Inverse formula for a function inside an integral.

Question

If I know that $$\int_0^\infty g(x)a^xdx=\frac{1}{{(2^c-a^c)}^{{b}/{c}}}$$ where $a\in [0,1]$ and $b$ and $c$ are fixed constant. Is it possible to get the expression for $g(x)$?

Answer 1

$$ \int_0^\infty g(x)a^xdx=\frac{1}{{(2^c-a^c)}^{{b}/{c}}} $$ Set $\displaystyle{g(x) = a^{-x}{ke^{-kx}\over{(2^c-a^c)}^{{b}/{c}}}}$

The integral becomes

$$ \int^\infty_0{ke^{-kx}\over{(2^c-a^c)}^{{b}/{c}}}dx={1\over{(2^c-a^c)}^{{b}/{c}}} $$

Since we can set any value for $k$ in $g(x)$, we can't expect for a unique solution

As Thomas Andrews said, in general, you can substitute any function $f(x)$ for $g(x)$ such that $$g(x)=f(x){a^{-x} \over{(2^c-a^c)}^{{b}/{c}} }$$ where $$\int_0^\infty f(x)\,dx=1 $$