Inverse of integral transform $f(s)=\int_0^\infty g(x) \exp(-s g(x)) \mathbb{d}x$

Given $g(x)$ defined for positive reals, say $f(s)$ is defined as below $$f(s)=\int_0^\infty g(x) \exp(-s g(x)) \mathbb{d}x.$$

Is there a relationship to named integral transforms, or a generic approach to obtain $g$ from $f$?

For instance, we can show the following holds through trial and error (notebook) $$ \begin{align} f(s) & = \frac{\sqrt{\pi }\ \text{erf}\left(\sqrt{s}\right)}{2 \sqrt{s}} &g(x)&=\frac{1}{(1+x)^2}\\ f(s)& = \frac{\Gamma \left(\frac{2}{3}\right)-\Gamma \left(\frac{2}{3},s\right)}{3 s^{2/3}} &g(x)&=\frac{1}{(x+1)^3}\\ f(s)& =\frac{-\cosh (s)+\sinh (s)+1}{s} &g(x)&=\frac{1}{e^{x}}\\ f(s)&=\frac{e^{\left.\frac{1}{4}\right/s} \sqrt{\pi } (2 s+1)}{4 s^{5/2}} &g(x)&=\log ^2(x)\\ f(s) & = \frac{e^{-s} \left(-s+e^s-1\right)}{s^2} & g(x)&=\text{max}(0, 1-x)\\ f(s) & = \frac{1}{(s-1)^2} &g(x)&=\log(1+x)\\ f(s)&=\frac{e^{-s} (s+1)}{s^2} &g(x)&=1+x\\ \end{align}.$$

Edit Sep 7 Partial solution below works for increasing $g(x)=x+1$, but not for decreasing ones like $g(x)=(1+x)^{-2}$

Motivation: if loss after running $s$ steps of gradient descent steps on quadratic $Q$ decays as $f(s)$, then $i$th eigenvalue of Q quadratic decays as $g(i)$, hence knowing shape of loss curve over time would let us infer shape of quadratic $Q$ (background)

Solution 1:

Partial solution: Under certain conditions, this can be done using the inverse Laplace transform.

Let's restrict $g$ to be an increasing bijection $g: [0,\infty) \to [0,\infty)$, such that the integral defining $f$ converges. By integrating both sides from $s=t$ to $\infty$, we obtain

$$\begin{align} \int_t^\infty f(s)ds &= \int_t^\infty \int_0^\infty g(x)e^{-sg(x)}dxds \\ &= \int_0^\infty \int_t^\infty g(x)e^{-sg(x)}dsdx \\ &= \int_0^\infty e^{-tg(x)}dx \\ &= \int_0^\infty (g^{-1})'(x) e^{-tx}dx \\ \end{align}$$

Thus, we have that $\int_t^\infty fds$ is the Laplace transform of $(g^{-1})'$. This means that $g^{-1}(y)$ is given by

$$\int_0^y \mathcal{L}^{-1}\bigg\{ \int_t^\infty f(s)ds\bigg\}(x) dx$$

from which the original $g$ can be recovered by inverting this expression.

Solution 2:

The answer of Franklin Pezzuti Dyer is good, however, it involves one unnecessary assumption, namely that $g: [0,\infty)\rightarrow [0,\infty)$. This is not required. Let us do the substitution $y=g(x)$ in the original integral. This means $\mathrm{d}y=g'(x) \mathrm{d}x$ and $a=g(0)$, $b=g(\infty)$. Thus we have $$f(s)=\int_{a}^{b}\mathrm{d}y\, \frac{1}{g'(x)}ye^{-s y} =\int_{a}^{b}\mathrm{d}y\, \left(g^{-1}\right)^\prime(y)\, y\,e^{-s y}, $$ where $g^{-1}(y)=x$, i.e., is the inverse of $g(x)$. In other words $g^{-1}(g(x))=x$. The derivative of the inverse function $g^{-1}(y)$ is denoted as $\left(g^{-1}\right)^\prime(y)$. What remains to be done is to use the Heaviside function $\theta(y)$ to bring the last integral to the standard Laplace form. Let us assume that $0\le a < b$. We can write then $$\int_{a}^{b} u(x)\mathrm{d}x = \int_{0}^{\infty} u(x)\left\{\theta(x-a)-\theta(x-b)\right\}\mathrm{d}x.$$ One should change a sign if $b>a$. With this in hands we can write $$f(s)=\int_{0}^{\infty}\mathrm{d}y\,\left\{\theta(y-a)-\theta(y-b)\right\} \left(g^{-1}\right)^\prime(y)\, y\,e^{-s y}.$$

Assume now that the inverse Laplace transform of $f(s)$ exists $\mathcal{L}^{-1}[f].$ Then we have $$\left\{\theta(y-a)-\theta(y-b)\right\} \left(g^{-1}\right)^\prime(y)\, y=\mathcal{L}^{-1}[f](y).$$ Of course, this is only a formal solution. It covers also the result of Franklin Pezzuti Dyer as a partial case of $a=0$ and $b=\infty$ considering that $\mathcal{L}^{-1}[f]\in [0,\infty).$

Now, how the determination of $g(x)$ should be performed? As the first step one performs the inverse Laplace transform and looks at the codomain of $\mathcal{L}^{-1}[f]$. This yields the constants $a$ and $b$. Next one solves the differential equation for $g(x)$ with these boundary conditions. Notice, that one interesting observation can be made about the table in OP. One can compute $\mathcal{L}^{-1}[f]$ for many presented functions, they indeed feature $\theta$-functions consistent with the formal solution.