Problem:
Evaluate $\displaystyle\int_0^1 (1-x^4)^{1/2} \mathrm{d}x$
My approach: I substitute $u=x^4$, then $\mathrm{d}u = 4x^3 \mathrm{d}x$, so $\mathrm{d}x = \dfrac{1}{4} u^{-3/4} \mathrm{d}u$, so I want to integrate
$$\frac 14 \int_0^1 (1-u)^{1/2} u^{-3/4} \mathrm{d}u$$
Is this integrable by Gamma functions? Do the identity
$\displaystyle\int_0^1 (1-u)^{a} u^b \mathrm{d}u = \dfrac{\Gamma(a) \Gamma(b)}{\Gamma(a+b+1)}$ hold in general?
Best Answer
Not quite. For complex $\alpha,\beta$ with $\Re\alpha,\beta>0$, we have: $$\mathscr{B}(\alpha,\beta):=\int_0^1t^{\alpha-1}(1-t)^{\beta-1}\,\mathrm{d}t=\frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}$$
So you want to compute $\frac{1}{4}\mathscr{B}(1/2+1,-3/4+1)=\frac{1}{4}\mathscr{B}(3/2,1/4)$. Note that shifting the index gives: $$\int_0^1t^{\alpha}(1-t)^{\beta}\,\mathrm{d}t=\frac{\Gamma(\alpha+1)\Gamma(\beta+1)}{\Gamma(\alpha+\beta+\color{red}{2})}$$
Using the Beta-Gamma identity and basic properties of $\Gamma$, this becomes: $$\frac{1}{4}\frac{\Gamma(3/2)\Gamma(1/4)}{\Gamma(7/4)}=\frac{1}{4}\frac{\frac{1}{2}\Gamma(1/2)\Gamma(1/4)}{\frac{3}{4}\Gamma(3/4)}=\frac{\sqrt{\pi}}{6}\cdot\frac{\Gamma(1/4)}{\pi\csc(\pi/4)/\Gamma(1/4)}=\frac{1}{6\sqrt{2\pi}}\Gamma^2(1/4)$$
No "closed" form of $\Gamma(1/4)$ in terms of more elementary constants or functions is known, as far as I know. $\Gamma$ seems only to be known "exactly" when evaluated at the half integers.