Evaluating Series Integral How do I show that $$\int_0^\infty\frac{(\ln x)^2dx}{1+x^2}$$=$$4(1-1/3^3+1/5^3-1/7^3...)$$
I expanded the integral to $$\int_0^\infty(\ln x)^2(1-x^2+x^4...)dx$$ using the power series for $$\frac{1}{1+x^2}$$ but I'm not sure how to continue from here.

\begin{align*}
J&=\int_0^\infty \frac{\ln^2 x}{1+y^2}\,dx\\
A&=\int_0^\infty \int_0^\infty \frac{\ln^2(xy)}{(1+x^2)(1+y^2)}\,dx\,dy\\
&=\int_0^\infty \int_0^\infty \frac{\ln^2(x)+\ln^2(y)}{(1+x^2)(1+y^2)}\,dx\,dy\\
&=\int_0^\infty \int_0^\infty \frac{2\ln^2(x)}{(1+x^2)(1+y^2)}\,dx\,dy\\
&=2J\Big[\arctan x\Big]_0^\infty\\
&=\pi J
\end{align*}
On the other hand,
\begin{align*}
A&\overset{u(x)=xy}=\int_0^\infty \left(\int_0^\infty \frac{\ln^2 u}{y(1+\left(\frac{u}{y}\right)^2)(1+y^2)}\,du\right)\,dy\\
&=\int_0^\infty \left(\int_0^\infty \frac{y\ln^2 u}{(u^2+y^2)(1+y^2)}\,du\right)\,dy\\
&=\frac{1}{2}\int_0^\infty \left[\frac{\ln\left(\frac{u^2+y^2}{1+y^2}\right)}{1-u^2}\right]_{y=0}^{y=\infty}\ln^2 u\,du\\
&=\int_0^\infty \frac{\ln^3 u}{u^2-1}\,du\\
&=\int_0^1 \frac{\ln^3 u}{u^2-1}\,du+\int_1^\infty \frac{\ln^3 x}{x^2-1}\,dx\\
&\overset{u=\frac{1}{x}}=2\int_0^1 \frac{\ln^3 u}{u^2-1}\,du\\
&=2\int_0^1 \frac{\ln^3 u}{u-1}\,du-\int_0^1 \frac{2x\ln^3 x}{x^2-1}\,dx\\
&\overset{u=x^2}=2\int_0^1 \frac{\ln^3 u}{u-1}\,du-\frac{1}{8}\int_0^1 \frac{\ln^3 u}{u-1}\,du\\
&=\frac{15}{8}\int_0^1 \frac{\ln^3 u}{u-1}\,du
\end{align*}
Moreover,
\begin{align*}
\int_0^1 \frac{\ln^3 u}{1-u}\,du&=\int_0^1 \left(\sum_{n=0}^\infty u^n\right)\ln^3 u\,du\\
&=\sum_{n=0}^\infty\left(\int_0^1 u^n\ln^3 u\,du\right)\\
&=-6\sum_{n=0}^\infty \frac{1}{(n+1)^4}\\
&=-6\zeta(4)
\end{align*}
Therefore,
\begin{align*}\
A&=\frac{15}{8}\times 6\zeta(4)\\
&=\frac{45}{4}\zeta(4)\\
J&=\frac{A}{\pi}\\
&=\frac{45\zeta(4)}{4\pi}\\
&=\frac{45\times\frac{\pi^4 }{90}}{4\pi}\\
&=\boxed{\frac{\pi^3}{8}}
\end{align*}
NB:
Observe that $\displaystyle \int_0^1 \frac{\ln x}{1+x^2}\,dx\overset{u=\frac{1}{x}}=-\int_0^1 \frac{\ln u}{1+u^2}\,du$
and, i  assume $\displaystyle \zeta(4)=\dfrac{\pi^4}{90}$
