Consider the following equation
$$ \int_0^\infty f(\sqrt{x(x-a)}) dx $$
For $a>0$ real and some analytic function $f(z)$ which dies off sufficiently fast for $\Re[z]>0$ and $\Im[z]>0$ so that the integral $\int_0^{\pi/2} R e^{i\theta} d\theta \, f(\sqrt{ R e^{i\theta}( R e^{i\theta} - a)}) \rightarrow 0$ as $R \rightarrow \infty$.
The square root's branch cut above is chosen such that when $0<x<a$, $\sqrt{x(x-a)} = i \sqrt{x(a-x)}$ (the "usual way" along the negative axis such that $\sqrt{-1} = i$).
My question is: How can I deform this integral to the imaginary axis while taking into account the branch cuts properly? Naively, I want to say:
$$ \int_0^\infty f(\sqrt{x(x-a)}) dx= i \int_0^\infty f(\sqrt{i x(i x-a)}) dx. $$
However, this seems to run afoul of the branch cut as I defined it since if I take a small $\delta>0$ and let $0<x<a$:
$$\sqrt{(x+i \delta)(x+i \delta - a)} \sim \sqrt{x(x- a) + i \delta(2x - a)} \sim \begin{cases} i \sqrt{x(a-x)}, & \mathrm{if}\; x>a/2 \\ -i\sqrt{x(a-x)} & \mathrm{if}\; x<a/2. \end{cases}$$
Thus, just rotating $x\rightarrow i x$ seems to not work. If possible, I'd like a solution that somehow curtails integrating around the branch cut and gives me an integral like $\int_0^\infty f(\sqrt{i x(i x-a)}) dx$.
Non-ideal solution (let me know if this is wrong or if its my only option): it seems like
$$\int_0^\infty f(\sqrt{x(x-a)}) dx= i \int_0^\infty f(\sqrt{i x(i x-a)}) dx+\int_0^{a/2} (f(i\sqrt{x(a-x)})-f(-i\sqrt{x(a-x)})) dx.$$
