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I haven't been able to find a good answer to this searching around online. There is a related old question here, but it never received much attention.

Suppose I have some physical property that I believe depends on $\int_{-\infty}^{\infty}xdx$. Ordinarily, we would say the integral is undefined, but we could also hypothetically take the Cauchy principal value and say the integral evaluates to zero.

My question is under what conditions should I let the integral remain undefined and when should it equal zero? It seems odd to me that based on different circumstances of the problem we could say that the same integral has different values. For pure math, as long as you work within a consistent framework, it doesn't really matter which you decide is true. But for physics, chemistry, etc if the integral relates to some property, it would seem there would be some definitive, empirical solution to the problem.

This question arises from some problems I have been doing with the Cauchy distribution. It seems that for certain physical examples, there is certain camp that treats the location parameter as the mean of the distribution, which is effectively saying the integral to determine the mean is equal to its principal value. I don't like this because it seems to suggest to me that all the higher order odd moments could also be argued to exist, at least in case where the distribution is symmetric about zero,

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    $\begingroup$ Basically it makes sense if symmetry of the integration domain is imposed by the model or by the math. For example, if the problem is due to insufficient decay at infinity, it may be that you are really not integrating from $-\infty$ to $\infty$, physically, but are rather integrating from $-M$ to $M$ and are estimating the result. If that symmetry of the integration domain comes from the physics, then PV is appropriate for your problem. $\endgroup$ – Ian Sep 29 '17 at 20:44
  • $\begingroup$ This example of the Cauchy distribution IMO is easily resolved by plotting sample means. Go into Matlab and run plot(cumsum(tan(pi*rand(10000,1)-pi/2))./(1:10000)') and compare the results to plot(cumsum(randn(10000,1))./(1:10000)'). There is something fundamentally different going on there. $\endgroup$ – Ian Sep 29 '17 at 20:54
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The Cauchy principal value is very important, especially in cases where the Lebesgue integral (which it seems you refer to as the improper integral) does not exist. The issue is that the Lebesgue integral doesn't deal too well with really big oscillations. Indeed, a measurable function $f$ is Lebesgue integrable if and only if $|f|$ is, so oscillations don't matter - only the magnitude does. This becomes problematic when dealing with functions such as $\frac{\sin x}{x}$ on $(0,\infty)$ or $\frac{1}{x}$ on $(-1,1)\setminus\{0\}$.

A purpose of the Cauchy principal value is to rectify this problem, to take into account oscillations like the Riemann integral does and give a meaningful number that represents the integral (i.e. scaled average) of the function in question. The Cauchy p.v. of $\frac{1}{x}$ is $\lim_{\epsilon \to 0} \int_{-1}^{-\epsilon} \frac{1}{x}dx+\int_{\epsilon}^1 \frac{1}{x}dx = 0$, which coincide with our intuition for what the average value of $\frac{1}{x}$ should be.

The most prominent use of the Cauchy principal value is the Hilbert transform, in which we study $\int \frac{f(x-y)}{y}dy$, which of course, needs to be defined properly. It is critical here that we don't just say the improper integral exists, but rather get a quantitative sense of the oscillation present.

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    $\begingroup$ I don't think I was entirely clear with my question. I recognize the use of the Cauchy principal value. My question is more in regards to what justifies the use of it or what is its meaning. Like you say the Cauchy principal value for that integral of 1/x is 0, but when should I just go ahead and say that is the value and when should I just leave it undefined. If there aren't rigorous criteria, it would seem I can arbitrarily decide the value of the integral to suit my need. $\endgroup$ – Tyberius Sep 29 '17 at 20:58
  • $\begingroup$ Of course you arbitrarily decide the value of the integral to suit your need. That's what all of mathematics is. The integral is not something that is just lurking in the air. We define it (and try to define it naturally) but of course I can't "prove" to you that the Cauchy p.v. is the correct value. What would you expect such a "proof" to entail $\endgroup$ – mathworker21 Sep 29 '17 at 21:03
  • $\begingroup$ I'm not sure if $\int_{-\infty}^\infty \frac{\sin x}x\,dx$ is the best example here, as the issue is Lebesgue integral vs Riemann integral, not Riemann integral vs Cauchy principal value. Cases of interest are when the Riemann (or Lebesgue) improper integral does not exist, but Cauchy principal value does. $\endgroup$ – mechanodroid Sep 29 '17 at 21:08
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    $\begingroup$ @mechanodroid The issue is essentially the same, it is all about the order in which positive and negative contributions to the integral are summed. $\endgroup$ – Ian Sep 29 '17 at 21:15
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    $\begingroup$ IMHO a good summary $\endgroup$ – Jyrki Lahtonen Sep 30 '17 at 5:33
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It isn't the "correct value" for the integral any more than the principal root is the correct value for a root or the principal logarithm is the correct value for a logarithm, or setting $C=0$ gives you the correct value of the antiderivative. There are plenty of other values the integral can take, depending on how you take the limit. See my answer to Why can't $\int_{-1}^1\frac{dx}x$ be evaluated?

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A physics related example may also be useful. Consider a potential of the form $V=-Log(\left|x\right|)$. Then the force is given by $$ F=-{d V \over d x}={1\over x}. $$ The work done on a particle which moves from $-a$ to $a$ (where $a>0$) is $$ W=\int_{-a}^a{1\over x} dx. $$ From conservation of energy, we would expect $W=0$ as $V(a)=V(-a)$ and so, in that case, the Cauchy principal value gives the correct answer. However, in general, we would expect that it is not physically possible to have an infinite potential and so some new physics would come in near to $x=0$ which would change the shape of the potential in that region. But, provided that energy is still conserved we would still get $W=0$ when moving the particle from $-a$ to $a$ and so the Cauchy principle value would still have given the correct answer even without accounting directly for the new physics.

However, if you examine the case of $V=1/x$, which gives $F=1/x^2$ then it can be seen that Cauchy principle value gives infinity whereas a contour based integration in the complex plane will give a result consistent with the conservation of energy. So, in general, the contour integration approach is the one best suited for physics.

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