mathematical physics - Physical significance of getting an non-integrable function in an equation

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).

Answer

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 "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.