<s>[INST] Computing $\sum\limits_{n=0}^\infty\frac{(-1)^nH_{n/2}}{(2n+1)^2}$ 
How to evaluate the following challenging sum:
  $$S=\sum_{n=0}^\infty\frac{(-1)^nH_{n/2}}{(2n+1)^2}=\frac74\zeta(3)+\frac3{32}\pi^3-\frac{\pi}{2}G-2\ln2G+\frac{\pi}{8}\ln^22-2\Im\operatorname{Li}_3(1+i)?$$

where $H_{n}$  is the harmonic number, $\operatorname{Li}_n$ is the polylogarithm function and $G$ is Catalan constant. 

This problem was proposed by a friend and here is my approach:
Using the identity 
$$\int_0^1\frac{x^n}{1+x}\ dx=H_{n/2}-H_n+\ln2, \quad x\mapsto x^2$$
$$2\int_0^1\frac{x^{2n+1}}{1+x^2}\ dx=H_{n/2}-H_n+\ln2$$
Multiply both sides by $\frac{(-1)^n}{(2n+1)^2}$ to get 
$$\sum_{n=0}^\infty (-1)^n\frac{H_{n/2}-H_n+\ln2}{(2n+1)^2}=2\int_0^1\frac{1}{1+x^2}\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{(2n+1)^2}\ dx\\=2\int_0^1\frac{1}{1+x^2}\Im\sum_{n=0}^\infty\frac{(i)^nx^n}{n^2}\ dx=2\Im\int_0^1\frac{\operatorname{Li}_2(ix)}{1+x^2}\ dx$$
Rearranging the terms to get
$$S=\sum_{n=0}^\infty\frac{(-1)^nH_{n}}{(2n+1)^2}-\ln2G+2\Im\int_0^1\frac{\operatorname{Li}_2(ix)}{1+x^2}\ dx$$
where 
$$\sum_{n=0}^\infty\frac{(-1)^nH_{n}}{(2n+1)^2}=-\int_0^1\ln x\sum_{n=0}^\infty(-1)^n x^{2n}H_n\ dx=\int_0^1\frac{\ln x\ln(1+x^2)}{1+x^2}\ dx$$
which is calculated here:
$$\int_0^1\frac{\ln x\ln(1+x^2)}{1+x^2}\ dx=\frac3{32}\pi^3+\frac{\pi}8\ln^22-\ln2~G-2\text{Im}\operatorname{Li_3}(1+i)$$

And the question here is how to evaluate the dilogarithm integral or a different way to compute $S$? Thank you.
 [/INST] We have that:$$\Im\operatorname{Li}_2 (ix)=\operatorname{Ti}_2 (x)=\int_0^x \frac{\arctan t}{t}dt=\sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)^2}$$
So let's rewrite the integral in terms of the inverse tangent function:$$\int_0^1\frac{1}{1+x^2}\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{(2n+1)^2}dx=\int_0^1 \frac{\operatorname{Ti}_2(x)}{1+x^2}dx$$
$$\overset{IBP}=\arctan x\operatorname{Ti}_2(x)\bigg|_0^1-\int_0^1 \frac{\arctan^2 x}{x}dx=\frac{\pi}{4}G-\mathcal J$$
$$\mathcal J=\int_0^1\frac{\arctan^2 x}{x}dx\overset{x=\tan t}=2\int_0^\frac{\pi}{4} \frac{t^2}{\sin(2t)}dt\overset{2t=x}=\frac14\int_0^\frac{\pi}{2}\frac{x^2}{\sin x}dx=\frac{\pi}{2}G-\frac{7}{8}\zeta(3)$$
The last integral is nicely evaluated here.
$$\Rightarrow \boxed{\Im\int_0^1 \frac{\operatorname{Li}_2(ix)}{1+x^2}dx=\int_0^1 \frac{\operatorname{Ti}_2(x)}{1+x^2}dx=\frac{7}{8}\zeta(3)-\frac{\pi}{4}G}$$
</s>