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.

Let $f(x)=\frac{\ln^n(x)}{1+x^2}$
$$I_n=\int_0^\infty f(x)\ dx=\int_0^1f(x)\ dx+\underbrace{\int_1^\infty f(x)\ dx}_{1/x\to x}=\int_0^1f(x)\ dx+\int_0^1(-1)^nf(x)\ dx$$
Clearly, for odd $n$, $I_n=0$, so for even $n$ , we have 
$$I_n=2\int_0^1\frac{\ln^n(x)}{1+x^2}\ dx=2\sum_{k=0}^\infty(-1)^k\int_0^1 x^{2k}\ln^n(x)\ dx$$
$$=2(-1)^nn!\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)^{n+1}}=2(-1)^nn!\beta(n+1)$$
and since $n$ is even, we better set $n=2m$ 
$$\int_0^\infty\frac{\ln^{2m}(x)}{1+x^2}\ dx=2(2m)!\beta(2m+1),\quad m=0,1,2,...$$
where $\beta(s)$ is The Dirichlet beta function.
