I'm currently struggling with something that came up in my studies. I'm trying to integrate a multivalued function like the square root on a given path, specifically a function with two branch points, with the integration path encircling the branch cut. I then want to use the theorem of residues and calculate the residue at infinity, employing the usual substitution $z \to \frac{1}{z}$ so that the new integrand is $-\frac{1}{z^2}f(\frac{1}{z})$, and then compute the residue at zero.
The problem with this procedure is that I often find the wrong sign. I suspect that this is because the substitution somehow leads me to the other sheet of the Riemann surface, because when I compare the new integrand in the limit of small $z$ with the old one for large $z$ i get opposite signs. I wonder if this is indeed what is happening, why it happens and if there is a way to choose the right sign in the first place.
Thank you very much in advance!
EXAMPLE: consider the integral $I=\int_0^1\sqrt{x(1-x)}dx$. Define $f(z)=\sqrt{z(1-z)}$ (taking the positive square root) above the branch cut on the interval $[0,1]$, and continue analytically so that below the cut the integral of $f(z)$ is $-I$. Therefore, if $\gamma$ is a clockwise path encircling the cut, $\int_\gamma f(z)dz=2I$. Now compute the integral by means of the residue at infinity, that is $2I=2\pi i Res(-\frac{1}{z^2}f(\frac{1}{z}),0)$, having made the substitution and noted that if we stretch $\gamma$ all the way to infinity we get a counterclockwise path around the point at infinity (I've checked with other simple integrals to make sure this consideration is correct). Now the new integrand is $-\frac{1}{z^3}\sqrt{z-1}=-\frac{i}{z^3}\sqrt{1-z}$, and its residue at zero is $\frac{i}{8}$. We see that in this way $I=-\frac{\pi}{8}$, which is obviously wrong, and in fact $f(z)$ for large $z$ goes something like $i$ times a large "positive" number (by which I mean if $z$ is a positive real number then the function is also positive), while $-\frac{i}{z^3}\sqrt{1-z}$ for small $z$ has an extra minus sign in front.
Best Answer
you used $\sqrt{wz} = \sqrt z \sqrt w$ and $\sqrt{z^2} = z$, and even $\sqrt{-1} = i$ when simplifying the new integrand and never checked which branches were involved along the way nor in the final result in $\sqrt{z-1}$, and thus had a 50% chance of making a sign mistake.
You defined $f(z)$ as the branch of $\sqrt{z(1-z)}$ that was positive just above the cut $[0;1]$. You can argue then that this is the branch where $f(z) \sim -i z$ as $|z| \to \infty$, and also this is the branch where $f(2)= -i\sqrt 2$ (where $\sqrt 2$ here means the good old positive real square root of $2$). But still, at this point you have to assume that $\sqrt z$ doesn't have any meaning and that you can only write $\sqrt{z(1-z)}$ as only this has a well-defined value.
Now when you talk about the new integrand, you have to define $g(z) = - \frac 1 {z^2} f(\frac 1 z)$. Then $g(1/2) = -4 f(2) = +4i\sqrt 2$.
Now if you were to let go of precautions, you can write $g(z) = - \frac 1 {z^2} \sqrt{\frac 1z(1-\frac 1z)} = - \frac 1 {z^2} (\pm \sqrt{\frac{z-1}{z^2}}) $.
This last square root function never got given a well-defined value among the possible twos, so I had to include a $\pm$ sign.
And then we continue and use the well-known wrong rules $\sqrt{wz} = \sqrt w \sqrt z, \sqrt{z^2} = z$ and $\sqrt{-1} = i$ (they are only true up to signs), to get $g(z) = - \frac i {z^3} (\pm \sqrt{1-z})$.
For $g$ we want the cut to be the image of $[0;1]$ (the cut of $f$) by the inversion, so $[1;+\infty)$.
Now I suppose an easy way to give meaning to $\sqrt{1-z}$ is to pick the branch whose value at $0$ is $1$, which actually corresponds to the principal square root of $1-z$.
When $z=\frac 12$ this $\sqrt{1-z}$ evaluates to $\sqrt{\frac 12} = \frac {\sqrt 2}2$ where that last square root is the good old positive real square root. So then we have $+4i\sqrt 2 = g(1/2) = - 4i(\pm \sqrt 2)$, and we find out that our $g(z)$ must incorporate a minus sign.
So we have found that if $\sqrt{\frac 1z(1-\frac 1z)}$ means the square root used in the definition of $f$ evaluated at $1/z$, and if $\sqrt{1-z}$ is the principal square root, then $\sqrt{\frac 1z(1-\frac 1z)} = - \frac {\sqrt{1-z}}z$