show that
by considering fourier transforms of
when x>b
My advice, is obviously to calculate the fourier transform of these functions, and then use the convolution formula, but in this case use the fact that exp (i b k)=cos(bk)+isin(bk) to replace this with sin(bk) in the integrand.
if f(k),g(k) are the fourier transforms of f(x),g(x), and h(z) is the convolution of f and g then, then h(x)=F^-1(f(k)g(k)) where F^-1 is the inverse fourier transform.
I think that the function g(x) ought to be a rectangular function. In other words, it is 1 on some interval ([–1,1], [–1/2,1/2] or [-π,π], depending on which definition of the Fourier transform you are using), and 0 elsewhere. (g obviously can't be 1 on the whole real line because then it would not have a Fourier transform.)
The integral represents the value at x of (a multiple of) the inverse Fourier transform of the product of two functions, namely and . But the sinc function is the Fourier transform of a rectangular function, and the function is the Fourier transform of . So the inverse Fourier transform of their product will be the convolution of the rectangular function and that exponential function.
If and then .
That looks promising, because it gets the right answer apart from a constant multiple. However, I have cheated by using the function instead of . That doesn't matter provided that x is positive. This means that, in the integral, we want x–t to be positive whenever ; and that in turn means that I am assuming that . I claim that this is a reasonable assumption. Indeed, if it does not hold then the function will not have a Fourier transform (in the conventional sense: as the Captain points out, it will have a distributional Fourier transform of some sort), and so this whole method of evaluating the integral will become suspect.