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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.

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    $\begingroup$ What do you mean by an "inversion formula or procedure"? The problem you pose is a natural infinite dimensional generalization of inverting a matrix. If there were an "inversion formula," that would be great, but... There is no such thing as an inverse for $\int_a^b f(x) dx$ (with fixed a and b), since a lot of functions have the same integral. $\endgroup$ Apr 9, 2012 at 22:29
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    $\begingroup$ If we ignore the second part, I still think the first part has value. What is the procedure for inverting a general integral transform? $\endgroup$ Apr 10, 2012 at 23:50
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    $\begingroup$ As previous commenters have mentioned, the question is asking for too much. But perhaps the following rephrasing of the question, which is moving the difficult part to something slightly more concrete will be of use. Given integral operators A and B with kernels a and b, i.e. (Af)(x) = int_R a(x,y)f(y)dy and (Bf)(x) = int_R b(x,y)g(y)dy, their composition is the integral operator AB with kernel c(x,y) = int_R a(x,z)b(z,y)dz. So long as c(x,y) = delta(x-y), then B is a (right) inverse of A. $\endgroup$ Apr 11, 2012 at 1:58
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    $\begingroup$ This is a very interesting question, but to broad to have a real answer. A suggestion is that you formulate is as big-list question, then you have to make it community-wiki. $\endgroup$
    – Marc Palm
    Apr 11, 2012 at 7:57
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    $\begingroup$ I would think of this as analogous to category theory and kan extensions and relational division. The residual $(Q/P)(x, y) = \forall t. P(x, t) \rightarrow Q(y, t)$ is adjoint to composition $(P \circ Q)(x, y) = \exists t. P(x, t) \wedge Q(t, y)$. You want something for $(A \circ B)(x, y) = \int A(x, t) B(t, y) \text{d}t$ integral transforms and not profunctors or spans or relations. Also IIRC you need very nicely behaving stuff to use Fubini's theorem and so on. I'm not good at calculus and got here looking for basically the same question $\endgroup$ Jul 16, 2022 at 18:48

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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.

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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.

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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.

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    $\begingroup$ Have you got references for this nice exercise? If I'm doing this right it ought to be the linear transformation that takes $1$ to the constant function $f:x\mapsto\frac{1}{b-a}$. $\endgroup$ Apr 11, 2012 at 11:55
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    $\begingroup$ Sorry, $\frac{1}{\sqrt{b-a}}$. $\endgroup$ Apr 11, 2012 at 11:57
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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.

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