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
Best Answer
\begin{align*} \sum_{n=1}^\infty\frac{x^{2n}}n\sum_{k=1}^{2n-1}\frac{(-1)^{k-1}}k &=\sum_{n=1}^\infty\frac{x^{2n}}{2n}\sum_{k=1}^{2n-1}\left(\frac{(-1)^{k-1}}{k}+\frac{(-1)^{2n-k-1}}{2n-k}\right) \\&=\sum_{n=1}^\infty x^{2n}\sum_{k=1}^{2n-1}\frac{(-1)^{k-1}}{k(2n-k)}=\frac12\big(f(x)+f(-x)\big), \end{align*} where $$ f(x)=\sum_{n=2}^\infty x^n\sum_{k=1}^{n-1}\frac{(-1)^{k-1}}{k(n-k)}=\color{blue}{-\log(1+x)\log(1-x)}=f(-x) $$ (as a Cauchy product). Hence the result is $f(x)$.