General procedure for inverse of an integral transform

Question

Is there a general inversion formula or procedure for an integral of the form (where f is the function being transformed and g depends on the type of transform) $\int^{a}_{b} f(x) g(x,\xi) dx $ ? Inverses are defined in the conventional ways for functionals and integral transforms, respectively.

For instance, for the Fourier transform. In the equation above, $a=∞,b=-∞$, $g(x,\xi)=e^{-2πix \xi }$. I know the inverse Fourier transform is simple but I am concerned with a general procedure or process.

I am most interested in the cases where $a,b=±∞$ although a simple inverse for $\int^{a}_{b} f(x) dx $ is also something I am curious about (as far as this part of the question is concerned, if the inverse of an indefinite integral is the derivative, what is the inverse of a definite integral— I am sorry if this is too elementary).

Feel free to use complex analysis or any other branch of math if it helps to answer the question. Also, you can repost on another site if it will help.

Answer 1

I believe OP is asking:

given a function $\varphi(\xi)$, can one find a function $f$ such that $\varphi(\xi) = \int_{a}^{b}f(x)g(x,\xi)dx$?

This is essentially the theory of integral equations. Tables of integral equations and their solutions can be found, a standard one is Polyanin and Manzhirov's. This book also has several chapters dedicated to methods for solving such equations. Many of the solutions are given in terms of series, but they are exact solutions.

Some special cases are easier. The Wikipedia article on the Fredholm integral equation gives the following example of a "general inversion formula or procedure" in the case when $a=-\infty,b=\infty,$ $g$ is continuous, and $g$ is a function of the difference of the arguments, so $g(x,\xi) = g(\xi-x)$. In this case $\varphi(\xi) = \int_{-\infty}^\infty f(x)g(\xi-x)dx$ is a convolution and can $f$ can therefore be written in terms of the Fourier transforms $\mathcal{F}(g)(t)$ and $\mathcal{F}(\varphi)(t)$: \begin{align*} f(x) = \int_{-\infty}^\infty \frac{\mathcal{F}(\varphi)(t)}{\mathcal{F}(g)(t)}e^{2\pi i t x}dt, \end{align*} if this exists. Existence is apparently a difficult question, see the discussions here and here.

Equations of the type $\varphi(\xi) = \int_a^b f(x)g(x,\xi)dx$ can sometimes be shown to be equivalent to differential equations, which is done by differentiating under the integral sign. Several examples of this for Fredholm and Volterra equations are done in the first chapter of Wazwaz.

Wazwaz, Abdul-Majid, A first course in integral equations, Singapore: World Scientific (ISBN 978-981-4675-11-6/hbk; 978-981-4675-12-3/pbk). xiv, 312 p. (2015). ZBL1327.45001.

Polyanin, Andrei D.; Manzhirov, Alexander V., Handbook of integral equations, Boca Raton, FL: CRC Press. 787 p. (1998). ZBL0896.45001.

Answer 2

I think what Michael is saying in his comment is that the question is way too general: you are trying to deal with arbitrary linear operators on an infinite dimensional vector space, so invertibility (let alone a simple formula) is a subtle question that can't be answered in a simple fashion. You wonder about the inverse operation of definite integration over a fixed interval $I$; integration yields a real number (assuming the integral even exists!), so for $a\in\mathbb{R}$, "$\int_I^{-1}a = $ any function $f$ with $\int_If=a$", so to speak (sorry for the horrible notation). This is satisfied by loads of functions, so the inverse doesn't exist.

So if you have a specific operator in mind, you could try to see if it's invertible and if you can find a nice expression for the inverse. The general question doesn't make much sense.

Answer 3

Probably a "general procedure" would be as follows: Find appropriate Hilbert spaces $X$ and $Y$ such that the operator $Lf(\xi) \int_a^b f(x) g(x,\xi) dx$ maps boundedly from $X$ to $Y$. Often the space $X = L^2([a,b])$ and $Y= L^2([c,d])$ works (e.g. if $g\in L^2([a,b]\times [c,d])$). Then there always exists the Moore-Penrose pseudoinverse of $L$. However, if the range of $L$ is not a closed subspace of $y$, it is an unbounded operator defined on $\text{range}L\oplus \text{range}L^\bot$ which is a dense subspace of $Y$.

This works for $Lf = \int_a^b f(x)dx$ as a mapping from $L^2([a,b])$ to $\mathbb{R}$ and it is a nice exercise to work out the pseudo-inverse.

Answer 4

If you can find an eigenfunction relation $\int^{a}_{b} f(x) g(x,\xi) dx = \lambda f(x)$ where $\lambda^n=1$ then you have an inverse by repeating the transformation n-1 times.