\int_0^{1/2}\frac{\ln^3(x)}{1-x}\,dx=-6\text{Li}_4\left(\frac12\right)-\frac{21\ln(2) \zeta(3)}{4}+\frac{\pi^2\ln^2(2)}{4}-\frac{\ln^4(2)}{2}

In today´s entry We will evaluate the following two integrals that look very similar, but are completely two different animals. They will turn out to be very useful in a future post that we will evaluate an alternating infinte sum involving the Harmonic number.

$\int_0^{1}\frac{\ln^3(x)}{1-x}\,dx=-\frac{\pi^4}{15}$

$\int_0^{1/2}\frac{\ln^3(x)}{1-x}\,dx=-6\text{Li}_4\left(\frac12\right)-\frac{21\ln(2) \zeta(3)}{4}+\frac{\pi^2\ln^2(2)}{4}-\frac{\ln^4(2)}{2}$

$\begin{aligned} \int_0^{1}\frac{\ln^3(x)}{1-x}\,dx&=\sum_{k=1}^{\infty}\int_0^1x^{k-1}\ln^3(x)\,dx\\ &=\sum_{k=1}^{\infty}\left(\frac{(-1)^3 3!}{k^4} \right)\\ &=-6\sum_{k=1}^{\infty}\frac{1}{k^4}\\ &=-6\zeta(4)\\ &=-6 \frac{\pi^4}{90}\\ &=-\frac{\pi^4}{15} \qquad \blacksquare \end{aligned}$

$\begin{aligned} \int_0^{1/2}\frac{\ln^3(x)}{1-x}\,dx&=\frac12\int_0^{1}\frac{\ln^3\left(\frac{x}{2}\right)}{1-\frac{x}{2}}\,dx \qquad \left(\text{by}\,\,\, x \to \frac{x}{2} \right)\\ &=\frac12\int_0^{1}\frac{\left(\ln(x)-\ln(2)\right)^3}{1-\frac{x}{2}}\,dx\\ &=\frac12\int_0^{1}\frac{\left(\ln^3(x)-3\ln(2)\ln^2(x)+3\ln^2(2)\ln(x)-\ln^3(2)\right)}{1-\frac{x}{2}}\,dx\\ &=\frac12 \left(\int_0^{1}\frac{\ln^3\left(x\right)}{1-\frac{x}{2}}\,dx-3\ln(2)\int_0^{1}\frac{\ln^2\left(x\right)}{1-\frac{x}{2}}\,dx+3\ln^2(2)\int_0^{1}\frac{\ln\left(x\right)}{1-\frac{x}{2}}\,dx-\ln^3(2)\int_0^{1}\frac{dx}{1-\frac{x}{2}} \right)\\ &=\frac12 \left(\text{I}_1-3\ln(2)\text{I}_2+3\ln^2(2)\text{I}_3-\ln^3(2)\text{I}_4 \right)\\ &=\frac12 \left(-12\text{Li}_4\left(\frac12\right)-12\ln(2)\left( \frac{\ln ^{3}(2)}{6}-\frac{\pi^{2} \ln (2)}{12}+\frac{7}{8} \zeta(3) \right)+3\ln^2(2)\left(\ln^2(2)-\zeta(2) \right)-2\ln^4(2) \right)\\ &=-6\text{Li}_4\left(\frac12\right)-\frac{21\ln(2) \zeta(3)}{4}+\frac{\pi^2\ln^2(2)}{4}-\frac{\ln^4(2)}{2} \qquad \blacksquare \end{aligned}$

$\begin{aligned} I_1&=\int_0^{1}\frac{\ln^3\left(x\right)}{1-\frac{x}{2}}\,dx\\ &=2\sum_{k=1}^\infty\frac{1}{2^k}\int_0^1x^{k-1}\ln^3(x)\,dx\\ &=2\sum_{k=1}^\infty\frac{1}{2^k}\left(\frac{(-1)^3 3!}{k^4} \right)\\ &=-12\sum_{k=1}^\infty\frac{1}{2^kk^4}\\ &=-12\text{Li}_4\left(\frac12\right) \qquad \blacksquare \end{aligned}$

$\begin{aligned} I_2&=\int_0^{1}\frac{\ln^2\left(x\right)}{1-\frac{x}{2}}\,dx\\ &=2\sum_{k=1}^\infty\frac{1}{2^k}\int_0^1x^{k-1}\ln^2(x)\,dx\\ &=2\sum_{k=1}^\infty\frac{1}{2^k}\left(\frac{(-1)^2 2!}{k^3} \right)\\ &=4\sum_{k=1}^\infty\frac{1}{2^kk^3}\\ &=4\text{Li}_3\left(\frac12\right) \\ &=4\left( \frac{\ln ^{3}(2)}{6}-\frac{\pi^{2} \ln (2)}{12}+\frac{7}{8} \zeta(3) \right)\qquad \blacksquare \end{aligned}$

See the evaluation of $\text{Li}_3\left(\frac12\right)$ in the appendix bellow

$\begin{aligned} I_3&=\int_0^{1}\frac{\ln\left(x\right)}{1-\frac{x}{2}}\,dx\\ &=2\sum_{k=1}^\infty\frac{1}{2^k}\int_0^1x^{k-1}\ln(x)\,dx\\ &=2\sum_{k=1}^\infty\frac{1}{2^k}\left(\frac{(-1) 1!}{k^2} \right)\\ &=-2\sum_{k=1}^\infty\frac{1}{2^kk^2}\\ &=-2\text{Li}_2\left(\frac12\right) \\ &=-2\left(\frac{\zeta(2)}{2}-\frac{\ln^2(2)}{2} \right)\\ &=\ln^2(2)-\zeta(2)\qquad \blacksquare \end{aligned}$

See the evaluation of $\text{Li}_2\left(\frac12\right)$ in the appendix bellow

$\begin{aligned} I_4&=\int_0^{1}\frac{1}{1-\frac{x}{2}}\,dx\\ &=2\sum_{k=1}^\infty\frac{1}{2^k}\int_0^1x^{k-1}\,dx\\ &=2\sum_{k=1}^\infty\frac{1}{2^k}\left(\frac{1}{k} \right)\\ &=2\sum_{k=1}^\infty\frac{1}{2^kk}\\ &=-2\ln\left(\frac12\right) \\ &=2\ln(2)\qquad \blacksquare \end{aligned}$

Appendix

An extremely useful integral

$\int_{0}^{1} x^{m} \ln ^{n}(x) d x=\frac{(-1)^{n} n !}{(m+1)^{n+1}}$

Proof:

$\begin{aligned} I &=\int_{0}^{1} x^{m} \ln ^{n}(x) d x \\ &=\int_{0}^{\infty} e^{-m x} \ln ^{n}\left(e^{-x}\right) e^{-x} d x \quad\left(x \rightarrow e^{-x}\right) \\ &=\int_{0}^{\infty} e^{-(m+1) x}(-1)^{n} x^{n} d x \\ &=\frac{(-1)^{n}}{(m+1)^{n+1}} \int_{0}^{\infty} e^{-x} x^{n} d x \quad((m+1) x \rightarrow x) \\ &=\frac{(-1)^{n} n !}{(m+1)^{n+1}} \quad \blacksquare \end{aligned}$

Recall the functional equation of the Dilogarithm proved here

$\text{Li}_{2}(x)+\text{Li}_{2}(1-x)=\zeta(2)-\ln(x)\ln(1-x)$

plugging $x=\frac{1}{2}$ in the above equation we obtain

$\text{Li}_{2}\left(\frac12\right)=\frac{\zeta(2)}{2}-\frac{\ln^2(2)}{2}$

Recall the functional equation of the Trilogarithm proved here

$\begin{aligned} \text{Li}_{3}(x)+\text{Li}_{3}(1-x)+\text{Li}_{3}\left(1-\frac{1}{x}\right)=\zeta(3)+\frac{\ln ^{3}(x)}{6}+\frac{\pi^{2} \ln (x)}{6}-\frac{\ln ^{2}(x) \ln (1-x)}{2} \end{aligned}$

plugging $x=\frac{1}{2}$ in the above equation we obtain

$\begin{aligned} \text{Li}_{3}\left(\frac{1}{2}\right)+\text{Li}_{3}\left(1-\frac{1}{2}\right)+\text{Li}_{3}\left(1-\frac{1}{2}\right)&=\zeta(3)+\frac{\ln ^{3}\left(\frac{1}{2}\right)}{6}+\frac{\pi^{2} \ln \left(\frac{1}{2}\right)}{6}-\frac{\ln ^{2}\left(\frac{1}{2}\right) \ln \left(1-\frac{1}{2}\right)}{2} \\ 2 \text{Li}_{3}\left(\frac{1}{2}\right)+\text{Li}_{3}(-1)&=\zeta(3)-\frac{\ln ^{3}(2)}{6}-\frac{\pi^{2} \ln (2)}{6}+\frac{\ln ^{3}(2)}{2} \end{aligned}$

Note that we got that $L i_{3}(-1)=-\eta(3)=-\frac{3}{4} \zeta(3)$ , then

$\begin{aligned} 2 \text{Li}_{3}\left(\frac{1}{2}\right)&=\frac{3}{4} \zeta(3)+\zeta(3)+\frac{\ln ^{3}(2)}{3}-\frac{\pi^{2} \ln (2)}{6} \\ \text{Li}_{3}\left(\frac{1}{2}\right)&=\frac{\ln ^{3}(2)}{6}-\frac{\pi^{2} \ln (2)}{12}+\frac{7}{8} \zeta(3) \qquad \blacksquare \end{aligned}$

Reference

Kam Cheong Au: Linear relations between logarithmic integrals of high weight and some closed-form evaluations, arXiv:1910.12113

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