Solve $\sum_{k=1}^\infty B(k,k)$

Question

I have solved$$\sum_{k=1}^\infty B(k,k)$$Like so: $$\sum_{k=1}^\infty B(k,k)=\int_0^1\sum_{k=0}^\infty t^k(1-t)^kdt=\int_0^1\frac1{1-t+t^2}dt=\int_0^1\frac1{(t-\frac12)+\frac34}dt=\int_{-\frac12}^\frac12\frac1{u^2+\frac34}du=\frac2{\sqrt{3}}\arctan\left(\frac{2x}{\sqrt3}\right)\bigg|_{-\frac12}^{\frac12}=\frac{2\pi}{3\sqrt3}$$I wonder if there is a way to solve this sum with the definition $$B(m,n)=2\int_0^\frac{\pi}2\sin^{2m-1}(x)\cos^{2n-1}(x)dx$$I couldn't see a way to sum the product of the trig functions.

Answer 1

You are missing a factor of 2 in your definition of the Beta function by the way. Its not too bad, summing the trig functions is just a geometric series. We will apply the double angle identity as well.

$$\sum_{k\geq 1} B(k,k) = 2\int_0^{\pi/2}\sum_{k\geq 0} (\sin(x)\cos(x))^{2k+1} dx = \int_0^{\pi/2}\frac{\sin(2x)}{1-(\tfrac{1}{2}\sin(2x))^2}dx$$

You can use the Pythagorean identity to get,

$$\int_0^{\pi/2}\frac{\sin(2x)}{\tfrac{\cos^2(2x)}{4}+ \tfrac{3}{4}} dx$$

Notice symmetry around the point $x = \pi/4$,and then use the usubstitution $u = \cos(2x)$.

$$\int_0^{1/2} \frac{2}{u^2 + 3/4}du$$

Which is identical to one of the integrals you found before.