The n-Category Café

I certainly don’t feel I found the best way to explain this stuff. It’s the first time I’ve tried to explain path integrals.

But, I said a lot of stuff that didn’t find its way into the notes, since I didn’t write it on the board.

Among other things, I said that path integrals are a bit mysterious — they really go beyond our standard textbook ideas on integration theory, which is why most quantum field theories haven’t been made rigorous. I said that in this simple problem, we see just a tiny taste of this: none of the integrals I wrote down converge absolutely! So the usual Lebesgue theory of integration does not apply. I explained that this is why we get, at the very end, an integral

$$\int_{-\infty}^\infty e^{-i u^2 / 2} d u$$

that does not converge absolutely. Like all real-time path integrals (as opposed to Wick-rotated path integrals), it involves an ‘infinite cancellation of phases’.

I mentioned that we could compute this integral using a step-function cutoff:

$$\lim_{c \to +\infty} \int_{-c}^c e^{-i u^2 / 2} d u$$

but said that it was easier to compute it using a Gaussian cutoff. That amounts to computing

$$\int_{-\infty}^\infty e^{- a u^2 / 2} d u$$

for $a$ with $Re(a) > 0$, and then taking the limit as $a \to i$.

Your analytic continuation looks very clever but of course a priori there is no reason why you can interchange taking the integral and doing the limit $k \to i$.

Just to be painfully clear: the issue of interchanging the limit and the integral works as follows. If we first do the integral and then the limit we get

$$lim_{a \to i} \int_{-\infty}^\infty e^{- a u^2 / 2} d u = \sqrt{2 \pi / i}$$

where of course I mean the limit as $a$ approaches $i$ while keeping $Re(a) > 0$, since the integral diverges wildly for $Re(a) < 0$.

By the way: we can even decide which branch of the square root function is the right one!

If on the other hand we try to do the limit first, we get

$$\int_{-\infty}^\infty \lim_{a \to i} e^{- a u^2 / 2} d u = \int_{-\infty}^\infty e^{-i u^2/2} d u$$

and now we’re trying to integrate a function that’s not Lebesgue integrable. In other words, this integral is not absolutely convergent. So, we need some other method of defining the integral. Any sensible method of defining the integral will give the answer

$$\int_{-\infty}^\infty e^{-i u^2 / 2} d u = \sqrt{2 \pi / i}$$

but Lebesgue integration is not one of these methods.

This issue afflicts all the path integrals in this week’s notes.

Robert wrote:

I agree that the oscillatory integral is not absolutely convergent. However (without remembering any good argument for it) I think it is still Lebesque integrable.

It’s not.

In Lebesgue integration theory, first we define the integral for nonnegative measurable functions $f : X \to [0,\infty)$. This integral always exists, but it can equal $+\infty$. We say that a nonnegative measurable function $f$ is integrable if $\int f < \infty$.

Then, for a general function $f: X \to \mathbb{R}$, we separate it into its positive and negative parts:

$$f = f_+ - f_-$$

with $f_+, f_- \ge 0$. We say $f$ is integrable if both $f_+$ and $f_-$ are integrable, and then we define

$$\int f = \int f_+ - \int f_-$$

Similarly, we say a complex-valued function $f : X \to \mathbb{C}$ is integrable if $Re(f)$ and $Im(f)$ are integrable, and in this case we again define the integral in the obvious way:

$$\int f = \int Re(f) + i \int Im(f)$$

So, any function $f$ for which $|f|$ is not integrable cannot itself be integrable, in the Lebesgue theory!

The point is that the integral we’re trying to do:

$$\int_\mathbb{R} e^{-i u^2/2} d u$$

has a sensible value if we ‘start integrating it near the middle of the real line and work our way out’, e.g by doing

$$\lim_{c \to + \infty} \int_{-c}^c e^{-i u^2/2} d u,$$

or using a Gaussian cutoff, or any sort of sensible cutoff function that ‘approaches 1 starting from the middle of the real line and working its way out’.

But, this notion relies on the domain $\mathbb{R}$ having structure beyond a mere measure space! It’s just like doing this sort of sum

$$1 - 1/2 + 1/3 - 1/4 + \cdots$$

If you sum from left to right, you get a well-defined answer, $ln(2)$. But if you allow arbitrary rearrangements, you can get any answer you want. The Lebesgue theory of integration applies to functions $f: X \to \mathbb{R}$ defined on a mere measure space $X$, so its answers, when well-defined, must be invariant under arbitrary ‘rearrangements’: measure-preserving transformations of $X$. So, any sort of integral that fails to converge absolutely is not Lebesgue integrable.

So, we need some other concept of integration to tackle even the simplest real-time path integrals! The whole point of these integrals is ‘infinite cancellation of phases’, and Lebesgue integration is blind to this.

Robert suspected:

I agree that the oscillatory integral is not absolutely convergent. However (without remembering any good argument for it) I think it is still Lebes[g]ue integrable.

Lebesgue integration is only about absolutely convergent integrals. (In fact, a sequence n: N → a_n: R defines a Lebesgue-integrable function x: R⁺ → a_[x], where [x] is the floor of x, if and only if it defines an absolutely convergent series.)

You may be thinking about (or want to look into) the Henstock integral (also known by many other names). Like the improper Riemann integral, it deals with integration over infinite intervals by working from the inside out; like the Lebesgue integral, it can handle all bounded measurable functions. (In fact, a function is Henstock integrable if it’s either Riemann integrable or Lebesgue integrable.)

A function is Lebesgue integrable if and only if it absolutely Henstock integrable. Thus Lebesgue : Henstock :: asbolutely convergent : convergent.

Of course, even the Henstock integral can’t handle series that require tricks like Abel summation. Is there a theory of Abel–Henstock integration? I don’t know.