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 relatively complex definite integrals, I always want to solve them using my own methods, which has reignited my passion for the "Feynman Integration Method—differentiation under the integral sign." Especially when I used the Feynman integration method to solve several interesting and complex definite integral problems, my sense of achievement soared. Therefore, I am summarizing it here to share with everyone.
What does this have to do with Eulerian mathematics? As mentioned before, Eulerian mathematics uses an unrigorous yet highly creative approach to give us an intuitive understanding of mathematics that lies between the emotional and the rational. I believe the Feynman integration method also belongs to this category; it focuses on solving problems from a special perspective while temporarily ignoring mathematical rigor. Reading about Feynman’s stories, I feel that this kind of thinking permeated his research throughout his life.
This article continues the study of the Feynman integration method, deriving some not-so-rigorous conclusions to lay the foundation for future applications.
Functions That Do Not Hold
First, let us reconsider \int_0^{\infty} \frac{\sin x}{x}dx. This time, we introduce 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
This time, readers can probably understand why we added the factor e^{-ax} when we first evaluated this integral, namely: F(a)=\int_0^{\infty} e^{-ax}\frac{e^{ix}}{x}dx
After differentiating with respect to a, it becomes: F'(a)=-\int_0^{\infty} e^{(-a+i)x}dx
Thus, we quickly obtain F'(a)=\frac{1}{-a+i}.
Integrating with respect to a, we get F(a)=-\ln(a-i)+C, where C is an undetermined constant. We know that as a \to +\infty, the value of the integral is 0. However, \lim_{a\to +\infty} \ln(a-i) \to \infty. Does this mean C \to \infty? What exactly went wrong in the above procedure?
Actually, the above reasoning is not wrong at all. The key to the problem is that \int_0^{\infty} \frac{\cos x}{x}dx tends to infinity!
However, we still have a compromise method: we treat infinity as if it were an ordinary number for calculation. The complex number a-i can be rewritten in the form re^{i\theta}, where r=\sqrt{1+a^2} and \theta=\arctan(-\frac{1}{a}), so: \begin{aligned}-\ln(a-i) &= -\frac{1}{2}\ln(1+a^2)-i\arctan\left(-\frac{1}{a}\right) \\ &= -\frac{1}{2}\ln(1+a^2)+i\text{arccot}(a)\end{aligned}
The infinite undetermined constant C is also composed of a real part and an imaginary part. The real part is infinite, but the imaginary part is finite. In fact, from the fact that the integral value is 0 when a \to +\infty, we conclude that the imaginary part is 0. Therefore, according to the principle of corresponding equality, we have: \int_0^{\infty} \frac{\sin x}{x}dx=\text{arccot}(0)=\frac{\pi}{2}
This is not the full conclusion of this article. If we directly replace x in the above integral with ax, we have: \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 as: \int_{-\infty}^{+\infty} \cos (ax) dx=0 \quad (a\neq 0).
It should be noted that this is incomprehensible within the scope of the calculus we study, because \lim_{x\to \infty}\sin(ax) does not exist. But this does not prevent us from using it. In fact, this is correct, but it is not a conventional function; it is a functional. For "Eulerian mathematics," we only need to obtain this result and apply it; its correctness will be verified in other derivations. Attentive readers will find that this is somewhat similar to the Fourier transform of the Dirac delta function; in fact, it is the real part of the Fourier transform of the Dirac delta function.
Highly Oscillatory Trigonometric Integrals
Given the integral \int_a^b f(x)\cos(\omega x)dx, where f(x) is a well-behaved function (its derivative exists and is bounded everywhere, and is independent of a). We consider 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\left(\frac{x}{h}\right)dx
Perhaps readers will think this depends on the form of f(x), but here I will tell you that this result must be 0.
How can we understand this situation? We might consider it this way: when \omega is very large, the period will be very small. At this time, for a very small integration interval, we have \int A\cos(\omega x)dx=\frac{A}{\omega}\sin(\omega x)=0, where A is an arbitrary constant. For \lim_{\omega \to \infty} \int_a^b f(x)\cos(\omega x)dx, because the period is so small, it is as if f(x) does not have time to change before its effect is canceled out by the periodicity of \cos(\omega x). Using the language Feynman used to describe quantum mechanics, the contribution of f(x_0)\cos(\omega x_0)dx from any point x_0 is canceled out by a nearby point f(x_0+\frac{\pi}{\omega})\cos(\omega (x_0+\frac{\pi}{\omega}))dx. Since \omega is very large (effectively infinite), f(x_0+\frac{\pi}{\omega}) has not yet changed significantly, so the total effect is 0.
Of course, this is just an intuitive understanding. Strictly speaking, it requires proof, but we will not do that because there is even more convincing evidence to make us certain that the above process is correct: Feynman used similar ideas to build a bridge between classical mechanics and quantum mechanics, creating the path integral formulation of quantum mechanics, "making quantum mechanics simpler than classical mechanics for the first time"! The way he transitioned from quantum mechanics to classical mechanics is exactly similar to our discussion above. It is equivalent to saying that in classical mechanics h is 0, while in quantum mechanics h is Planck’s constant.
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