# How do Integral Transforms work

It's been a while since I have learnt Laplace's Transform and I am taking a look at Fourier's. But I feel I know nothing about them, just how to use in calculations. So I would like to have

Any explanations or books on why and how Integral Transforms work. I do know most of applications, however I'm tired of repeating steps. I want to try things out....

Thanks

• Related question at Mathoverflow; the top answer links to a video that you might find helpful. Feb 7, 2020 at 19:46
• I'm gonna watch it, thanks! Feb 7, 2020 at 19:48
• I have watched. So summarizing, if I want create one, just set things like he showed and that is it? Feb 8, 2020 at 20:48

## 1 Answer

There are many classes of problems that are difficult to solve - or at least quite unwieldy algebraically - in their original representations. An integral transform "maps" an equation from its original "domain" into another domain. Manipulating and solving the equation in the target domain can be much easier than manipulation and solution in the original domain. The solution is then mapped back to the original domain with the inverse of the integral transform. They have been successfully used for almost two centuries in solving many problems in applied mathematics, mathematical physics, and engineering science.

General formula: An integral transform is any transform $$~\text T~$$ of the following form: $$F(u)={\displaystyle (Tf)(u)=\int _{t_{1}}^{t_{2}}f(t)\,K(t,u)\,dt}$$ The input of this transform is a function $$~f~$$, and the output is another function $$~Tf~.~$$ An integral transform is a particular kind of mathematical operator.

There are numerous useful integral transforms. Each is specified by a choice of the function $$~K~$$ of two variables, the kernel function, integral kernel or nucleus of the transform.

Of course, the interpretation of this new function $$~F(u)~$$ will depend on what the function $$~K(t,u)~$$ is. Choosing $$~K(t,u)=0~$$, for example, will mean that $$~F(u)~$$ will always be zero. But this tells us nothing about $$~f(t)~$$. Whereas choosing $$~K(t,u)=t^u~$$ will give us the $$~u^\text{th}~$$ moment of $$~f(t)~$$ whenever $$~f(t)~$$ is a probability density function. For $$~u=1~$$ this is just the mean of the distribution $$~f(t)~$$. Moments can be really handy.

A particularly interesting class of functions $$~K(t,u)~$$ are ones that produce invertible transformations (which implies that the transform destroys no information contained in the original function). Some kernels have an associated inverse kernel $$~K^{−1}(u, t)~$$ which (roughly speaking) yields an inverse transform: $${\displaystyle f(t)=\int_{u_{1}}^{u_{2}}(Tf)(u)\,K^{-1}(u,t)\,du}$$

Whenever this is the case, we can view our operation as changing the domain from $$~t~$$ space to $$~u~$$ space. Each function $$~f~$$ of $$~t~$$ becomes a function $$~F~$$ of $$~u~$$ that we can convert back to $$~f~$$ later if we so choose to. Hence, we’re getting a new way of looking at our original function!

• The Fourier transform :

It turns out that the Fourier transform, which is one of the most useful and magical of all integral transforms, is invertible for a large class of functions. We can construct this transformation by setting:

$$K(t,u) = e^{-i t u}\qquad\text{and}\qquad K(t,u) = e^{i t u}$$

which leads to a very nice interpretation for the variable $$~u~$$. We call $$~F(u)~$$ in this case the “Fourier transform of $$~f~$$”, and we call $$~u~$$ the frequency. Why is $$~u~$$ frequency ? Well, we have Euler’s famous formula: $$e^{i t u} = \cos(t u) + i \sin(t u)$$ so modifying $$~u~$$ modifies the oscillatory frequency of $$~\cos(tu)~$$ and $$~\sin(tu)~$$ and therefore of $$~K(t,u)~$$. There is another reason to call $$~u~$$ frequency though. If $$~t~$$ is time, then $$~f(u)~$$ can be thought of as a waveform in time, and in this case $$~|F(u)|~$$ happens to represent the strength of the frequency $$~u~$$ in the original signal. You know those bars that bounce up and down on stereo systems ? They take the waveforms of your music, which we call $$~f(t)~$$, then apply (a discrete version of) the Fourier transform to produce $$~F(u)~$$. They then display for you (what amounts to) the strength of these frequencies in the original sound, which is $$~|F(u)|~$$. This is essentially like telling you how strong different notes are in the music sound wave.

• The Laplace transform : $$k(t,u) = e^{-tu}$$ This is handy for making certain differential equations easy to solve.

• The Hilbert transform : $$k(t,u) = \frac{1}{\pi} \frac{1}{t-u}$$ This has the property that (under certain conditions) it transforms a harmonic function into its harmonic conjugate, elucidating the relationship between harmonic functions and holomorphic functions, and therefore connecting problems in the plane with problems in complex analysis.

• The identity transform : $$k(t,u) = \delta(t-u)$$ Here $$~\delta~$$ is the dirac delta function. This is the transformation that leaves a function unchanged, and yet it manages to be damn useful.

References:

"Integral Transforms and Their Applications", by Lokenath Debnath and Dambaru Bhatta

"Mathematics for Physical Science and Engineering" by Frank E. Harris

https://www.askamathematician.com/2011/07/q-what-are-integral-transforms-and-how-do-they-work/

https://en.wikipedia.org/wiki/Integral_transform

How to learn Integral Transform?

https://mathoverflow.net/questions/2809/intuition-for-integral-transforms

• First, thanks for your answer! Surely, I'll get these books to read. Secondly, let's suppose I had set $K(t,u) = \sqrt{tu}$, how can I find the inverse Kernel? Feb 14, 2020 at 13:39
• @Mr. N. This is another kind of question you asked. I think you should have to ask it as an another different new question. Feb 14, 2020 at 14:42
• Fine, no problem. Thanks Feb 14, 2020 at 17:04
• You are welcome @Mr.N Feb 14, 2020 at 17:27