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.

Starting with Integrand's observation in the comments, we have $$
   \begin{array}{rcl} \displaystyle I = 2\int_{0}^{1}{\displaystyle\frac{\log^{2}{x}}{1+x^{2}}\,\mathrm{d}x}=2 \int_0^1 \log^2{x} \sum_{k \ge 0} (-1)^k x^{2k}\,\mathrm{d}x = 2\sum_{k \ge 0} (-1)^k\int_0^1 x^{2k}\log^2{x}\,\mathrm{d}x\end{array}
   $$
Now, consider $\displaystyle f(\alpha) = \int_0^1 x^{2\alpha}\,\mathrm{dx} = \frac{1}{1+2\alpha}$. Then finding $f''(\alpha)$ for both sides gives:
$$\displaystyle  \int_0^1 x^{2\alpha} \log^2{x}\,\mathrm{dx} = \frac{2}{(1+2\alpha)^3}$$
Therefore:
$$I = 4 \sum_{k \ge 0} \frac{(-1)^k}{(2k+1)^3} = 4 \cdot \frac{\pi^3}{32} = \frac{\pi^3}{8}. $$
