Puzzle in a Calculation

Question

I want to calcalate $a$ from $$\frac{d a}{d u} = \frac{\sqrt{2}}{4\pi} \oint_\alpha \frac{dx}{\sqrt{(x-1)(x+1)(x-u)}}$$ where $\alpha$ is the contour only around cut between -1 and 1. First I want to integrate $u$ and I find some trouble.

Directly integrating $u$ gives $$ a= \frac{-2\sqrt{2}}{4\pi} \oint_\alpha \frac{dx \sqrt{(x-u)}}{\sqrt{(x-1)(x+1)}}$$

If we first change the sign of $u$ $$ \frac{d a}{d u} = \frac{\sqrt{2}}{4\pi} \oint_\alpha \frac{dx}{\sqrt{(x-1)(x+1)(x-u)}} \\ =\frac{\sqrt{2}}{4\pi} \oint_\alpha \frac{dx}{\sqrt{(-x+1)(x+1)(-x+u)}}$$ and then integrating $u$ gives $$ a= \frac{2\sqrt{2}}{4\pi} \oint_\alpha \frac{dx \sqrt{(-x+u)}}{\sqrt{(-x+1)(x+1)}}$$

The answers seem different.

What is wrong with the above calculation?

Thanks in advance.

Answer 1

EDIT: "Fuller" answer at asker's request

Well, let's look at an example $1 = \sqrt{1\cdot 1}=\sqrt{-1\cdot -1} = \sqrt{-1} \cdot \sqrt{-1} = i^2 = -1$, which is ridiculous. Where in this calculation did I go wrong? What have you implicitly assumed in your calculations? We have a branch point of $f(z) = z^{1/2}$ at $z=0$ perhaps?

The fact that $\sqrt{ab} = \sqrt{a}\cdot \sqrt{b}$ is only an identity on $\mathbb{R}^+$; it does not hold when you extend your number system to the complex plane. See the following for more details:

Why $\sqrt {-1}\cdot \sqrt{-1}=-1$ rather than $\sqrt {-1}\cdot \sqrt{-1}=1$. Pre-definition reason!