Integral to Gamma function

Question

For this function i need to convert it to either a Gamma or a Beta function $$\int_0^1 \frac{3}{(1-x^3)^\frac{1}{3}} \,dx$$

I know I need to make the substitution $x^3=z$ but I am unsure where to go from here

Answer 1

For simplicity, define

$$\mathcal I := \int_0^1 \frac{3}{(1-x^3)^\frac{1}{3}} \,dx$$

Make the $u$-substitution $u=x^3$ as you prescribed. Then:

• $x = u^{1/3}$
• $dx = \frac 1 3 u^{-2/3} du$

Insert both of these definitions into the integral and pull out constants via linearity. (They happen to cancel.) We find that

$$\mathcal I = \int_0^1 u^{-2/3} (1-u)^{-1/3}du$$

Recall that the definition of the beta function is given by

$$\text B(x,y) := \int_0^1 t^{x-1}(1-t)^{y-1}dt$$

$\mathcal I$ now matches this same form for appropriate $x,y$:

• $-2/3 = x-1 \implies x = 1/3$
• $-1/3 = y-1 \implies y = 2/3$

Thus,

$$\mathcal I = \text B \left( \frac 1 3, \frac 2 3 \right) \approx 3.6276$$

WolframAlpha numerically confirms this answer: their answer for the integral and for the beta function are at the links. If you want to convert to a gamma function version, note the identity

$$\text B(x,y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$$

In this case, then (since $\Gamma(1/3 + 2/3) = \Gamma(1) = 1$),

$$\mathcal I = \Gamma \left( \frac 1 3 \right) \Gamma \left( \frac 2 3 \right)$$

We can even simplify this result further by use of Euler's reflection formula and get a better closed form. The reflection formula is given by

$$\Gamma(z)\Gamma(1-z) = \frac{\pi}{\sin(\pi z)}$$

Take $z = 1/3$ and how it might be utilized for $\mathcal I$ is clear. Then

$$\mathcal I = \frac{\pi}{\sin(\pi/3)} = \frac{\pi}{\sqrt{3} / 2} = \frac{2 \pi}{\sqrt 3}$$

(Credits to user Ty who pointed out the simplification owing to Euler's reflection formula in the comments!)