Question

Tentei avaliar o integral
$I=\int_0^\infty \frac{\ln^2x}{(x+1)^2+1} \, dx$
e conseguiu mostrar que
$\begin{align} I &= \int_0^1 \frac{\ln^2x}{(x+1)^2+1} \, dx + \int_0^1 \frac{\ln^2x}{(x+1)^2+x^2} \, dx\\ &= \int_0^1 \frac{\ln^2x}{(x+1+i)(x+1-i)} \, dx + \int_0^1 \frac{\ln^2x}{(x+1+ix )(x+1-ix )} \, dx\\ &= -2\operatorname{Im}\operatorname{Li}_3\left(-\frac{1+i}2\right) -2\operatorname{Im} \operatorname{Li}_3(-1-i) \end{align}\ \textless \ br /\ \textgreater$
$que \: é \: igual \: a \: \frac{5\pi^3}{64}+\frac\pi{16}\ln^22$
Talvez seja desnecessário, entretanto, recorrer à avaliação em um espaço complexo. Eu gostaria de trabalhar em uma derivação elementar desse resultado integral.
​

Answer 1

Resposta:

$\begin{align}J&=\int_0^\infty \frac{\ln^2x}{(x+1)^2+1} \, dx\\ &=\int_0^\infty \frac{\ln^2x}{x^2+2x+2} \, dx\\ &\overset{x=\sqrt{2}t}=\frac{1}{\sqrt{2}}\int_0^\infty \frac{\ln^2(\sqrt{2}t)}{t^2+\sqrt{2}t+1} \, dt\\ &=\frac{1}{\sqrt{2}}\underbrace{\int_0^\infty \frac{\ln^2(\sqrt{2})}{t^2+\sqrt{2}t+1} \, dt}_{=\dfrac{\pi\ln^2 2}{8\sqrt{2}}}+\underbrace{\sqrt{2}\int_0^\infty \frac{\ln(\sqrt{2})\ln t}{t^2+\sqrt{2}t+1} \, dt}_{=0}+\frac{1}{\sqrt{2}}\int_0^\infty \frac{\ln^2 t}{t^2+\sqrt{2}t+1} \, dt\\ &=\dfrac{\pi\ln^2 2}{16}+\frac{1}{\sqrt{2}}\int_0^\infty \frac{\ln^2 t}{t^2+\sqrt{2}t+1} \, dt\\ K&=\int_0^\infty \int_0^\infty \frac{\ln^2 (tx)}{(t^2+\sqrt{2}t+1)(x^2+\sqrt{2}x+1)} \, dt\,dx\\ &\overset{u(x)=tx}=\int_0^\infty \int_0^\infty \frac{t\ln^2 u}{(t^2+\sqrt{2}t+1)(u^2+\sqrt{2}tu+t^2)} \, dt\,du\\ &=\int_0^\infty \ln^2 u\left[\frac{(u+1)\ln\left(\frac{t^2+\sqrt{2}tu+u^2}{t^2+\sqrt{2}t+1}\right)}{2(1-u)(1+u^2)}-\frac{\arctan\left(\frac{u+\sqrt{2}t}{u}\right)}{1+u^2}-\frac{\arctan\left(\sqrt{2}t+1\right)}{1+u^2}\right]_{t=0}^{t=\infty}du\\ &=\int_0^\infty \ln^2 u\left(-\frac{\pi}{1+u^2}-\frac{(1+u)\ln u}{(1-u)(1+u^2)}+\frac{\frac{\pi}{2}}{1+u^2}\right) du\\ &=-\frac{\pi}{2}\int_0^\infty \frac{\ln^2 u}{1+u^2}du-\int_0^\infty\frac{(1+u)\ln^3 u}{(1-u)(1+u^2)}du\\ &=-\frac{\pi}{2}\int_0^\infty \frac{\ln^2 u}{1+u^2}du-\int_0^1\frac{(1+u)\ln^3 u}{(1-u)(1+u^2)}du-\int_1^\infty\frac{(1+v)\ln^3 v}{(1-v)(1+v^2)}dv\\ &\overset{u=\frac{1}{v}}=-\frac{\pi}{2}\int_0^\infty \frac{\ln^2 u}{1+u^2}du-\int_0^1\frac{(1+u)\ln^3 u}{(1-u)(1+u^2)}du-\int_0^1\frac{(1+u)\ln^3 u}{(1-u)(1+u^2)}du\\ &=-\frac{\pi}{2}\int_0^\infty \frac{\ln^2 u}{1+u^2}du-2\int_0^1\frac{(1+u)\ln^3 u}{(1-u)(1+u^2)}du\\ &=-\frac{\pi}{2}\int_0^\infty \frac{\ln^2 u}{1+u^2}du-2\int_0^1 \frac{\ln^3 u}{1-u}du-\int_0^1 \frac{2v\ln^3 v}{1+v^2}dv\\ &\overset{u=v^2}=-\frac{\pi}{2}\int_0^\infty \frac{\ln^2 u}{1+u^2}du-2\int_0^1 \frac{\ln^3 u}{1-u}du-\frac{1}{8}\int_0^1 \frac{\ln^3 u}{1+u}du\\ &=-\frac{\pi}{2}\int_0^\infty \frac{\ln^2 u}{1+u^2}du-2\int_0^1 \frac{\ln^3 u}{1-u}du-\frac{1}{8}\left(\int_0^1 \frac{\ln^3 u}{1-u}du-\int_0^1 \frac{2v\ln^3 v}{1-v^2}dv\right)\\ &\overset{u=v^2}=-\frac{\pi}{2}\int_0^\infty \frac{\ln^2 u}{1+u^2}du-2\int_0^1 \frac{\ln^3 u}{1-u}du-\frac{1}{8}\left(\int_0^1 \frac{\ln^3 u}{1-u}du-\frac{1}{8}\int_0^1 \frac{\ln^3 u}{1-u}du\right)\\ &==-\frac{\pi}{2}\int_0^\infty \frac{\ln^2 u}{1+u^2}du-\frac{135}{64}\int_0^1 \frac{\ln^3 u}{1-u}du\\ &=-\frac{\pi}{2}\times \frac{\pi^3}{8}-\frac{135}{64}\times -\frac{\pi^4}{15}\\ &=\frac{5\pi^4}{64} \end{align}$

Por outro lado,

$\begin{align}K&=2\underbrace{\left(\int_0^\infty\frac{\ln x}{x^2+\sqrt{2}x+1}\right)^2}_{=0}+2\sqrt{2}\left(J-\frac{\pi\ln^2 2}{16}\right)\underbrace{\int_0^1 \frac{1}{x^2+\sqrt{2}x+1}dx}_{=\frac{\pi}{2\sqrt{2}}}\end{align}$

Portanto,

$\displaystyle \boxed{J=\frac{5\pi^3}{64}+\frac{\pi\ln^2 2}{16}}$

NB: Presumo que,

$\begin{align} \int_0^\infty \frac{\ln^2 x}{1+x^2}dx&=\frac{\pi^3}{8}\\ \int_0^\infty \frac{\ln^3 x}{1-x}dx&=-6\zeta(4)\\ \displaystyle\zeta(4)&=\frac{\pi^4}{90}\end{align}$

Espero ter ajudado!

Bons estudos!