In the previous article, I provided a brief introduction to “Feynman’s Integration Method” and demonstrated its operational steps through an example. However, to truly understand a method, beyond knowing what it can do, one must also understand its principles and mechanics. This allows us to master it more effectively. Therefore, we need to establish the general theory of “differentiation under the integral sign” to lay the foundation for further applications.
General Principle
Let us denote: G(a) = \int_{m(a)}^{n(a)} f(x,a) dx
Here, f(x,a) is a function of x containing a parameter a, and the limits of integration are two functions of the parameter a. Such an integral is also called an integral with variable limits, which can be understood as a generalization of the ordinary definite integral. Let F(x,a) be the antiderivative of f(x,a) with respect to x, meaning \frac{\partial F(x,a)}{\partial x} = f(x,a). Then, according to the Fundamental Theorem of Calculus, we have: G(a) = F(n(a),a) - F(m(a),a)
Differentiating with respect to a: \begin{aligned} \frac{d G(a)}{da} = & \frac{\partial F(n(a),a)}{\partial n(a)} \times \frac{d n(a)}{da} + \frac{\partial F(n(a),a)}{\partial a} \\ & - \frac{\partial F(m(a),a)}{\partial m(a)} \times \frac{d m(a)}{da} - \frac{\partial F(m(a),a)}{\partial a} \end{aligned}
This long expression essentially tells us: G'(a) = \int_{m(a)}^{n(a)} \frac{\partial f(x,a)}{\partial a} dx + f(n(a),a) \times \frac{d n(a)}{da} - f(m(a),a) \times \frac{d m(a)}{da}
This is the rule for “differentiation under the integral sign”! If m and n are constants, it simplifies directly to: G'(a) = \int_{m}^{n} \frac{\partial f(x,a)}{\partial a} dx
The reason Feynman’s integration method works is that the step of “differentiation” might seem to complicate things, but in reality, it allows many cumbersome parts to temporarily disappear during the process, yielding a relatively simple result which is then integrated back. In a sense, this is the idea of step-by-step processing.
More Examples
To further illustrate the application of Feynman’s integration method, BoJone will present more examples. It should be noted that the example shown previously involved a relatively low amount of computation, whereas general definite integral calculations are often quite complex. Regardless of the method used, one must rely on certain transformations to turn the integral into a familiar form. Therefore, it is necessary to accumulate a certain number of definite integral results and know some basic integrable functions. (Of course, one can also consult an Integral Table when necessary.)
On the other hand, the core of Feynman’s integration method lies in the choice of parameters. Not every integral to be solved naturally contains a parameter, such as \int_0^{\frac{\pi}{2}} \frac{x}{\tan x} dx. Even if a parameter is present, it may not necessarily meet the requirements. Therefore, we need to gradually learn how to “retrofit” the integrand into a form with parameters, where the original definite integral corresponds to a case where the parameter takes a specific value. As for how to add parameters and how to determine their form, BoJone also has only a partial understanding and needs further improvement.
Example 1:
\int_0^{\frac{\pi}{2}} \frac{x}{\tan x} dx
This is a fairly common form, with the basic shape \int \frac{x}{f(x)} dx. Our approach is to try (every calculation is an attempt) to adapt it into \int \frac{f^{-1}(a \times f(x))}{f(x)} dx, where f^{-1}(x) is the inverse function of f(x), and the original integral is equivalent to the case where a=1. Why do this? You will understand as soon as you differentiate it ^_^
For this problem, we transform it into: G(a) = \int_0^{\frac{\pi}{2}} \frac{\arctan(a \tan x)}{\tan x} dx, \quad f(x,a) = \frac{\arctan(a \tan x)}{\tan x}
Then: \begin{aligned} \frac{\partial f(x,a)}{\partial a} = & \frac{1}{a^2 \tan^2 x + 1} \\ = & \frac{\cos^2 x}{a^2 \sin^2 x + \cos^2 x} = \frac{\cos^2 x}{a^2 + (1-a^2) \cos^2 x} \\ = & \frac{1}{1-a^2} \left[1 - \frac{2a^2}{(1+a^2) + (1-a^2) \cos 2x}\right] \end{aligned}
Wait, wait, wait... first, we must clarify what we are actually doing and what we want to do. The next step after differentiation is integration. What kind of integral can be easily solved? By checking an Integral Table, we can see that integrals of the form \int \frac{1}{a+b \cos x} dx can be solved. Therefore, we should try to bring it as close to this form as possible!
Ultimately, we get: \begin{aligned} G'(a) = & \frac{1}{1-a^2} \int_0^{\frac{\pi}{2}} \left[1 - \frac{2a^2}{(1+a^2) + (1-a^2) \cos 2x}\right] dx \\ = & \frac{1}{1-a^2} \Big[x - a \arctan(a \tan x)\Big] \Big|_0^{\pi/2} \\ = & \frac{1}{1-a^2} \left(\frac{\pi}{2} - a \times \frac{\pi}{2}\right) \\ = & \frac{\pi}{2(1+a)} \end{aligned}
Integrating again, we get: G(a) = \frac{\pi}{2} [\ln(1+a) + C]
When a=0, f(x,a) = f(x,0) = 0, so G(0) = \int_0^{\frac{\pi}{2}} 0 dx = 0. From this, we find C=0.
Incidentally, regarding the result of the first integration step (with respect to the variable x), an experienced reader might find inspiration from: \frac{1}{a^2 \tan^2 x + 1} and guess that the antiderivative would be in the form of \arctan(a \tan x). Differentiating it, we find: \begin{aligned} \frac{d \arctan(a \tan x)}{dx} = & \frac{a}{a^2 \tan^2 x + 1} \times \frac{1}{\cos^2 x} \\ = & \frac{a}{a^2 \tan^2 x + 1} \times (1 + \tan^2 x) \\ = & \frac{a}{a^2 \tan^2 x + 1} \times \left(\frac{1}{a^2} + \tan^2 x + 1 - \frac{1}{a^2}\right) \\ = & \frac{1}{a} + \frac{a - 1/a}{a^2 \tan^2 x + 1} \end{aligned}
This differs from the required expression only by a constant, so the antiderivative \frac{1}{1-a^2} [x - a \arctan(a \tan x)] can be quickly obtained. Of course, this requires a certain level of mathematical intuition.
Note: Readers might feel a headache after seeing the above process and wonder: how is this simple? It is clearly so complex! In fact, regardless of the method used, the process of calculus is quite complicated. The key is the operability and feasibility of the algorithm. In other words, once we know it is feasible, we can follow the steps to calculate it. The tediousness of the calculation process is not where the difficulty of a problem lies. The difficulty of a good problem lies in: how to obtain the inspiration that leads us to this process!
To keep the article from becoming too long, more examples will be placed in the next post.
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