Let $p\in[0,1]$ and $\rho(x): [0,1] \rightarrow [0,\infty)$ such that
$$\int_0^1 dx \rho(x) = 1.$$ I'd like to investigate the following transcendental equation:
$$\frac{1}{2p} = \int_0^{1} dx x^{\gamma -1} \rho(x)$$
In particular I wish to find bounds for $\gamma$ and say something about the uniqueness of the solution. I tried to re-formulate the problem by employing the first mean value theorem for integration, i.e.
$$\int_0^{1} dx x^{\gamma -1} \rho(x)= \xi^{\gamma-1} \int_0^{1} dx \rho(x)=\xi^{\gamma - 1}$$
with $\xi\in [0,1]$. That is I'd like to know something about the uniqueness and possible bounds for $\gamma$ in
$$\frac{1}{2p} = \xi^{\gamma - 1}$$
Is this a achievable target? Any ideas? I suppose when restricting $\gamma$ to the real numbers one has
$$\gamma = 1- \frac{\log{(2p)}}{\log{(\xi)}}$$
But what about the uniqueness of $\xi$ in the first mean value theorem for integration or $\gamma$ can also be complex...
Fix $p$ and $\rho(x)$. By uniqueness, I think you are asking if you can have two distinct values $a,b$ such that $\int_0^1 x^a\rho(x)dx = \int_0^1 x^b\rho(x)dx = 1/(2p)$.
Restricting to the reals: If a function $f(x)$ is non-negative and satisfies $\int_0^1f(x)dx = 0$, then $f(x)=0$ almost everywhere. So, for example, suppose $0\leq a < b$. Then $(x^a - x^b)\rho(x) \geq 0$ for all $x \in [0,1]$. Thus, if $\int_0^1(x^a-x^b)\rho(x)dx = 0$, we know $(x^a-x^b)\rho(x)=0$ almost everywhere. But $x^a=x^b$ only for $x=0$ or $x=1$. Thus $\rho(x)=0$ almost everywhere, which contradicts $\int_0^1\rho(x)dx=1$.
In the complex case: If $z$ is a complex number with a nonzero imaginary part, and if $\int_0^1 x^z\rho(x)dx = 1/(2p)$, then this equality also holds for the complex conjugate $\overline{z}$, and so the answer is not unique. But that requires finding out if there is such a $z$. This can depend on $\rho(x)$.
Example 1: Suppose $\rho(x)=1$ for $x \in [0,1]$. Then for any real number $c\neq 0$, the equation $\int_0^1 x^z\rho(x) dx = c$ has one and only one solution $z$, and that solution is real. Specifically, $z=1/c - 1$.
Example 2: Suppose $\rho(x) = 2$ for $x \in [0,1/2]$, $\rho(x)=0$ else. Then $\int_0^1 x^z\rho(x) dx = \frac{(1/2)^{z}}{z+1}$ can be the same real number for multiple complex solutions $z$. For example, if $z=a+ib$ then $\int_0^1 x^z\rho(x) dx$ is real whenever $a = -b\cot(b\log(2))-1$. If $b=5$ then $\int_0^1 x^z\rho(x)dx$ is real and satisfies $\int_0^1 x^z\rho(x)dx = \int_0^1x^{\overline{z}}\rho(x)dx > 3844$.