# Inverse formula for a function inside an integral.

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 $$g(x)$$?

• Certainly, there is no reason to think $g$ is unique. Sep 20, 2021 at 17:08
• Looks like a Laplace transform to me - and there is an inverse. Sep 20, 2021 at 17:09
• Is that true for all $a$ and some $b$ and $c$ that depend on $a$, or a particular set of values of $a,b,c$? Sep 20, 2021 at 17:09
• It would be better to change variable; let $t=-\log a$, so that $t\in (0, \infty)$ and the integral reads $$\int_0^\infty g(x)e^{-xt}\, dx,$$ which is the one-sided Laplace transform of $g$. Sep 20, 2021 at 17:12

$$\int_0^\infty g(x)a^xdx=\frac{1}{{(2^c-a^c)}^{{b}/{c}}}$$ 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 $$\int_0^\infty f(x)\,dx=1$$

• Or you can use any $f$ with $\int_0^\infty f(x)\,dx=1$ instead if $ke^{-kx},$ of course. Sep 20, 2021 at 17:23
• Thank you for your answer. The expression of g doesn't coincide with the Laplace's antitransformation of the function $\frac{1}{(2^c-a^c)^{\frac{b}{c}}}$? Is this correct? Sep 21, 2021 at 9:32
• @m91c a special case would be using the Dirac delta distribution, letting $g(x)=\delta (x)k$ where you can set $k$ as $\frac{1}{(2^c-a^c)^{\frac{b}{c}}}$, in fact you can also set $g(x)=h(x)\delta(x)k$ as long as $h(0)=1$, so I wouldn't say that the inverse Laplace as you've said, coincide Sep 21, 2021 at 10:30