Getting a finite result from a non-converging integral

In summary, the author solves the equation $$\int_0^\infty dx \: e^{-iax} - e^{iax}$$ by taking the Fourier transform of a distributional limit. This is valid in the same sense that the Fourier transform of a constant is proportional to a delta function. However, the author stretches the theory a bit by using the theory of generalized functions (aka distributions) without telling the reader. If you get picky, you will notice that the result is singular at ##a=0##.
  • #1
hideelo
91
15
I am looking at the integral

$$\int_0^\infty dx \: e^{-iax} - e^{iax}$$

I know that this does not converge for many reasons, but most obviously because I can rewrite it as

$$2i \int_0^\infty dx \: sin(ax) = -2i a [\cos(ax)]_0^\infty$$

which does not converge to anything.

However the book I'm following (Solutions to Kardar's Statistical Mechanics of Particles; Chapter 5, Problem 1) has a "clever" solution for this

$$\int_0^\infty dx \: e^{-iax} - e^{iax} = \lim_{\epsilon \to 0} \int_0^\infty dx \: e^{-iax-\epsilon x} - e^{iax-\epsilon x}$$

$$= \lim_{\epsilon \to 0} \left[\frac{e^{-iax-\epsilon x}}{-ia-\epsilon} - \frac{e^{iax-\epsilon x}}{ia-\epsilon}\right]_0^\infty = \frac{2i}{a}$$How the hell is this allowed? You can't just swap limits and integrals like that, or am I missing something?
 
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  • #2
Mathematically, you are correct, but physicists like to stretch things a bit.
 
  • #3
To me it looks like the author is using the theory of generalized functions (aka distributions) without telling you. It is valid in the same sense that the Fourier transform of a constant is proportional to a delta function:
[tex] \int_{-\infty}^\infty dt \, e^{-i \omega t} = 2\pi \delta(\omega) [/tex]
Here the integral sign is just a symbol, and what you are really doing is taking a Fourier transform in a distributional sense. The result is a distribution, not a function, so the important properties are how it behaves when multiplied by 'nice' functions ##\phi## and 'integrated',
[tex] \int_{-\infty}^\infty d\omega \, \delta(\omega) \, \phi(\omega) = \phi(0).[/tex]

One nice feature of distribution theory is that interchange of limiting processes is easy to establish and the manipulations the author performed are valid. In your example, you are finding the Fourier transform of the ##\mathrm{sign}(x)## function, where ##\mathrm{sign}(x)=-1## when ##x<0## and ##\mathrm{sign}(x)=1## when ##x>0##. If we want to take the Fourier transform in the distributional sense we can define a sequence of functions ##f_\epsilon(x) = \mathrm{sign}(x) \, e^{-\epsilon |x|}##. Clearly ##\lim_{\epsilon \rightarrow 0^+} f_\epsilon(x) = \mathrm{sign}(x)##. Then we use the fact that limits and Fourier transforms commute in the theory of distributions so that
$$\mathcal{F} \left\{ \mathrm{sign}(x)\right\}=\mathcal{F} \left\{ \lim_{\epsilon \rightarrow 0^+} f_\epsilon(x) \right\} = \lim_{\epsilon \rightarrow 0^+} \mathcal{F}\left\{f_\epsilon(x)\right\}$$
where ##\mathcal{F}## denotes the Fourier transform operator. This is the interchange you are concerned with, and of course if you are dealing with regular functions and integration then it is not valid.

If you get picky you will notice that your result is singular at ##a=0##. It turns out that if you are careful with the way you take the distributional limits then you find out that the action of multiplying it by a nice function ##\phi## and integrating should be interpreted as,
$$ \lim_{\epsilon \rightarrow 0^+} \left[\int_{-\infty}^{-\epsilon} da \, \frac{2i\, \phi(a)}{a} + \int_{\epsilon}^\infty da \, \frac{2i\, \phi(a)}{a} \right] $$
where this is an ordinary improper Riemann integral. But worrying about this level of detail is often not required in applications.

jason
 
Last edited:

What is a non-converging integral?

A non-converging integral is an integral that does not have a finite value when evaluated. This can happen when the function being integrated has an infinite or undefined value at one or more points within the interval of integration.

Why is it important to get a finite result from a non-converging integral?

A finite result from a non-converging integral is important because it allows us to accurately calculate and analyze physical phenomena in the real world. Without a finite result, our calculations would be incorrect and could lead to inaccurate predictions and conclusions.

What are some techniques for obtaining a finite result from a non-converging integral?

Some techniques for obtaining a finite result from a non-converging integral include using transformation methods, such as contour integration, and applying regularization techniques, such as the Riemann zeta function or Hadamard regularization.

Can a non-converging integral ever have a finite result?

Yes, a non-converging integral can have a finite result if certain conditions are met. For example, if the function being integrated has a removable singularity within the interval of integration, the integral can be redefined to give a finite result. Additionally, using transformation methods or regularization techniques can also lead to a finite result.

What are some real-world applications of obtaining a finite result from a non-converging integral?

Obtaining a finite result from a non-converging integral has many real-world applications, including in physics, engineering, and economics. For example, in physics, calculating the finite result of a non-converging integral can help us understand the behavior of particles in quantum mechanics. In engineering, it can be used to design more efficient structures. In economics, it can be used to model and predict economic growth and stability.

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