Deux intégrales pour ce jeudi 19/01/2023

Pour terminer le calcul de $J$ il faut que je calcule aussi \begin{align}\int_0^1 \frac{t\ln^2 t\ln\left(\frac{1-t}{1+t}\right)}{1-t^2}dt\end{align}
J'ai peur que le résultat soit une expression avec des valeurs de fonctions polygammas. $K$ est nettement plus facile à calculer moins d'une dizaine de lignes de calculs.

Une quasi synchronicité https://les-mathematiques.net/vanilla/index.php?p=/discussion/2333039/calcul-dune-integrale-generalisee  (en lien avec mon intégrale $K$ )

Un pas de plus, sans doute le plus difficile, \begin{align*} J&=\int_0^\infty \frac{\arctan^2 x\ln^2 x}{1+x^2}dx\\ &\overset{\text{IPP}}=\left[\left(\int_0^x \frac{\ln^2 t}{1+t^2}dt\right)\tan^2 x\right]_0^\infty-2\int_0^1 \left(\int_0^\infty\frac{x\ln^2(tx)\arctan x}{(1+x^2)(1+t^2x^2)}dx\right)dt\\ &=\frac{\pi^5}{32}-\int_0^1\left(\frac{2}{1-t^2}\underbrace{\int_0^\infty \frac{\ln^2(tx)\arctan x}{x(1+t^2x^2)}dx}_{=K(t)}-\frac{2}{1-t^2}\underbrace{\int_0^\infty \frac{\ln^2(tx)\arctan x}{x(1+x^2)}dx}_{=L(t)} \right)dt\\ K(t)&\overset{u(x)=tx}=\int_0^\infty \frac{\ln^2 u\arctan\left(\frac{u}{t}\right)}{(1+u^2)u}du\\ K^\prime(t)&=-\int_0^\infty \frac{\ln^2 u}{(1+u^2)(t^2+u^2)}du=\frac{1}{1-t^2}\int_0^\infty \frac{\ln^2 u}{1+u^2}du-\frac{1}{1-t^2}\underbrace{\int_0^\infty \frac{\ln^2 u}{t^2+u^2}du}_{z(u)=\frac{u}{t}}\\ &=\frac{\pi^3}{8(1-t^2)}-\frac{1}{t(1-t^2)}\int_0^\infty \frac{\ln^2(tz)}{1+z^2}dz\\ &=\frac{\pi^3}{8(1-t^2)}-\frac{1}{t(1-t^2)}\left(\int_0^\infty \frac{\ln^2 z}{1+z^2}dz+\int_0^\infty \frac{\ln^2 t}{1+z^2}dz+2\underbrace{\int_0^\infty \frac{\ln t\ln z}{1+z^2}dz}_{=0}\right)\\ &=-\frac{\pi^3}{8t(1+t)}-\frac{\pi\ln^2 t}{2t(1-t^2)} \end{align*} Puisque, \begin{align*} \lim_{t\rightarrow \infty}K(t)=0 \end{align*} alors, \begin{align*} K(t)&=-\left(K(\infty)-K(t)\right)=-\int_t^\infty K^{\prime}(s)ds=\int_t^\infty\left(\frac{\pi^3}{8s(1+s)}+\frac{\pi\ln^2 s}{2s(1-s^2)}\right)ds\\ &=\frac{\pi}{2}\int_t^\infty\frac{\ln^2 s}{s(1-s^2)}ds-\frac{\pi^3}{8}\ln\left(\frac{t}{1+t}\right)\\ &=\frac{\pi}{2}\int_t^1\frac{\ln^2 s}{s(1-s^2)}ds+\frac{\pi}{2}\underbrace{\int_1^\infty\frac{\ln^2 s}{s(1-s^2)}ds}_{w=\frac{1}{s^2}}-\frac{\pi^3}{8}\ln\left(\frac{t}{1+t}\right)\\ &=\frac{\pi}{2}\int_t^1\frac{\ln^2 s}{s(1-s^2)}ds-\frac{\pi}{16}\int_0^1 \frac{\ln^2 w}{1-w}dw-\frac{\pi^3}{8}\ln\left(\frac{t}{1+t}\right)\\ &=\frac{\pi}{2}\int_t^1\frac{\ln^2 s}{s(1-s^2)}ds-\frac{\pi\zeta(3)}{8}-\frac{\pi^3}{8}\ln\left(\frac{t}{1+t}\right)\\ \end{align*} \begin{align*} L(t)&=\ln^2 t\underbrace{\int_0^\infty \frac{\arctan x}{x(1+x^2)}dx}_{=A}+2\ln t\underbrace{\int_0^\infty \frac{\arctan x\ln x}{x(1+x^2)}dx}_{=B}+\underbrace{\int_0^\infty \frac{\arctan x\ln^2 x}{x(1+x^2)}dx}_{=C}\\ A&\overset{x=\tan u}=\int_0^{\frac{\pi}{2}} \frac{u}{\tan u}du\overset{\text{IPP}}=\underbrace{\Big[u\ln(\sin u)\Big]_0^{\frac{\pi}{2}}}_{=0}-\int_0^{\frac{\pi}{2}}\ln(\sin u)du=\frac{\pi}{2}\ln 2\\ B(a)&=\int_0^\infty \frac{\arctan(ax)\ln x}{x(1+x^2)}dx\\ B^\prime(a)&=\int_0^\infty \frac{\ln x}{(1+x^2)(1+a^2x^2)}dx=\frac{1}{1-a^2}\underbrace{\int_0^\infty \frac{\ln x}{1+x^2}dx}_{=0}-\frac{a}{1-a^2}\underbrace{\int_0^\infty \frac{a\ln x}{1+a^2x^2}dx}_{u(x)=ax}\\ &=-\frac{a}{1-a^2}\int_0^\infty \frac{\ln u-\ln a}{1+u^2}du=\frac{\pi a\ln a}{2(1-a^2)}\\ \end{align*} Puisque, \begin{align*} B(0)=0,B(1)=B \end{align*} alors, \begin{align*} B=\frac{\pi}{2}\int_0^1 \frac{s\ln s}{1-s^2}ds\overset{w=s^2}=\frac{\pi}{8}\int_0^1 \frac{\ln w}{1-w}dw=-\frac{\pi^3}{48} \end{align*} \begin{align*} C(a)&=\int_0^\infty \frac{\arctan(ax)\ln^2 x}{x(1+x^2)}dx\\ C^\prime(a)&=\int_0^\infty \frac{\ln^2 x}{(1+x^2)(1+a^2x^2)}dx\\ &=\frac{1}{1-a^2}\int_0^\infty \frac{\ln^2 x}{1+x^2}dx-\frac{a}{1-a^2}\underbrace{\int_0^\infty \frac{a\ln^2 x}{1+a^2x^2}dx}_{u(x)=ax}\\ &=\frac{\pi^3}{8(1-a^2)}-\frac{a}{1-a^2}\int_0^\infty \frac{\ln^2 u}{1+u^2}du+\frac{2a\ln a}{1-a^2}\int_0^\infty \frac{\ln u}{1+u^2}du-\frac{a\ln^2 a}{1-a^2}\int_0^\infty \frac{1}{1+u^2}du\\ &=\frac{\pi^3}{8(1+a)}-\frac{\pi a\ln^2 a}{2(1-a^2)} \end{align*} Puisque, \begin{align*} C(0)=0,C(1)=C \end{align*} alors, \begin{align*} C&=\int_0^1 \left(\frac{\pi^3}{8(1+s)}-\frac{\pi s\ln^2 s}{2(1-s^2)}\right)ds=\frac{\pi^3\ln 2}{8}-\frac{\pi}{2}\underbrace{\int_0^1\frac{s\ln^2 s}{1-s^2}ds}_{w=s^2}=\frac{\pi^3\ln 2}{8}-\frac{\pi}{16}\int_0^1 \frac{\ln^2 w}{1-w}dw\\&=\frac{\pi^3\ln 2}{8}-\frac{\pi\zeta(3)}{8} \end{align*} Donc, \begin{align*} L(t)&=A\ln^2 t+B\ln t+C=\frac{\pi\ln 2}{2}\ln^2 t-\frac{\pi^3}{24}\ln t+\frac{\pi^3\ln 2}{8}-\frac{\pi\zeta(3)}{8}\\ J&=\frac{\pi^5}{32}-2\int_0^1 \left(\frac{K(t)-\frac{\pi^3\ln 2}{8}+\frac{\pi\zeta(3)}{8}}{1-t^2} \right)dt+\pi\ln 2\underbrace{\int_0^1 \frac{\ln^2 t}{1-t^2}dt}_{=\frac{7\zeta(3)}{4}}-\frac{\pi^3}{12}\underbrace{\int_0^1 \frac{\ln t}{1-t^2}dt}_{=-\frac{\pi^2}{8}}\\ &=\frac{\pi^5}{24}-2\int_0^1 \left(\frac{K(t)-\frac{\pi^3\ln 2}{8}+\frac{\pi\zeta(3)}{8}}{1-t^2} \right)dt+\frac{7\pi\zeta(3)\ln 2}{4}\\ &=\frac{\pi^5}{24}+\frac{\pi^3}{4}\underbrace{\int_0^1 \frac{\ln t}{1-t^2}dt}_{=-\frac{\pi^2}{8}}-\underbrace{\pi\int_0^1 \frac{1}{1-t^2}\left(\int_t^1\frac{\ln^2 s}{s(1-s^2)}ds+\frac{\pi^2\ln (1+t)}{4}-\frac{\pi^2\ln 2}{4}\right)dt}_{\text{IPP}}+\\&\frac{7\pi\zeta(3)\ln 2}{4}\\ &=\frac{\pi^5}{96}+\frac{7\pi\zeta(3)\ln 2}{4}+\frac{\pi}{2}\int_0^1 \frac{\ln^2 t\ln\left(\frac{1-t}{1+t}\right)}{t(1-t^2)}dt-\frac{\pi^3}{8}\underbrace{\int_0^1 \frac{\ln\left(\frac{1-t}{1+t}\right)}{1+t}dt}_{u=\frac{1-t}{1+t}}\\ &=\frac{\pi^5}{96}+\frac{7\pi\zeta(3)\ln 2}{4}+\frac{\pi}{2}\int_0^1 \frac{\ln^2 t\ln\left(\frac{1-t}{1+t}\right)}{t(1-t^2)}dt-\frac{\pi^3}{8}\underbrace{\int_0^1 \frac{\ln u}{1+u}du}_{=-\frac{\pi^2}{12}}\\ &=\frac{\pi^5}{48}+\frac{7\pi\zeta(3)\ln 2}{4}+\frac{\pi}{2}\underbrace{\int_0^1 \frac{\ln^2 t\ln\left(\frac{1-t}{1+t}\right)}{t(1-t^2)}dt}_{=D} \end{align*} \begin{align*} D&=\int_0^1 \frac{t\ln^2 t\ln\left(\frac{1-t}{1+t}\right)}{1-t^2}dt+\underbrace{\int_0^1 \frac{\ln^2 t\ln\left(\frac{1-t}{1+t}\right)}{t}dt}_{\text{IPP}}\\ &=\int_0^1 \frac{t\ln^2 t\ln\left(\frac{1-t}{1+t}\right)}{1-t^2}dt+\frac{2}{3}\underbrace{\int_0^1 \frac{\ln^3 t}{1-t^2}dt}_{=-\frac{\pi^4}{16}}=\int_0^1 \frac{t\ln^2 t\ln\left(\frac{1-t}{1+t}\right)}{1-t^2}dt-\frac{\pi^4}{24}\\ D&\overset{\text{IPP}}=\underbrace{\left[\left(\int_0^t\frac{s\ln^2 s}{1-s^2}ds-\int_0^1\frac{s\ln^2 s}{(1-s^2)}ds\right)\ln\left(\frac{1-t}{1+t}\right)\right]_0^1}_{=0}+\\&2\int_0^1 \frac{1}{1-t^2}\left(\underbrace{\int_0^t\frac{s\ln^2 s}{1-s^2}ds}_{z(s)=\frac{s}{t}}-\int_0^1\frac{s\ln^2 s}{1-s^2}ds\right)dt-\frac{\pi^4}{24}\\ &=2\int_0^1 \frac{1}{1-t^2}\left(\int_0^1 \frac{t^2z\ln^2(tz)}{1-t^2z^2}dz-\int_0^1\frac{s\ln^2 s}{1-s^2}ds\right)dt-\frac{\pi^4}{24}\\ &=2\int_0^1\int_0^1\left(\frac{s\ln^2(st)}{(1-s^2)(1-t^2)}-\frac{s\ln^2(st)}{(1-s^2)(1-s^2t^2)}-\frac{s\ln^2 s}{(1-t^2)(1-s^2)}\right)dsdt-\frac{\pi^4}{24}\\ &=4\left(\underbrace{\int_0^1 \frac{s\ln s }{1-s^2}ds}_{=-\frac{\pi^2}{24}}\right)\left(\underbrace{\int_0^1 \frac{\ln t}{1-t^2}dt}_{=-\frac{\pi^2}{8}}\right)+2\int_0^1\int_0^1\left(\frac{s\ln^2 t}{(1-s^2)(1-t^2)}-\frac{s\ln^2(st)}{(1-s^2)(1-s^2t^2)}\right)dsdt-\\&\frac{\pi^4}{24}\\ &=2\underbrace{\int_0^1\frac{1}{1-s^2}\left(\int_s^1 \frac{\ln^2 t}{1-t^2}dt\right)ds}_{\text{IPP}}+2\int_0^1\int_0^1\left(\frac{s\ln^2 t}{(1-s^2)(1-t^2)}-\frac{\ln^2 t}{(1-s^2)(1-t^2)}\right)-\frac{\pi^4}{48}\\ &=-\int_0^1 \frac{\ln\left(\frac{1-s}{1+s}\right)\ln^2 s}{1-s^2}ds-2\left(\int_0^1 \frac{1}{1+s}ds\right)\left(\underbrace{\int_0^1 \frac{\ln^2 t}{1-t^2}dt}_{=\frac{7\zeta(3)}{4}}\right)-\frac{\pi^4}{48}\\ &=\boxed{-\int_0^1 \frac{\ln\left(\frac{1-s}{1+s}\right)\ln^2 s}{1-s^2}ds-\frac{7\zeta(3)\ln 2}{2}-\frac{\pi^4}{48}} \end{align*} Donc, \begin{align*}\boxed{J=\frac{\pi^5}{96}-\frac{\pi}{2}\int_0^1 \frac{\ln\left(\frac{1-s}{1+s}\right)\ln^2 s}{1-s^2}ds}\end{align*}