Due to a bicycle trip, this article was delayed for a week; in fact, it should have been written a week ago. This article continues to discuss some examples of the Feynman integration method. Readers may discover the basic directions and methods of its application from these different types of examples, thereby enhancing their understanding of it.
Example 2:
\int_0^{\infty} \frac{\sin x}{x}dx
This is a relatively common type of integral, taking the form \int \frac{f(x)}{x}dx. For this form, our first instinct might be to rewrite it in a parametric form \int \frac{f(ax)}{x}dx. The purpose of this is simple: to eliminate the denominator, which is consistent with the integration strategy for \int \frac{x}{f(x)}dx. However, upon deeper investigation, one finds that even if this eliminates the denominator and makes the first integration simpler, during the second integration, it often reverts to the form \int \frac{f(x)}{x}dx, preventing us from proceeding. Therefore, this method of introducing parameters fails in most cases.
There is a magical method where we transform it into:
\begin{aligned}G(a)&=\int_0^{\infty} e^{-ax}\frac{\sin x}{x}dx \\ f(x,a)&=e^{-ax}\frac{\sin x}{x}\end{aligned}
By adding the term e^{-ax} to the original expression, we are able to eliminate the denominator because:
\frac{\partial f(x,a)}{\partial a}=-e^{-ax} \sin x
Thus, \begin{aligned}G'(a)&=\int_0^{\infty} -e^{-ax} \sin x dx \\ &=\frac{1}{a^2+1} e^{-ax}(a \sin x+\cos x)|_0^{\infty} \\ &=-\frac{1}{a^2+1}\end{aligned} (Refer to the Exponential Function Integral Table)
Finally, we obtain: G(a)=-\int \frac{1}{a^2+1} da = -\arctan a + C
As a \to \infty, f(x,a) \to 0, so G(a) \to 0, which gives C = \frac{\pi}{2}. The final result for the parametric integral is:
\int_0^{\infty} e^{-ax}\frac{\sin x}{x}dx = -\arctan a + \frac{\pi}{2}
Therefore: \int_0^{\infty} \frac{\sin x}{x}dx = G(0) = \frac{\pi}{2}
Example 3:
In the book Advanced Calculus, which Feynman studied, there is a very typical example whose solution process combines knowledge of differential equations.
Given \int_0^{\infty} e^{-x^2}dx = \frac{\sqrt{\pi}}{2}, find the integral: u = \int_0^{\infty} e^{-x^2-\frac{a^2}{x^2}}dx
Since this integral already provides a parameter a, we might as well differentiate directly with respect to this parameter and see what happens.
\begin{aligned}\frac{du}{da} &= \int_0^{\infty} \frac{\partial (e^{-x^2-\frac{a^2}{x^2}})}{\partial a}dx \\ &= -2 \int_0^{\infty} \frac{a}{x^2} e^{-x^2-\frac{a^2}{x^2}} dx \\ &= 2\int_0^{\infty} e^{-x^2-\frac{a^2}{x^2}}d\left(\frac{a}{x}\right)\end{aligned}
Let t = \frac{a}{x}, then the expression becomes: \int_0^{\infty} e^{-x^2-\frac{a^2}{x^2}}d\left(\frac{a}{x}\right) = -\int_0^{\infty} e^{-t^2-\frac{a^2}{t^2}}dt
The form of this integral is exactly the same as the integral we are seeking, because the variable symbols (x, t) are merely notation; the integral itself is independent of the choice of variable. Thus, we can confidently say: \int_0^{\infty} e^{-t^2-\frac{a^2}{t^2}}dt = u
In summary: \frac{du}{da} = -2u
Solving this differential equation, we get: u = C \times e^{-2a}. Using the given initial condition (when a=0), we can determine the constant C = \frac{\sqrt{\pi}}{2}. Therefore:
\int_0^{\infty} e^{-x^2-\frac{a^2}{x^2}}dx = \frac{\sqrt{\pi}}{2} e^{-2a}
Summary
The transformation forms of the Feynman integration method are diverse and unpredictable. In practical applications, one must dare to think and experiment, and often a "flash of inspiration" is required to produce a brilliant solution! Perhaps it is precisely because of this flexibility, which often brings unexpected surprises, that Feynman was so fond of it.
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