Evaluate $\int_0^1 \frac{dx}{(x^2-x^3)^\frac13}$ using contour integration

Question

Evaluate $\int_0^1 \frac{dx}{(x^2-x^3)^\frac13}$ using contour integration

I try to proceed by first calculating residues of $$f(x) = \frac{1}{(x^2-x^3)^\frac13} = \frac{1}{x\left(\frac1x-1\right)^\frac13}$$

Then answer to integral can be provided by Cachy's Residue theorem = $2i\pi(\text{Residue at } x=0 +\text{ Residue at } x=1)$

Above function has simple pole at $x=0$ but I am unable to find out what kind of singularity $f$ has at $x= 1$.

Is this the right approach? Or is there any way to expand given function(Laurent's series) to find out all residues at once?

Answer 1

This doesn't work. For one thing, the residue theorem is only valid around closed contours. But the bigger problem is that this function is not analytic near $0$ or $1$.

There is a sneaky way to solve this by making a branch cut along the interval $[0,1]$ and using the residue at $\infty$. Here, you would integrate over a very narrow oval encircling the interval $[0,1]$ and pay close attention to the fact that the values approaching $[0,1]$ on either side of the interval are not quite the same. (They differ by multiplication by $e^{2\pi i/3}$ or $e^{4\pi i /3}$, depending on which way you look at it.)

But it's important that you apply the residue theorem to the outside of this contour, since the function has a branch cut within it.

The easier way is to use the beta function $$B(s,t) = \int_0^1 x^{s-1} (1-x)^{t-1} \, \mathrm{d}x = \frac{\Gamma(s)\Gamma(t)}{\Gamma(s+t)}$$ and the Euler reflection formula $$\Gamma(s)\Gamma(1-s) = \frac{\pi}{\sin(\pi s)},$$ and notice that your integral is $$\int_0^1 x^{-2/3} (1-x)^{-1/3} \, \mathrm{d}x = \frac{\Gamma(1/3) \Gamma(2/3)}{\Gamma(1)} = \frac{\pi}{\sin(\pi/3)}.$$

Answer 2

Another variation on a solution theme in Finding the integral $I=\int_0^1{x^{-2/3}(1-x)^{-1/3}}dx$

Use the logistic map $x=\frac{1}{1+e^w}$, $1-x=xe^w$, $x^2-x^3=x^3e^w$, to map the interval $w\in(-∞,∞)$ to $x\in(0,1)$, moving the branch points to $w=\pm\infty$. Then $dx=x^2e^w\,dw$ so that $$ \int_0^1\frac{dx}{(x^2-x^3)^{1/3}}=\int_{-∞}^∞\frac{x^2e^wdw}{xe^{w/3}}=\int_{-∞}^∞\frac{e^{2w/3}}{1+e^w}dw $$

Now consider $w=u+iv$ restricted to the rectangle bounded by the lines $u=\pm R$, $v=0$, $v=2\pi$ which has one pole at $w=i\pi$ inside with residue $$ \lim_{w\to i\pi}\frac{e^{2w/3}(w-i\pi)}{1+e^w} =\lim_{h\to 0}\frac{e^{2(i\pi+h)/3}h}{1-e^h}=-e^{i2\pi/3} $$ The contributions from the sides $\pm R+i[0,2\pi]$ go to zero with $R\to\infty$ like $e^{-2R/3}$ resp. $e^{-R}$. The integral over the top segment $[-R,R]+i2\pi$ gives a phase-shifted multiple of the integral of the bottom segment $[-R,R]+i0$, $$ \int_{-R+i2\pi}^{R+i2\pi}\frac{e^{2w/3}}{1+e^w}dw =e^{i4\pi/3}\int_{-R}^{R}\frac{e^{2w/3}}{1+e^w}dw $$ so that the residual theorem gives in the limit $R\to\infty$ $$ -i2\pi\,e^{i2\pi/3}=(1-e^{i4\pi/3})\int_{-∞}^∞\frac{e^{2w/3}}{1+e^w}dw $$ or $$ \int_0^1\frac{dx}{(x^2-x^3)^{1/3}}=\frac{i2\pi}{e^{i2\pi/3}-e^{-i2\pi/3}}=\frac{\pi}{\sin(2\pi/3)}. $$