Feynman's Integration Method (5): The Legacy of Euler-style Mathematics

费曼积分法(5)：欧拉数学的传承 苏剑林 · 2013-03-24 · Mathematics integral euler mathematics feynman functional

In the second semester of university, our Mathematical Analysis course finally crawled, at a snail's pace, to the chapter on definite integrals. For some of the more complicated definite integrals, I always wanted to solve them my own way, which reignited my enthusiasm for "Feynman's integration method — differentiating under the integral sign." In particular, after using Feynman's method to crack a few interesting and fairly intricate definite integrals, my sense of achievement soared, so I'm writing up a summary here to share with everyone.

What does this have to do with Euler-style mathematics? As I've mentioned before, Euler-style mathematics gives us an intuitive understanding of mathematics that sits somewhere between the intuitive and the rigorous, achieved through methods that are creative but not rigorous. I think Feynman's integration method belongs to this same category: it focuses on solving problems from a particular angle while temporarily setting aside mathematical rigor. Reading about Feynman's life, I feel that this way of thinking runs through his research throughout his career.

This post continues my exploration of Feynman's integration method, arriving at some conclusions that aren't entirely rigorous, laying the groundwork for future applications.

I. A Function That Doesn't Hold

Let's first reconsider $\int_0^{\infty} \frac{\sin x}{x}dx$. This time we bring it into the realm of complex numbers, and consider:

$$\int_0^{\infty}\frac{\cos x+i \sin x}{x}dx=\int_0^{\infty}\frac{e^{ix}}{x}dx$$more

This time readers can probably see why, when we originally computed this integral, we had to attach the factor $e^{-ax}$ to it, namely

$$F(a)=\int_0^{\infty} e^{-ax}\frac{e^{ix}}{x}dx$$

After differentiating with respect to a, this becomes

$$F'(a)=-\int_0^{\infty} e^{(-a+i)x}dx$$

and from this we quickly get $F'(a)=\frac{1}{-a+i}$.

Integrating with respect to a gives $F(a)=-ln(a-i)+C$, where C is an undetermined constant. We know that when $a\to +\infty$ the value of the integral is 0, yet $\lim_{a\to +\infty} ln(a-i) \to \infty$ — does that mean $C \to \infty$? Where exactly did the above derivation go wrong??

Actually, the reasoning above isn't wrong at all. The crux of the matter is that $\int_0^{\infty} \frac{cos x}{x}dx$ tends to infinity!

Even so, we have a compromise: we can treat infinity as though it were an ordinary number and carry out arithmetic with it. The complex number $a-i$ can be rewritten in the form $re^{i\theta}$, with $r=\sqrt{1+a^2}$ and $\theta=arcrtan(-\frac{1}{a})$, so that

$$\begin{aligned}-ln(a-i)=-\frac{1}{2}ln(1+a^2)-iarctan(-\frac{1}{a}) \\ =-\frac{1}{2}ln(1+a^2)+iarccot(a)\end{aligned}$$

The undetermined constant C, being infinite, also has a real part and an imaginary part: the real part is unbounded, but the imaginary part is finite. In fact, from the fact that the integral value is 0 when $a\to +\infty$, we deduce that the imaginary part is 0. So, matching corresponding terms, we get

$$\int_0^{\infty} \frac{\sin x}{x}dx=arccot(0)=\frac{\pi}{2}$$

But this isn't the whole story of this post. If we replace x with ax directly in the integral above, we get:

$$\int_0^{\infty} \frac{\sin (ax)}{ax}d(ax)=\frac{\pi}{2}$$

that is,

$$\int_0^{\infty} \frac{\sin (ax)}{x}dx=\frac{\pi}{2}$$

Differentiating both sides with respect to a gives:

$$\int_0^{\infty} \cos (ax) dx=0$$

or, written differently,

$\int_{-\infty}^{+\infty} cos (ax) dx=0(a\neq 0)$.

Note that this is not something we can make sense of within the calculus we normally learn, since $\lim_{x\to \infty}sin(ax)$ doesn't exist as an ordinary function. But that doesn't stop us from using it. In fact, this is correct — it just isn't a function in the conventional sense, but rather a kind of functional (or generalized function). And as far as "Euler-style mathematics" is concerned, all we need is to obtain this result and put it to use; its validity can be checked later through other derivations. Sharp readers will notice that this bears some resemblance to the Fourier transform of the Dirac delta function — indeed, it's precisely the real part of the Fourier transform of the Dirac delta function.

II. Integrals of Rapidly Oscillating Trigonometric Functions

Consider the integral $\int_a^b f(x)cos(\omega x)dx$, where f(x) is a well-behaved function (its derivative exists everywhere and is bounded, and it does not depend on a). We are interested in the limiting case $\lim_{\omega \to \infty} \int_a^b f(x)cos(\omega x)dx$, or written as:

$$\lim_{h\to 0} \int_a^b f(x)\cos(\frac{x}{h})dx$$

Readers might think the result depends on the specific form of f(x), but here I'll tell you: the result is necessarily 0.

How should we understand this? Let's think about it this way: when $\omega $ is very large, the period becomes very small, so that even over a tiny interval of integration we already have $\int Acos(\omega x)dx=\frac{A}{\omega}sin(\omega x)=0$, where A is an arbitrary constant. As for $\lim_{\omega \to \infty} \int_a^b f(x)cos(\omega x)dx$, because the period is so tiny that f(x) doesn't even have time to change appreciably, the periodicity of $cos(\omega x)$ has already cancelled out its effect. To put it in the language Feynman used to describe quantum mechanics: the contribution of $f(x_0)cos(\omega x_0)dx$ from any given $x_0$ is cancelled out by a neighboring point $f(x_0+\frac{\pi}{\omega})cos(\omega (x_0+\frac{\pi}{\omega}))dx$, and because $\omega$ is extremely large (indeed, infinite), $f(x_0+\frac{\pi}{\omega})$ has not yet changed appreciably, so the net effect is 0.

Of course this is just an intuitive picture — strictly speaking it would require a proof, but we won't bother with that here, because there's an even more convincing piece of evidence that assures us the reasoning above is sound: Feynman used exactly this kind of thinking to build a bridge connecting classical mechanics and quantum mechanics, creating the path-integral formulation of quantum mechanics — an approach that "for the first time made quantum mechanics simpler than classical mechanics"! And the way he transitioned from quantum mechanics to classical mechanics is precisely analogous to the discussion we just had above. It amounts to saying that in classical mechanics h is 0, while in quantum mechanics h is Planck's constant.