Getting a finite result from a non-converging integral

[hideelo]

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?

 

[mathman]

Mathematically, you are correct, but physicists like to stretch things a bit.

 

[jasonRF]

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:
\int_{-\infty}^\infty dt \, e^{-i \omega t} = 2\pi \delta(\omega)
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',
\int_{-\infty}^\infty d\omega \, \delta(\omega) \, \phi(\omega) = \phi(0).

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