Closed form of a series involving the squared Beta function

Question

Is there a closed form for the expression $$\sum_{n=1}^{\infty}\left(n+\frac12\right)B^2\left(n,\frac32\right)$$ where $B(x,y)$ is the Beta function? Here is what I've done.

I wrote:

$$\sum_{n=1}^{\infty}\left(n+\frac12\right)B^2\left(n,\frac32\right)=\sum_{n=1}^{\infty}\left(n+\frac12\right)\int_0^1\int_0^1(xy)^n\sqrt{(1-x)(1-y)}dxdy$$ which gives $$\int_0^1\int_0^1\frac{(3-xy)xy}{(xy-1)^2}\sqrt{(1-x)(1-y)}dxdy$$ And unfortunatly I stuck here.

Answer 1

You may exploit: $$ \sum_{n\geq 1}(n+1/2) B(n,3/2) z^{n-1} = \frac{\arcsin(\sqrt{z})}{\sqrt{z(1-z)}}\tag{1} $$ to get that your series is simply related with: $$ \int_{0}^{1}\frac{\arcsin(\sqrt{z})}{\sqrt{z}}\,dz \tag{2}$$ that is not a difficult integral to compute (through the substitution $z=\sin^2\theta$).

Answer 2

Yes, the sum is $\pi-2$. The method shown by @Jack Aurizio shows that.

BTW, your integral has a mistake in it; the integral you present has value $\frac{176}{9}-6\pi$.

Start from the definition of the beta function $$ (n+\frac12)B^2(n,\frac32) = (n+\frac12) \frac{\Gamma^2(n)\Gamma^2(\frac32)}{\Gamma^2(n+\frac32)}= (n+\frac12) \frac{(n-1)!)^2\frac{\pi}{4}}{(n+\frac12)^2\Gamma^2(n+\frac12)}= \frac{\pi(n-1)!)^2}{4(n+\frac12)\Gamma^2(n+\frac12)} $$ Now apply the formula $$\Gamma(n+\frac12) = \frac{(2n)!}{4^n n!}\sqrt{\pi}$$ to obtain $$ \frac{\pi(n-1)!)^2}{4(n+\frac12)\Gamma^2(n+\frac12)} = \frac{1}{2}\frac{2^{4n}(n!)^2(n-1)!)^2}{(2n+1)(2n)!(2n)!} $$ Next use $\frac{(2n)!}{2^n}= n!(2n-1)!!$ to get $$ \frac{\pi(n-1)!)^2}{4(n+\frac12)\Gamma^2(n+\frac12)} = \frac{1}{2}\frac{2^{2n}(n!)^2(n-1)!)^2}{(2n+1)(n!)^2((2n-1)!!)^2} = \frac{1}{2}\frac{2^{2n}(n-1)!)^2}{(2n+1)((2n-1)!!)^2} $$ This is a fairly simple seires: The first term is $\frac23$ and the term ratio for $n\geq 1$ is $$ \frac{a_{n+1}}{a_n} = \frac{4n^2}{(2n+3)(2n+1)} $$ So now we know that $$ \pi = 2+\frac23 \sum_{n=1}^\infty \prod_{k=1}^{n-1} \frac{4k^2}{(2k+3)(2k+1)} $$ This is not among the relations I can find in Mathworld, Abramowitz and Stegun, Rektorys, or various lists of series for $\pi$ I can find on the web.

And now we have this (round-about) way of proving it.