How can we compute a binomial series of the form $\sum_{n=0}^\infty {-{3}/{2} \choose n}\frac{(-1)^n}{(2n+1)^3}$

Question

I have been trying to find problems involving binomial series and then converting the series into an integral, but this particular series, $$S = \sum_{n=0}^\infty {-{3}/{2} \choose n}\frac{(-1)^n}{(2n+1)^3}$$ has stumped me. My initial approach was showing that $$S = \int_0^1 \frac{\log^2{x}}{\sqrt[3]{1-x^2}}\ \mathrm{d}x$$ which I am not sure how to solve. An easier case would be if we had $-\frac{1}{2}$ instead of $-\frac{1}{3}$. Any ideas would be greatly appreciated. Thanks!

Answer 1

Hint. Concerning the evaluation of $$ S = \int_0^1 \frac{\log^2{x}}{\sqrt[3]{1-x^2}}\ \mathrm{d}x $$ one may recall the Euler beta result $$ \int_0^1 t^{a-1}(1-t)^{b-1}\ \mathrm{d}t=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)},\quad a>0,\,b>0, $$ giving with $t=x^2$, $\,dt=2xdx$, $$ 2\int_0^1 x^{2a-1}(1-x^2)^{-1/3}\ \mathrm{d}x=\frac{\Gamma(a)\Gamma\left(\frac23\right)}{\Gamma\left(a+\frac23\right)},\quad a>0, $$ then by differentiating twice one gets $$ S = \left.\partial_a^2\left(\frac{\Gamma(a)\Gamma\left(\frac23\right)}{8\Gamma\left(a+\frac23\right)}\right)\right|_{a=\frac12} $$ a result which one may simplify in terms of the polygamma function.