Recently I have come across this book of Integral Transforms. At this point I realized that there are so many transforms with different kernel out there. Now to provide a little bit of context I would summarize my thoughts concerning Integral Transforms:
First of all there are I would call the classical transforms such as the Fourier Transform or the Laplace Transform, given by
$$\mathcal{F}\{f(t)\}(\omega)~=~\int_{-\infty}^{\infty}f(t)e^{-i\omega\cdot t}~\mathrm{d}t~~~~~~\mathcal{L}\{f(t)\}(s)~=~\int_{0}^{\infty}f(t)e^{-st}~\mathrm{d}t,$$
which are widely used within the theory of differential equations, probability or at the analysis of signal processing. Beside them there are transforms like the Mellin Transform or the Hankel Transform, given by
$$\mathcal{M}\{f(t)\}(s)~=~\int_{0}^{\infty}f(t)t^{s-1}~\mathrm{d}t~~~~~~H_{\nu}\{f(t)\}(u)~=~\int_{0}^{\infty}f(t)J_{\nu}(ut)t~\mathrm{d}t,$$
which are there for some more specific manner since I did not encountered them so often and it is even harder to find transform tables, like this one which contains a lot of transformed functions, than e.g. for the Fourier Transform. To go even further lately I have found out that there are some transform I honestly speaking never heard of before like the Wavelet Transform
$$\mathcal{W}_{\psi}\{x(t)\}(a,b)~=~\frac1{a}\int_{-\infty}^{\infty}\overline{\psi\left(\frac{t-b}{a}\right)}f(t)\mathrm{d}t,$$
where $\psi$ denotes a wavelet as far as I understand, or the two types of the Narain Transform
$$g(z)~=~\int_{0}^{\infty}k(z,y)f(y)~\mathrm{d}y~~~~~~f(z)~=~\int_{0}^{\infty}h(z,y)g(y)~\mathrm{d}y$$
where the kernels $k(z,y)$ and $h(z,y)$ are given by
$$\begin{align} k(z,y)~&=~2\gamma(zy)^{\gamma-1/2}G^{m,p}_{p+q,m+n}\left(\left.\begin{array} \text{\textbf{a}_\textbf{p}},\textbf{b}_\textbf{q}\\\textbf{c}_\textbf{m},\textbf{d}_\textbf{n}\end{array}\right|(zy)^{2\gamma}\right)\\ k(z,y)~&=~2\gamma(zy)^{\gamma-1/2}G^{n,q}_{p+q,m+n}\left(\left.\begin{array} \text{-}\textbf{b}_\textbf{q},-\textbf{a}_\textbf{p}\\-\textbf{d}_\textbf{n},-\textbf{c}_\textbf{m}\end{array}\right|(zy)^{2\gamma}\right) \end{align}$$
where $G^{m,n}_{p,q}$ denotes the Meijer G-function. Hence the complexity of the kernel is widely scattered I am not sure why there are so many different transforms.
Since some of them can written in terms of other transforms, for example the Mellin Transform and the Fourier Transform
$$\mathcal{M}_f(is)~=~\sqrt{2\pi}\mathcal{F}^{-1}(s)~~~~~~\mathcal{B}_f(i\omega)~=~\mathcal{F}_f(\omega)$$
I cannot justify the variety of transforms for myself. Surely some of them are way more useful hence one can transform nearly every single function because the properties of the kernel allow it. On the other hand some of them seem to be to specific for a small area of application to be useful somewhere else.
I am interested in a more general approach to this whole topic of integral transforms. Therefore I would be satisfied if someone could explain to me why there are so many different integral transforms and not one single generalized which somehow contains all the other ones as special cases.
Thanks in advance!
I don't know what sort of answer you expect to such a broad question. I'm about to answer a few of the questions you ask. I suspect in each case when you read my answer you'll be inclined to say "Well duh, I knew that...". You might want to refrain from saying that out loud, because if you do my reply will be "Fine. That is the answer to the question - if you already knew the answer then why did you ask?".
A: Each of the transforms you mention is used for something. Something that the other transforms on the list don't do.
Of course you knew that.
A: They are. By definition if $T$ is an "integral transform" then $T$ is a special case of $$Tf(x)=\int K(x,t) f(t)\,dt.$$
In that much generality there's nothing interesting that can be said about $T$.
Of course you knew that.