How to Find closed form :$$\sum_{n=1}^{\infty}\frac{x^{2n}}{n}a_n$$ where $a_n=\sum_{k=1}^{2n-1}\frac{(-1)^{k-1}}{k}$
$$S(x)=\sum_{n=1}^{\infty}\frac{x^{2n}}{n}a_n=\sum_{n=1}^{\infty}\int^1_0y^{n-1}x^{2n}a_n dy =\int^1_0 \frac{1}{y}\sum_{n=1}^{\infty}a_n(yx^2)^n dy =x^2\int^1_0 \sum_{n=1}^{\infty}a_{n+1}(yx^2)^n dy$$
we have : $$\boxed{\sum_{n=1}^{\infty}a_nx^n=\dfrac{1}{1-x}\sum_{n=1}^{\infty}(a_n-a_{n-1})x^n}$$
therfore: $$S(x)=x^2\int^1_0 \frac{1}{1-yx^2}\sum_{n=1}^{\infty}(a_{n+1}-a_n)(yx^2)^n dy=\int^1_0\frac{-x^2}{1-yx^2}\sum_{n=1}^{\infty}\frac{(yx^2)^n}{2n(2n+1)}dy$$ Let :$$S_1=\sum_{n=1}^{\infty}\frac{(yx^2)^n}{2n(2n+1)}=\frac{1}{x\sqrt y}\sum_{n=1}^{\infty}\frac{(x\sqrt y)^{2n+1}}{2n(2n+1)}$$ let :$t:=x\sqrt y$ therfore: $$S_1=\frac{1}{2t}\int^t_0\sum_{n=1}^{\infty}\frac{(z^2)^n}{n} dz =-\frac{1}{2t}\int^t_0\ln(1-z^2)dz=-\frac{1}{2t}\left({t\ln(1-t^2)-2t+\ln(1+t)-\ln(1-t)}\right)$$ Then: $$S_1=1-\frac{1}{2}\ln(1-yx^2)-\frac{1}{2x \sqrt y}\ln(1+x \sqrt y)+\frac{1}{2x \sqrt y}\ln(1-x \sqrt y)$$
we have : $$S(x)=\int^1_0\frac{-x^2}{1-yx^2}S_1dy=I_1+I_2+I_3+I_4$$
where : $$I_1=\int^1_0\frac{-x^2}{1-yx^2}dy=\ln(1-x^2)$$ $$I_2=\frac{1}{2}\int^1_0\frac{x^2\ln(1-yx^2)}{1-yx^2}dy=-\frac{1}{4}\ln^2(1-x^2)$$ $$I_3=\frac{1}{2}\int^1_0\frac{x\ln(1+x \sqrt y)}{ \sqrt y(1-yx^2)}dy$$ $$I_4=-\frac{1}{2}\int^1_0\frac{x\ln(1-x \sqrt y)}{ \sqrt y(1-yx^2)}dy$$
How To Evaluate $I_3$ and $I_4$ ??
Is there a better way than this??
Thank you very much for your interest