Ever since I learned about the Feynman integration method, I have always wanted to use it to evaluate the Gaussian integral \int_0^{\infty} e^{-x^2}dx = \frac{\sqrt{\pi}}{2}, this magical integral, but I had no success. In the book Mathematical Bridge, the author solves it by transforming it into a double integral, which is concise and ingenious. However, to demonstrate the power of the Feynman integration method, I have always wanted to find other ways to evaluate the Gaussian integral. Last week, in Methods of Mathematical Physics, I saw the author evaluate this integral using the Laplace transform, which was eye-opening; however, that belongs to the realm of integral transforms and the technique of "integrating within the integral sign," which will not be discussed here. Today, while searching for materials online, I saw a method within the scope of the Feynman integration method in a paper by Zhao Jie titled Integrals with Parameters, which I would like to share with everyone.
From a "post-hoc analysis" perspective, the result of the Gaussian integral involves the quantity \sqrt{\pi}. Generally speaking, many common formulas we see contain \pi, but almost none contain \sqrt{\pi}. Therefore, we usually square it. We introduce: f(x) = \left(\int_0^x e^{-t^2}dt\right)^2
This greatly expanded my horizons regarding the Feynman integration method. I had almost never thought that the parameter could be placed in the integration limits. What is the benefit of doing this? Let us differentiate with respect to x: f'(x) = 2\int_0^x e^{-t^2}dt \times e^{-x^2} = 2\int_0^x e^{-(t^2+x^2)}dt
It turns out that after differentiating this way, the product of two integrals is transformed into an integral multiplied by a single term! This serves to simplify things. Making the substitution t = ux, then dt = x du, and the limits change from t \in [0, x] to u \in [0, 1]: f'(x) = \int_0^1 2x e^{-(1+u^2)x^2}du
Now we work backwards: f(x) = \int \left(\int_0^1 2x e^{-(1+u^2)x^2}du\right)dx = \int_0^1 \left[ \int(2x e^{-(1+u^2)x^2}dx)\right]du
That is: f(x) = -\int_0^1 \left(\frac{e^{-(1+u^2)x^2}}{1+u^2}\right)du + C
When x=0, we can find f(0)=0. At this point, the integral term becomes \int_0^1 \frac{1}{1+u^2}du = \frac{\pi}{4}, so C = \frac{\pi}{4}. Thus: f(x) = -\int_0^1 \left(\frac{e^{-(1+u^2)x^2}}{1+u^2}\right)du + \frac{\pi}{4}
The Gaussian integral corresponds to x \to +\infty. In that case: \lim_{x\to +\infty} \int_0^1 \left(\frac{e^{-(1+u^2)x^2}}{1+u^2}\right)du = 0
Therefore f(+\infty) = \frac{\pi}{4}, which leads to: \int_0^{\infty} e^{-x^2}dx = \frac{\sqrt{\pi}}{2}
Wonderful!! This combines various techniques into one. It greatly expands our view of Feynman’s integration method. Note that previously we used Feynman’s method to find the specific form of an integral containing a parameter; here we didn’t, we only took the values of two specific points! But this is enough! This gives us a brand new perspective, and we will be able to use this technique more powerfully in the future!
The solution in Mathematical Bridge: Let I = \int_{-\infty}^{+\infty} e^{-x^2}dx, then \begin{aligned}I^2 &= \int_{-\infty}^{+\infty} e^{-x^2}dx \times \int_{-\infty}^{+\infty} e^{-y^2}dy \\ &= \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} e^{-(x^2+y^2)}dxdy\end{aligned} Converting to polar coordinates, we have: \begin{aligned}I^2 &= \int_{0}^{2\pi} \int_{0}^{\infty} e^{-r^2}r dr d\theta \\ &= \int_{0}^{2\pi} \left(-\frac{1}{2} e^{-r^2}\Big|_0^{\infty}\right)d\theta \\ &= \int_0^{2\pi} \frac{1}{2} d\theta \\ &= \pi\end{aligned} So I = \sqrt{\pi}. The entire process flows like water and clouds—truly a masterpiece!
When reposting, please include the original address of this article: https://kexue.fm/archives/1967
For more detailed information regarding reposting, please refer to: Scientific Space FAQ