Cauchy Principal Value IntegralDate: 09/17/2002 at 23:33:36 From: John Harter Subject: Improper Integral Theory Why isn't an integral from negative infinity to positive infinity defined as the limit as t (or any other variable) approaches infinity of the integral from -t to t? In other words, why must we split an integral into two halves and evaluate each as a one-sided improper integral (as the standard definition of such an integral states)? For example, it would be intuitively pleasing for the integral of x^3 (or another odd function like x / (x^2+1), x, etc.) from negative infinity to positive infinity to be zero. Yet this integral diverges under the present definition because each side diverges! The fact that the sides are equal and opposite doesn't even come into play. This does not make sense to me. If someone has some insight into why mathematicians haved defined integrals from negative infinity to positive infinity this way, I would greatly appreciate it if they would share their knowledge with me. I know there must be some good reason for it. Thank you. Date: 09/18/2002 at 03:13:50 From: Doctor Pete Subject: Re: Improper Integral Theory Hi John, Your question raises a very good point, one that is often inadequately discussed in undergraduate calculus courses. However, it is important to realize that what may make intuitive sense is not necessarily what is consistent with other mathematical concepts. In calculus, we must be careful in deciding what it really means to compute an improper integral, and more generally, the related notions of convergence and indeterminate forms. But should you feel like the type of improper integral you are describing does not have its own meaning or significance in mathematics, do not worry. You have just described what is known as the Cauchy principal value integral. There are essentially two types: the first is the doubly infinite improper integral, e.g., G = Integrate[F[x], {x,-Infinity,Infinity}] for which the Cauchy principal value is defined as PV(G) = Limit[Integrate[F[x], {x,-R,R}], R -> Infinity]. The second is where F has a discontinuity at some point c on the interval [a,b]. Then the improper integral H = Integrate[F[x], {x,a,b}] has Cauchy principal value PV(H) = Limit[Integrate[F[x], {x,a,c-e}] + Integrate[F[x], {x,c+e,b}], e -> 0]. In both cases, the idea is to consider the improper integrals as limits that are symmetrically computed about the point of discontinuity (in the first case, the point of discontinuity is at infinity). In light of this, you are probably wondering why the improper integral itself is not defined to be the Cauchy principal value. After all, why talk about the two things as if they were separate? The simple answer is that they are indeed not the same thing at all. The key difference here is that the improper integral must satisfy a more strict criterion for convergence than the CPV, as required under the definition of integral as a Riemann sum. We need to distinguish these two things because not every function has a CPV integral, and those that do, do not necessarily have a convergent improper integral. So a distinction is made between the two because there are contexts in which it is inappropriate to give an integral a CPV, and others in which it is necessary. We have now arrived at the heart of your question: Why not give a function like F[x] = x^3 the Cauchy principal value when integrating over the entire real line? Well, the function itself increases (decreases) without bound as x tends to infinity (-infinity), and therefore talking about the value of the improper integral is like saying Infinity - Infinity = 0, which is not necessarily the case as a discussion of indeterminate forms and L'Hopital's Rule would reveal. In fact, we don't even need to consider a function that is unbounded at infinity, only that the improper integral over a half-infinite subinterval is unbounded. Simply put, the symmetry of the function is unimportant in light of the fact that we are dealing with an indeterminate form. Another way to look at it: What if you shifted the point at which you partition the real line? You would not want the value of the improper integral to depend on where you choose this point. And while it may be the case that there is an "obvious" choice for functions like F[x] = x^3, what about arbitrary polynomials of odd degree, which are not necessarily symmetric about the origin or any other point? It is pretty obvious that if you shift this point, the value of the improper integral will change. You could always choose to split the real line at the origin, but this merely amounts to computing the Cauchy principal value. Hence the CPV of doubly infinite integrals lacks a very important quality: independence from the underlying coordinate system. Therefore, we want to maintain a sufficiently strict interpretation of the improper integral so that when it exists, it is independent of the choice of limits of integration. Violation of this property is considered inconsistent with the development of calculus, more so than trying to preserve some aspect of symmetry in the integrand. That is the crux of the matter. Still, the CPV exists and is not without its uses, for example, in the computation of Hilbert transforms, or other more advanced topics in real and complex analysis. But it is not generally needed in elementary calculus, so it is not discussed. I hope I've answered your question to your satisfaction, long as my response may be. You've asked an excellent question, one that needs careful explanation and is not easily dismissed. If I may say so, it reveals your potential as a mathematician! If you want further clarification, or have any other questions, feel free to write. - Doctor Pete, The Math Forum http://mathforum.org/dr.math/ Date: 09/18/2002 at 11:51:49 From: John Harter Subject: Thank you (Improper Integral Theory) Doctor Pete, Thank you very much for your answer. It was very clear and helpful (and I much prefer long responses to short ones!). I knew there was some important reason for defining an improper integral this way, and I now know that reason! Thank you again, John Harter. |
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