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I just found out during my Calculus course in High School, that there exist functions which cannot be integrated.

Then I thought that I come across a lot of integrals while solving Physics questions.

Although I haven't come across such a non-integrable function yet, what physical significance would such an equation hold?

Has anyone come across such functions while solving a physics problem? Some examples would help a lot(High School level examples would be even better).

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There is a nice series of math books with titles Counterexamples in X that illustrate pathological and counter-intuitive things in various fields like analysis and topology. I think there are cheap Dover editions of the books now. Anyway I doubt any of this is really relevant to physics though since physical systems tend to be "smooth" and "well behaved" in that vague sense that drives mathematicians mad. But I too would be interested in hearing of any examples like this in physics. –  Michael Brown Aug 15 '13 at 13:44

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First off I'm not sure exactly what you mean be non-integral function. John Dodson discussed two interpretations that commonly appear in physics: non-elementary functions and singularities. The only other possibility I can think of are integrals like $$ \int_{-1}^{1}\mathrm{sin}(\frac{1}{x})dx $$ This oscillates infinitely many times near the origin, such that it is basically undefined at x = 0. As far as I know, these "pathological functions" don't appear in physics.

That being said, I'll try and give a more satisfactory answer by going over an example from electromagnetism: the Green's Function. So that you know where this is going, I'll give an overview: the Green's Function is important for solving Maxwell's equations for arbitrary charge distributions, but trying to evaluate it naively leads to an integration similar to $\int \frac{dx}{x}$ - but for the equations to have any meaning, we need this integral to be finite. To accomplish this we will need to integrate through the complex plane, a strange process that yields physically interpretable results. The math of this example, will be a little sophisticated, but only a vague understanding of the concepts is required to see what's going on.

Simplified Maxwell's Equations

Maxwell's equations talk about the interactions of the electric and magnetic field through differential equations. It turns out we can represent the electric and magnetic fields by introducing a single potential vector $A$ with four components: one "time" component $\phi/c$ and three space components represented by the vector $\bf{\vec{A}}$. Then the electric and magnetic fields are given by $$ \bf{\vec{E}}=\bf{\vec{\nabla}}\phi,\quad \bf{\vec{B}}=\bf{\vec{\nabla}}\times\bf{\vec{A}} $$ and Maxwell's equations are $$ \frac{1}{c^2}\frac{\partial^2\phi(\bf{x},t)}{\partial t^2} - \nabla^2\phi(\bf{x},t) = \frac{\rho}{\epsilon_0} \\ \frac{1}{c^2}\frac{\partial^2\bf{\vec{A}}(\bf{x},t)}{\partial t^2} - \nabla^2\bf{\vec{A}} (\bf{x},t) = \mu_0\bf{\vec{J}} $$ (You can go here to learn more about Maxwell's equations, and here to learn more about the operator $\bf{\vec{\nabla}}$).

Green's Function

Let's focus on the first of Maxwell's equations, the one for $\phi$. One way to find solutions for any given $\rho$ (that is, the density of charge) is to start with the solutions for $\rho=0$ - that is, solutions to: $$ \frac{1}{c^2}\frac{\partial^2\phi}{\partial t^2} = \nabla^2\phi $$ If you're currently in Calculus, this equation may look daunting; however, the solution turns out to be trivial: see wave equation on Wikipedia. If we can relate solutions to Maxwell's equations for any $\rho$ to these known solutions for $\rho=0$, then we're good! The way to do this is to use a Green's function for the wave equation: a function such that $$ \phi_\rho(\bf{x},t) = \phi_0(\bf{x},t) + \int G(\bf{x-x'},t-t')\rho(\bf{x'},t')d^4 x $$ where $\phi_\rho$ is a solution to Maxwell's equation, and $\phi_0$ is a solution to the $\rho=0$ case. The integral is taken over all four dimensions of spacetime (hence the symbol $d^4x$). There is a special differential equation that will give us Green's function, and its solution can be represented as:

$$ G(\textbf{x},t) = \int \frac{e^{i(\omega t - \textbf{k}\cdot\textbf{x})}d^4k}{\omega^2 - \textbf{k}^2} $$

This integral is carried out over four variables: the "frequency" variable $\omega$ and the "wavenumber" vector $\textbf{k}$. These correspond to the various wave solutions to the $\rho=0$ equation.

Non-integrable function

So what does this have to do with your question? Let's say we try to integrate over the variable $\omega$. This looks like $$ \int\left(\int_{-\infty}^{\infty} \frac{e^{i(\omega t - \textbf{k}\cdot\textbf{x})}}{\omega^2 - \textbf{k}^2}d\omega\right) d^3\textbf{k} $$

If we partial-fraction decompose, we have $$ \int\frac{e^{i(\textbf{k}\cdot\textbf{x})}}{2|\textbf{k}|}\left(\left(\int_{-\infty}^{\infty} \frac{e^{i\omega t}}{\omega - |\textbf{k}|}d\omega\right) - \left(\int_{-\infty}^{\infty} \frac{e^{i\omega t}}{\omega + |\textbf{k}|}d\omega\right)\right) d^3\textbf{k} $$ Both of these internal integrals are obviously problematic: integrating straight from $-\infty$ to $\infty$, we run over two singularities when $\omega = |\textbf{k}|$ or $\omega = -|\textbf{k}|$!

When we encounter integrals like these, we have to do some tricks to get meaningful information out of them. The trick we use here is not to integrate straight from one end to the other, but to take a curvy path through the complex plane that avoids the non-integrable singularities. The possible paths include but are not limited to:

"Advanced" solution "Advanced" solution "Retarded" solution "Retarded" solution

The retarded solution is so named because it accounts for waves that travel forwards in time (which means the charge $\rho$ can only affect the future) and the advanced solution accounts for waves that travel backwards in time (meaning the charge $\rho$ can affect the past). Since we have not yet witnessed any ability to send messages into the past, we typically interpret the retarded solutions as being the only "allowed" ones; the advanced solutions are "banned" (though Feynman-Wheeler absorber theory offers another interpretation).

The retarded solution is: $$ G_R(\textbf{x},t) = \Theta(t)\frac{\delta(t-\frac{r}{c})}{4\pi r} $$ The step function $\Theta(t)$ is zero for negative $t$, while the Dirac delta function $\delta(x)$ is zero whenever $x$ is nonzero. $r$ represents the radius of the point $\textbf{x}$. Thus, this equation corresponds to waves travelling forwards in time with velocity $c$.

Recap

To summarize - there are certain times that non-integrable functions enter into physical calculations, and the attempt to intepret these integrals and find useful solutions can lead us to a broader understanding of the physical situation.

One other significant way this happens is in quantum field theory. The calculation of interacting quantum fields can lead to severely diverging integrals, none of which are as easily avoided as the one we looked at. A class of techniques known as "Renormalization" were created to deal with these infinities, centering around the idea that field theories are not valid on small scales, pointing us towards a more accurate high-energy theory that will eliminate the infinities. Though initially controversial, the modern understanding of such theories is well-founded and has uses beyond infinity-avoidance. (Thanks, Michael Brown!)

Thus one could argue that non-integrable functions not only occur in our understanding of nature, but they augment our understanding of it by leading us to new methods of interpretation.

I hope that I have provided you with a satisfactory example and answer to your question.

EDIT: Initially I called renormalization techniques "controversial" - as Michael Brown pointed out, this is a bit of a mischaracterization. When they were first introduced, many prominent physicists expressed discomfort with these techniques, but a better understanding of the science behind renormalization has led to a well-founded theory and today most quantum field theories are judged by their ability to be renormalized.

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Fantastic answer! Thank you. –  udiboy1209 Aug 17 '13 at 7:38
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The modern understanding of renormalization (due to Kadanoff, Wilson and others) is hardly controversial and has nothing really to do with infinities. It is necessary even in completely finite theories, but the fact that it fixes infinities in QFT is a bonus. –  Michael Brown Aug 17 '13 at 8:04

Well an easy example (although maybe not satisfying) would be the integral of 1/x from [0, a].

$\int\limits_0^a \frac{1}{x}dx$

If you integrate this, it will diverge to infinity. This kind of equation pops up extremely often in physics. Quantum mechanics has many examples of this. Often times in quantum mechanics, solutions such as these are "thrown out" and are said to be meaningless, while other solutions that make sense are kept (that was a crude way of putting it, but common books such as Griffiths or Townsend teach the mathematics like this). You can imagine this can come up with repulsive forces/fields where the radius goes from [0, a] and you must integrate over the radius.

You may be talking about indefinite integrals that are non-integrable though, which is much different than the relatively simple example I just showed. For these types of examples, check out the wiki-page, it's pretty interesting.

http://en.wikipedia.org/wiki/Nonelementary_integral

These are much rarer to find in physics and often are just mathematical anomalies...But from looking at the ones on the wiki-page I do recognize one in particular which is:

$e^{-x^2/2}$

This is also extremely common in quantum mechanics when you start using guassian integrals. This can be evaluated given certain limits, but sometimes you don't have those particular limits and need to use other methods to evaluate the integral. Keep in mind, this is very real physically and is even used by particle physicists at CERN all the time when analyzing data. What you need to do is use a Taylor Series Expansion or the Error Function to approximate the answer....no exact answer can be found, but pretty precise answers can be found using those two methods.

Hope that answers your question!

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I think the first part of your answer is really about singularities in physics rather than non-integrable functions. –  John Rennie Aug 15 '13 at 16:21
    
Well I think you are mostly right, which is why I included the second part and also said the first part probably isn't satisfying. But I will say, in physics, we usually like to talk about things that have limits if we're actually seeking an answer. So I thought it was an interesting example about the consequences of when integrating a function and evaluating it gives a "bad answer" –  Spaderdabomb Aug 15 '13 at 17:20
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@JohnDodson, My main question wasn't whether these kind of integrals pop up in physics, It was what can you imply physically when such integrals are obtained. And I have seen $\int\frac 1x dx$ many times, and the physical meaning of getting $\infty$ can be explained clearly in each case. –  udiboy1209 Aug 16 '13 at 11:58
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I took the OP to be refering to non-integrable functions (and similar discussions apply for non-Riemann integrals) rather than nonelementary integrals. In other words I though the OP was asking about integrals which don't exist rather than ones which merely cannot be expressed in terms of elementary functions but are perfectly well defined and could be evaluated numerically. –  Michael Brown Aug 16 '13 at 12:57
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Fair point, but that never happens in physics unless you are not careful. An example of the kind of thing I'm talking about is the Dirichlet function which is not Riemann integrable but is Lebesque integrable. I'd imagine something like that coming into physics through some weird number theory/string theory connection. That would be conceivable, but very bizarre - something that would have to have some deep meaning to it if true. But integrating into a singularity just means you haven't regulated your calculation properly. ;) –  Michael Brown Aug 16 '13 at 13:10

I would like to summarize what satisfied me from each answer to this question.

It is true that non-integrable or non-elementary functions frequently occur during calculations. They are a kind of feedback on our theory, as to how accurately it can depict Nature.

We have to renormalize our theory, more correctly, our understanding of that situation to avoid a diverging solution.

I thank @FrancisFlute, for that excellent example of Green's Function, to show that such integrals do occur in physics, and it leads to us changing our understanding of Nature, eventually forming better theories!

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Your summary is incorrect regarding non-elementary functions. When an integral is nonelementary, as defined at en.wikipedia.org/wiki/Nonelementary_integral , that fact has absolutely no physical significance. –  Ben Crowell Aug 17 '13 at 22:18
    
@BenCrowell, could you provide a more detailed answer. I was pretty much satisfied with FrancisFlute's answer. Is that wrong or is my summary wrong? –  udiboy1209 Aug 18 '13 at 5:16
    
I don't see anythig wrong with FrancisFlute's answer, but FrancisFlute's answer doesn't have anything to do with nonelementary integrals, as defined in the WP link. –  Ben Crowell Aug 18 '13 at 16:33

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