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 $g(x)$?

  • $\begingroup$ Certainly, there is no reason to think $g$ is unique. $\endgroup$ Sep 20, 2021 at 17:08
  • 1
    $\begingroup$ Looks like a Laplace transform to me - and there is an inverse. $\endgroup$
    – Ron Gordon
    Sep 20, 2021 at 17:09
  • $\begingroup$ Is that true for all $a$ and some $b$ and $c$ that depend on $a$, or a particular set of values of $a,b,c$? $\endgroup$
    – Henry
    Sep 20, 2021 at 17:09
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    $\begingroup$ 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 $g$. $\endgroup$ Sep 20, 2021 at 17:12

1 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 $$\int_0^\infty f(x)\,dx=1 $$

  • $\begingroup$ Or you can use any $f$ with $\int_0^\infty f(x)\,dx=1$ instead if $ke^{-kx},$ of course. $\endgroup$ Sep 20, 2021 at 17:23
  • $\begingroup$ 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? $\endgroup$
    – user268193
    Sep 21, 2021 at 9:32
  • $\begingroup$ @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 $\endgroup$
    – wd violet
    Sep 21, 2021 at 10:30

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