# 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)

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.

• Interesting! I wanted to try this in simulation, but having trouble finding any $g$ such that $\int_t^\infty f(s) ds$ is finite. Sep 6, 2021 at 18:11
• @YaroslavBulatov Try $g(x) = x+1$ or $g(x) = x + \alpha$ for any positive real constant $\alpha$. Sep 6, 2021 at 18:56
• Pretty neat trick! Leaving this up for a few days in case someone can discover the equivalent that works for $g(x)=\frac{1}{(1+x)^2}$ Sep 7, 2021 at 8:42
• Do you mean to have $\mathcal{L}^{-1}$ at the end? Sep 7, 2021 at 15:19
• Indeed, thanks for catching that @ChipHurst Sep 7, 2021 at 16:40

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.