I'm interested in evaluating $\displaystyle \int_{0}^{1} \frac{1}{(x^2-x^3)^{1/3}} \ dx $ using contour integration and the residue at infinity.
But I'm not sure how to define $\displaystyle f(z) = \frac{1}{(z^2-z^3)^{1/3}}$ so that it is well-defined on the complex plane if the line segment $[0,1]$ is omitted.
And then once defined, how do you determine the value of $f(z)$ just above the branch cut and just below the branch cut?
Best Answer
Let us analyze the function for each of the branch points $z=0$ and $z=1$ separately:
Around $z=0$, the function $$f_0(z) = z^{-2/3} + O(z^{1/3}).$$ We choose the branch cut such that it extends to $z>0$. Thus, we have $$f_0(z) = |z|^{-2/3} e^{-i (2/3)\arg z} + O(z^{1/3})$$ with $0 <\arg z < 2\pi$. Just above the the branch cut (at $z=0^+ + i 0^+$), we have $\mathop{\rm arg} z = 2\pi$ and thus $f_0(z) = |z|^{-2/3} e^{-i 4\pi/3} = |z|^{-2/3} e^{i 2\pi/3}$. Just below the branch cut (at $z=0^+ - i 0^+$), we have $\arg z =0$ and thus $f_0(z) = |z|^{-2/3} .$
At $z=1$, we have a somewhat simpler structure. We can write the function $$f_1(z) = |z^2- z^3|^{-1/3} e^{-i(1/3) [\arg (z^2 -z^3) + 2\pi n]}$$ which has a branch cut along the real line for $z<1$. Here, $n\in\mathbb{Z}$ will be chosen below such that the two expressions $f_0$ and $f_1$ coincide for $z$ close to 0.
So, let us see what happens at $z=0$:
Just above the branch cut, the function $f_1(z)$ assumes the form $$f_1(x+i 0^+) = |x^2 - x^3|^{-1/3} e^{-i 2\pi n/3} $$
Just below the branch cut, the function $f_1(z)$ assumes the form $$f_1(x-i 0^+) = |x^2 - x^3|^{-1/3} e^{-i 2\pi/3 - i 2\pi n/3} $$
We observe that for $n=-1$ we have found a consistent branch cut structure, where the branch cut only extends between $z=0$ and $z=1$.
The function just above the branch cut reads $$f(x + i 0^+) = | x^2 -x^3|^{-1/3} e^{i 2\pi /3} .$$ Just below the branch cut, we have $$f(x - i 0^+) = | x^2 -x^3|^{-1/3} .$$ For $x>1$, we have $$f(x)= f_1(x) = |x^2 -x^3|^{-1/3} e^{-i(1/3)(\pi -2\pi)} = |x^2 -x^3|^{-1/3} e^{i \pi/3}.$$ For $x<1$, we have $$f(x)= |x^2 - x^3|^{-1/3} e^{-i 2\pi/3}.$$
In general for $|z| \to \infty$ the function assumes the form $$f(z) = |z|^{-1} e^{-i\arg (z)+i \pi/3} = \frac{e^{i \pi/3}}{z}$$ which gives you the residue $e^{i\pi/3}$ at infinity.
Thus, we have $$\int_{0}^{1} [f(x - i 0^+) - f(x + i 0^+) ] \ dx = -2 \pi i e^{i \pi/3}$$ or in other words $$\int_{0}^{1} \frac{1}{(x^2-x^3)^{1/3}} (1- e^{i 2\pi/3}) \ dx = -2 \pi i e^{i \pi/3}$$ from which you obtain (e.g., taking the imaginary part) the result $$\int_{0}^{1} \frac{1}{(x^2-x^3)^{1/3}} = \frac{2\pi}{\sqrt{3}}.$$