Integral transforms, like the Fourier and Hankel transforms, are not always easy to compute (at least not in closed form). However when dealing with the case of the "dirac delta" (aka "unit impluse" function) then any integral transform becomes a very easy operation. If you fully understand what the dirac delta function is and how it works then you will understand why it is the most easy of all functions to integral transform.
So the short anwser is: Given another function other than the impluse and the Hankel transform may well be difficult to compute (due to the fact that it gets multiplied by a Bessel function). But for the dirac delta function it's easy all the same.
Why is it easy? Well in general all integral transforms take the form of :
F(w) = Integral(x= -infty..+infty, f(x) * K(w,x) dx),
where F(w) is the transform of f(x).
In the specific case where f(x) is a dirac delta, eg f(x) = delta(x-a), then irrespective of the complexity or otherwise of K(w,x), the integral is still very easy to find. It follows directly from the properties of the delta function that the above integral is simply equal to K(w,a). That is, F(w) = K(w,a) is the transform of f(x) = delta(x-a), and that is true regardless of what particular type of integral transform that you're doing ( that is, the exact nature of K(w,x) ).
If you're still unsure then I'd recommend a little bit of reading and studying of the properties of the dirac Delta function, particularly those relating to it's integration, to become fully familiar with why it is so easy to integral transform.
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