Encontrar la integral $\int_{0}^{1}\frac{\operatorname{Li_2}\left ( x^2-x \right )}{x}dx$

Question

Encuentre $$\int\limits_{0}^{1}\frac{\operatorname{Li_2}\left ( x^2-x \right )}{x}dx$$ $$ \lim_{a\rightarrow 0^+}\int\limits_{a}^{1}\frac{\operatorname{Li_2}\left ( x^2-x \right )}{x}dx=\lim_{a\rightarrow 0^+}-\int\limits_{0}^{1}\left ( 1-x \right ){}_3F_2\left ( 1,1,1;2,2;x^2-x \right )dx=$$ $$= -\sum\limits_{n=0}^{\infty }\frac{\Gamma \left ( n+1 \right )\Gamma \left ( n+1 \right )\Gamma \left ( n+1 \right )}{\Gamma \left ( n+2 \right )\Gamma \left ( n+2 \right )}\frac{\left ( -1 \right )^n}{n!}\int\limits_{0}^{1}x^n\left ( 1-x \right )^{n+1}dx=$$ $$\bigstar \; \int\limits_{0}^{1}x^n\left ( 1-x \right )^{n+1}dx=B\left ( n+1,n+2 \right )=\frac{1}{2}\frac{\left ( 1 \right )_n}{2^{2n}\left ( \frac{3}{2} \right )_n}$$ $$=-\frac{1}{2}\sum\limits_{n=0}^{\infty }\frac{\left ( 1 \right )_n\left ( 1 \right )_n\left ( 1 \right )_n\left ( 1 \right )_n}{\left ( \frac{3}{2} \right )_n\left ( 2 \right )_n\left ( 2 \right )_n}\frac{\left ( -1/4 \right )^n}{n!}=-\frac{1}{2}{}_4F_3\left ( 1,1,1,1;\frac{3}{2},2,2;-\frac{1}{4} \right )$$ $$\bigstar \; {}_4F_3\left ( 1,1,1,1;\frac{3}{2},2,2;-\frac{1}{4} \right )=\frac{4}{5}\zeta \left ( 3 \right )$$ $$\lim_{a\rightarrow 0^+}\int\limits_{a}^{1}\frac{\operatorname{Li_2}\left ( x^2-x \right )}{x}dx=-\frac{2}{5}\zeta \left ( 3 \right )$$ Dudo que haya calculado correctamente la integral... No he encontrado otra manera que resolver a través de series hipergeométricas ...

¿He tomado la decisión correcta? ¿Puede resolverlo más fácilmente?

Answer 1

Una forma más sencilla de evaluarlo es utilizar la siguiente identidad: $$\boxed{\int _0^1\frac{x\ln ^n\left(t\right)}{1-xt}\:dt=\left(-1\right)^nn!\operatorname{Li}_{n+1}\left(x\right)}$$

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Aplicándolo a la integral deseada se obtiene:

$$\int _0^1\frac{\operatorname{Li}_2\left(x\left(x-1\right)\right)}{x}\:dx=\int _0^1\ln \left(t\right)\int _0^1\frac{1-x}{1+x\left(1-x\right)t}\:dx\:dt$$ $$=-2\int _0^1\frac{\ln \left(t\right)\arctan \left(\frac{\sqrt{t}}{\sqrt{-4-t}}\right)}{\sqrt{-4-t}\sqrt{t}}dt=2\int _0^1\frac{\arctan ^2\left(\frac{\sqrt{t}}{\sqrt{-4-t}}\right)}{t}\:dt=-2\int _0^1\frac{\operatorname{arctanh} ^2\left(\sqrt{\frac{t}{4+t}}\right)}{t}\:dt$$ $$=-4\int _0^{\frac{1}{\sqrt{5}}}\frac{\operatorname{arctanh} ^2\left(x\right)}{x\left(1-x^2\right)}\:dx=-\frac{1}{2}\int _{\frac{1}{\phi ^2}}^1\frac{\ln ^2\left(x\right)}{x}\:dx-\int _{\frac{1}{\phi ^2}}^1\frac{\ln ^2\left(x\right)}{1-x}\:dx$$ $$=-\frac{4}{3}\ln ^3\left(\phi \right)-2\sum _{k=1}^{\infty }\frac{1}{k^3}+2\sum _{k=1}^{\infty }\frac{1}{k^3\:\phi ^{2k}}+4\ln \left(\phi \right)\sum _{k=1}^{\infty }\frac{1}{k^2\:\phi ^{2k}}+4\ln ^2\left(\phi \right)\sum _{k=1}^{\infty }\frac{1}{k\:\phi ^{2k}}$$ $$=\frac{8}{3}\ln ^3\left(\phi \right)-2\zeta \left(3\right)+2\operatorname{Li}_3\left(\frac{1}{\phi ^2}\right)+4\ln \left(\phi \right)\operatorname{Li}_2\left(\frac{1}{\phi ^2}\right)$$ Mediante el uso de ecuaciones funcionales estándar tenemos: $$=-\frac{4}{3}\ln ^3\left(\phi \right)-2\zeta \left(3\right)+\frac{8}{5}\zeta \left(3\right)+\frac{4}{3}\ln ^3\left(\phi \right)-\frac{8}{5}\ln \left(\phi \right)\zeta \left(2\right)+\frac{8}{5}\ln \left(\phi \right)\zeta \left(2\right)$$

Y así: $$\boxed{\int _0^1\frac{\operatorname{Li}_2\left(x\left(x-1\right)\right)}{x}\:dx=-\frac{2}{5}\zeta \left(3\right)}$$