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

Since you already received a good answer, another possible solution using the substitution proposed by @gt6989b in comments. $$\int \left(\frac1{u^3} - \frac1{u^2}\right)^{1/4}\, du=2 \sqrt[4]{(1-u) u}+2 \sqrt[4]{u} \,\,\, _2F_1\left(\frac{1}{4},\frac{3}{4};\frac{5}{4};u\right)$$ $$\int_0^1 \left(\frac1{u^3} - \frac1{u^2}\right)^{1/4}\, du=2\,_2F_1\left(\frac{1}{4},\frac{3}{4};\frac{5}{4};1\right)=\frac{2 \sqrt{2 \pi }\, \Gamma \left(\frac{5}{4}\right)}{\Gamma \left(\frac{3}{4}\right)}=\frac{\left(\Gamma(1/4)\right)^2}{2\sqrt{\pi}}$$

Answer 2

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 $(x,y) = (1/4, 5/4)$.

Answer 3

Enforce the substitution $\sin(\theta)=\sin^2(t)$ to find that

$$\begin{align} \int_0^{\pi/2} \left( \frac1{\sin^3(\theta)}-\frac1{\sin^2(\theta)}\right)^{1/4}\,\cos(\theta)\,d\theta&=2\int_0^{\pi/2}\cos^{3/2}(t)\sin^{-1/2}(t)\,dt\\\\ &=B(5/4,1/4)\\\\ &=\frac{\Gamma(5/4)\Gamma(1/4)}{\Gamma(3/2)}\\\\ &=\frac{\frac14 \Gamma^2(1/4)}{\frac12 \Gamma(1/2)}\\\\ &=\frac{\Gamma^2(1/4)}{2\sqrt{\pi}} \end{align} $$

as was to be shown!

Answer 4

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[15px,#ffd]{\int_{0}^{\pi/2}\bracks{{1 \over \sin^{3}\pars{\theta}} - {1 \over \sin^{2}\pars{\theta}}}^{1/4}\cos\pars{\theta}\,\dd\theta} \\[5mm] \stackrel{t\ =\ \sin\pars{\theta}}{=}\,\,\,& \int_{0}^{1}\pars{{1 \over t^{3}} - {1 \over t^{2}}}^{1/4}\,\dd t = \int_{0}^{1}t^{-3/4}\pars{1 - t}^{1/4}\,\dd t = {\Gamma\pars{1/4}\Gamma\pars{5/4} \over \Gamma\pars{3/2}} \\[5mm] = &\ {\Gamma\pars{1/4}\bracks{\pars{1/4}\Gamma\pars{1/4}} \over \Gamma\pars{1/2}/2} = \bbox[15px,#ffd,border:1px solid navy]{{\Gamma^{2}\pars{1/4} \over 2\root{\pi}}}\ \approx\ 3.7081 \end{align} Note that $\ds{\Gamma\pars{1/2} = \root{\pi}}$.