An integral related to $\Gamma(\tfrac14)^2$

Question

I converted $\Gamma(\tfrac14)^2=16\Gamma(\tfrac54)^2$ to a double integral by definition of Gamma function. By using polar coordinates I eliminated one integral and found $12\sqrt{\pi}$ times the integral $$\int_0^{\tfrac\pi 2}\frac{(\sin\theta\cos\theta)^{\tfrac14}}{(\cos\theta+\sin\theta)^{\tfrac52}}d\theta\tag1$$ I couldn't proceed further to show that $(1)$ is equal to $\frac13K(\tfrac1{\sqrt2})$ where $K(k)$ is the complete elliptic integral of the first kind.

Can anybody proceed further?

Answer 1

The integral in question: $$I=\frac{2^{3/4}}{2}\int_{0}^{\pi/2}\frac{\sin{(2x)}^{1/4}}{(\cos{x}+\sin{x})^{5/2}}dx \\\overset{x=x/2}=\frac{2^{3/4}}{4}\int_{0}^{\pi}\frac{\sin{(x)}^{1/4}}{(\cos({x/2})+\sin({x/2}))^{5/2}}dx \\ \overset{x=2\arctan{t}}= \frac{1}{12}\int_{0}^{\infty}\frac{dt}{t^{3/4}\sqrt{1+t}}.\tag{1}$$ In the other hand: $$K(\frac{1}{\sqrt{2}})=\sqrt{2}\int_{0}^{\pi/2}\frac{dx}{\sqrt{1+\cos^2{x}}}\\\overset{x=2\arctan{t}}=2\int_{0}^{1}\frac{dt}{\sqrt{1+t^4}}\\\overset{t=t^{1/4}}=\frac{1}{2}\int_{0}^{1}\frac{dt}{t^{3/4}\sqrt{1+t}}\\\overset{t=\frac{1}{t}}=\frac{1}{2}\int_{1}^{\infty}\frac{dt}{t^{3/4}\sqrt{1+t}}\tag{2}.$$ Then with $(1)$ and making the sum of last two equalities in $(2)$ we have: $$\int_{0}^{\infty}\frac{dt}{t^{3/4}\sqrt{1+t}}=4K(\frac{1}{\sqrt{2}})=12I \implies I=\frac{1}{3}K(\frac{1}{\sqrt{2}}).\tag{3}$$

Remarks

• In the first step we have used: $$\sin({2x})=2\sin{x}\cos{x}\tag{a}$$.
• To get the integrals in $t$ we have used: $$\cos({\arctan{t}})=\frac{1}{\sqrt{1+t^2}}\tag{b},$$ and $$\sin({\arctan{t}})=\frac{t}{\sqrt{1+t^2}}\tag{c}.$$ With $(a)$, $(b)$ and $(c)$ then: $$\sin({2\arctan{t}})=\frac{2t}{1+t^2}\tag{d},$$ and $$\cos({2\arctan{t}})=\frac{2}{1+t^2}-1\tag{e}.$$