Show that $\int_{0}^{\pi/2} (\frac{1}{\sin^3{\theta}} – \frac{1}{\sin^2{\theta}})^{1/4} \cos{\theta} d\theta = \frac{(\Gamma(1/4))^2}{2\sqrt{\pi}}$

Question

Well, I have shown that $B(n, n+1) = \frac{(\Gamma(n))^2}{2\sqrt{2n}}$

From there I could deduce that $B(1/4, 5/4) = \frac{(\Gamma(1/4))^2}{2\sqrt{\pi}}$, then $n=1/4$.

I also know that $B(x, y) = \frac{(\Gamma(x))(\Gamma(y))} {\Gamma(x+y)} = 2 \int_{0}^{\pi/2} \sin^{2x-1}{\theta} \cos^{2y-1}{\theta} d\theta$.

So I suppose I should reduce the given integral to a form similar to the one above. Is there any trigonometric property that can help me that or am I seeing it wrong?

Answer 1

We find \begin{align*} \int_{0}^{\pi/2} \left( \frac{1}{\sin^3{\theta}} - \frac{1}{\sin^2{\theta}} \right)^{1/4} \cos{\theta}\, d\theta &= 2\int_0^{\pi/2} \sin^{-1/2}t \cos^{3/2}t \, dt, & \theta\rightarrow \arcsin \sin^2 t \end{align*} which is of the form $$2\int_0^{\pi/2} \sin^{2x-1}t \cos^{2y-1}t \, dt,$$ for $(x,y) = (1/4, 5/4)$.