Approximating Integrals with Deriv & Discont: Help Needed!

[zhermes]

I'm trying to do an integral, in which the integrand is composed of arbitrary functions and contains a derivative of the variable of integration. Further still, the integral is over a discontinuity. Because these are arbitrary functions I assume there is no exact solution, but I'm looking for an approximation.

Anyway, there is a discontinuity at 'x'
$$\lim_{\Delta x \to \infty} \int_{x-\Delta x}^{x+\Delta x} \frac{1}{g(r)} f'(r) dr$$

The functions are defined in the limits on each side of the discontinuity, i.e.
$$\lim_{\Delta x \to 0} \hspace{0.3in} f(x+\Delta x) = A \hspace{0.3in} f(x-\Delta x) = B \hspace{0.3in}<br /> g(x+\Delta x) = C \hspace{0.3in} g(x-\Delta x) = D$$

Thus, if the integrand did not contain g(r), the result would be simple (from Leibniz's rule)
$$\lim_{\Delta x \to \infty} \int_{x-\Delta x}^{x+\Delta x} f'(r) dr = A - B$$

Any help, tips, or pointers would be greatly appreciated!
Thanks,
Z

 

[pmsrw3]

I don't think there's enough information to answer. Obviously, the Riemann integral formally doesn't exist, since f' doesn't exist at x. If one imagines f as the limit of a series of functions in which the region within which the change from A to B occurs gets smaller and smaller, then the value of the integral will depend on the nature of the limiting series. This is obvious, because one could have all the change occur on the minus side, in which case the integral will be (A-B)/D, or you could have it all occur on the plus side, in which case the integral will be (A-B)/C. Yet in both cases the limit of the series is the same discontinuous function f.

 

[zhermes]

I don't think there's enough information to answer.

I think I understand the basis of your point, but it seems as though it should apply to the integral over f'(r) itself [i.e. without the 1/g(r)]; but that integral is indeed defined [i.e. (A-B)].

Further, the only difference is that g is not a derivative. But I think, given the particular situation, one could say that:
$$g(x) \equiv h'(x)$$ $$\lim_{\Delta x \to 0} h(x+\Delta x) = h(x - \Delta x) = h_0$$
Does that help / change anything?

 

[pmsrw3]

I think I understand the basis of your point, but it seems as though it should apply to the integral over f'(r) itself [i.e. without the 1/g(r)]; but that integral is indeed defined [i.e. (A-B)].

Further, the only difference is that g is not a derivative.

"The only difference" between what and what?

The key difference between f' and f'/g is that g is discontinuous, and it is discontinuous at the exact same point as f. My argument doesn't apply to f' (or even to g, if C = D), because a limiting series of functions that converges to f will give the same integral for f', whether f changes from A to B below x or above x. It it only when f is multiplied by another function that is discontinuous exactly at x that this limit fails.

But I think, given the particular situation, one could say that:
$$g(x) \equiv h'(x)$$ $$\lim_{\Delta x \to 0} h(x+\Delta x) = h(x - \Delta x) = h_0$$
Does that help / change anything?

Not in any way that I can see.