-\frac12\,\operatorname{Li}_{4}{\left(-\frac13\right)}+\ln{(3)}\,\operatorname{Li}_{3}{\left(\frac13\right)}-\frac{\ln{(3)}}{2}\,\operatorname{Li}_{3}{\left(-\frac13\right)}}\\ &~~~~~\small{+\frac{\ln^{2}{(3)}}{2}\,\operatorname{Li}_{2}{\left(\frac13\right)}-\frac{\ln^{2}{(3)}}{4}\,\operatorname{Li}_{2}{\left(-\frac13\right)}+\frac{\ln^{4}{(3)}}{16}+\operatorname{Li}_{4}{\left(-1\right)}}\\ &=\small{\operatorname{Li}_{4}{\left(\frac13\right)}-\frac12\,\operatorname{Li}_{4}{\left(-\frac13\right)}+\ln{(3)}\left[\operatorname{Li}_{3}{\left(\frac13\right)}-\frac12\,\operatorname{Li}_{3}{\left(-\frac13\right)}\right]}\\ &~~~~~\small{+\frac{\ln^{2}{(3)}}{2}\left[\operatorname{Li}_{2}{\left(\frac13\right)}-\frac12\,\operatorname{Li}_{2}{\left(-\frac13\right)}\right]+\frac{\ln^{4}{(3)}}{16}-\frac78\,\operatorname{Li}_{4}{\left(1\right)}}\\ &=\operatorname{Li}_{4}{\left(\frac13\right)}-\frac12\,\operatorname{Li}_{4}{\left(-\frac13\right)}\\ &~~~~~+\ln{(3)}\left[\frac{13}{12}\,\zeta{(3)}-\frac12\zeta{(2)}\ln{(3)}+\frac{1}{12}\ln^3{(3)}\right]\\ &~~~~~+\frac{\ln^{2}{(3)}}{2}\left[\frac12\,\zeta{(2)}-\frac14\ln^2{(3)}\right]+\frac{\ln^{4}{(3)}}{16}-\frac78\,\zeta{(4)}\\ &=\operatorname{Li}_{4}{\left(\frac13\right)}-\frac12\,\operatorname{Li}_{4}{\left(-\frac13\right)}+\frac{13}{12}\ln{(3)}\zeta{(3)}\\ &~~~~~-\frac14\ln^{2}{(3)}\zeta{(2)}+\frac{1}{48}\ln^{4}{(3)}-\frac78\,\zeta{(4)}.\\ \end{align}

• Good! (+1) It wasn't that bad with the proper tools. :-) Jul 2, 2015 at 18:42
• @Chris'ssistheartist Thanks. The pair of quadrilog terms are still bugging me though. I have a hunch they can be simplified to $Li_4(1/2)$ similar to the lower order terms, but I'm not sure how to show it without a corresponding Landen identity... Jul 2, 2015 at 20:23