Solving for $f\left(\sqrt{x^2+z^2}\right)$ given $g(z)=\int_{-\infty}^{+\infty}{\left(f\left(\sqrt{x^2+z^2}\right)\frac{1}{\sqrt{x^2+z^2}}\right)}dx$

Question

How would I solve for a function $f\left(\sqrt{x^2+z^2}\right)$ given $g(z)=\int_{-\infty}^{+\infty}{\left(f\left(\sqrt{x^2+z^2}\right)\frac{1}{\sqrt{x^2+z^2}}\right)}dx$? I understand that derivatives are the inverse operation of indefinite integral but I'm not sure what the inverse operation of a definite integral from $-\infty$ to $\infty$ is. How would I solve for $f\left(\sqrt{x^2+z^2}\right)$ given a certain function of $g(z)$?

Answer 1

A hint.

$$g(z)=\int_{-\infty}^{\infty} f \left( \sqrt{x^2+z^2} \right) \frac{dx}{\sqrt{x^2+z^2}}=$$

$$=2\int_{0}^{\infty} f \left( \sqrt{x^2+z^2} \right) \frac{dx}{\sqrt{x^2+z^2}}=\int_{z}^{\infty} f \left( r \right) \frac{dr}{\sqrt{r^2-z^2}}$$

Where $r=\sqrt{x^2+z^2}$.

We obtained an integral equation, namely a linear Volterra equation of the first kind, because one of the integration limits is variable.

$$\frac{1}{2}g(z)=\int_{z}^{\infty} f \left( r \right) \frac{dr}{\sqrt{r^2-z^2}} \tag{1}$$

Where integral kernel is:

$$K(z,r)=\frac{1}{\sqrt{r^2-z^2}}$$.

The methods for solving Volterra equation and other integral equations are present in various literature. You can look something up here for example.

If think the problem can be reduced to Abel equation which has an analytic solution.

Here's how you can do it.

Make a change of variable:

$$r=\frac{1}{\sqrt{y}}$$

Then (1) transforms to:

$$\frac{1}{2}g(z)=\frac{1}{2} \int_{0}^{1/z^2} \frac{1}{y} f \left( \frac{1}{\sqrt{y}} \right) \frac{dy}{\sqrt{1-z^2 y}}$$

$$z g(z)=\int_{0}^{1/z^2} \frac{1}{y} f \left( \frac{1}{\sqrt{y}} \right) \frac{dy}{\sqrt{1/z^2-y}}$$

Now set:

$$p=\frac{1}{z^2}$$

$$G(p)=z g(z)$$

$$F(y)=\frac{1}{y} f \left( \frac{1}{\sqrt{y}} \right)$$

Finally you obtain:

$$G(p)=\int_{0}^{p} F(y) \frac{dy}{\sqrt{p-y}} \tag{2}$$

Which is exactly Abel equation. The solution is provided in the linked file.