Prove that $\int^2_0 x(8-x^3)^\frac{1}{3}dx=\frac{16\pi}{9\sqrt{3}}$

Question

I have to prove that:

$$\int^2_0 x(8-x^3)^\frac{1}{3}dx=\frac{16\,\pi}{9\sqrt{3}}.$$

I tried substituting $x^3=8u$ but I just got stuck.

Any help would be appreciated.

Answer 1

By setting for first $x=2z$ we have: $$I=\int_{0}^{2}x(8-x^3)^{1/3}\,dx = 8\int_{0}^{1}z(1-z^3)^{1/3}\,dz \tag{1}$$ and by setting $z=t^{1/3}$ we have: $$I=\frac{8}{3}\int_{0}^{1}t^{-1/3}(1-t)^{1/3}\,dt = \frac{8}{3}\cdot\frac{\Gamma\left(\frac{2}{3}\right)\Gamma\left(\frac{4}{3}\right)}{\Gamma(2)}\tag{2}$$ through Euler's Beta function. Exploiting the reflection formula for the $\Gamma$ function: $$ I = \frac{8}{9}\,\Gamma\left(\frac{1}{3}\right)\Gamma\left(\frac{2}{3}\right)=\frac{8\pi}{9\sin\frac{\pi}{3}}=\color{red}{\frac{16\,\pi}{9\sqrt{3}}}\tag{3}$$ as wanted.

Answer 2

Just for fun, i will give a solution using tools from complex analysis

Let's use a substitution $x=2y^{\frac{1}{3}}$. We get $$ I=\frac{8}{3}\int_{0}^{1}y^{\frac{-1}{3}}(1-y)^{\frac{1}{3}}dy $$

Now (we interpret $y$ as a complex variable), we can choose the principal branch of $\log(y)$ to be the positive real axis. Then carefully adding the phases coming from both parts of the product, we see that effectivly we have created a branch cut in the interval $[0,1]$ . Furthermore in this region it holds that:

$\text{Arg}(\Re[y]+i\delta)=0$ and $\text{Arg}(\Re[y]-i\delta)=2 \pi $ ,where $\delta \rightarrow 0^+$ .

Applying the residue theorem to a contour which encloses the branch cut counter clockwise, we get

$$ I\times(1-e^{ 2 \pi i /3})=-2 \pi i\times \text{res}(y=\infty) $$

With our choice of branches $\text{res}(y=\infty)=-\frac{e^{i \frac{\pi}{3}}}{3}$

Putting everything together this gives us

$$ I=\frac{16 \pi}{9\sqrt3{}} $$