I came across the term "integral transform", and I'm curious what exactly this means, and whether integration by parts would be considered an integral transform.
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2$\begingroup$ Examples of integral transforms include Laplace, Fourier, Mellin, Hankel, and Bessel transforms. $\endgroup$ – Mark Viola Jul 29 '16 at 20:41
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1$\begingroup$ No, int by parts is not considered a transform. At best it's kind of the analogue to the product rule for derivatives. $\endgroup$ – user307169 Jul 29 '16 at 20:46
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Integral transforms look like
$$ \int_a ^b K(s, t) f(t) \ dt $$
where the function $K$ is called the kernel. The function $K$ acts as a catalyst, changing the form of the function $f$ fed into the function.
As @Dr. MV has said, the Laplace and Fourier transforms are common examples. The Laplace transform is
$$ \mathcal{L}(f) = \int_0 ^{\infty} e^{-st}f(t) \ dt $$
and the Fourier (well, the convention used in the text I'm reading) is
$$ \mathcal{F}(f) = \frac{1}{2\pi}\int_{-\infty} ^{\infty} e^{i\omega x} f(x) \ dx $$
Integration by parts is not an integral transform; it is a method we use to turn hard integrals into easier ones. You'll usually use integral transforms in differential equations. They're very powerful there.
answered Jul 29 '16 at 20:46
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Just a comment:
One thing is what a "integral transform" is. Another, for what it is useful to. The beauty of Laplace Transform for example, is to change a "differential equation" problem into a "algebraic" one. So, instead of solving the differential equation directly, you first transfom it (to the "s" variable), solve the algebraic equation, and transform back by a Laplace Inverse transformation to get the solution in the original variable. Very common aplication over electrical circuits on steady-state analysis: instead of analyzing the problem over "time" (using t as a variable) you transform it over the "frecuency space" (using s as a variable) and make the analysis as you need.
