Recently, my primary focus has been on researching path integrals and deriving a new transformation method (or perhaps a new perspective) for them. However, there is still a hurdle I cannot overcome, which is why updates on this blog have been sparse. Speaking of integration and differentiation, these two are inherently inverse operations, but under the unification of complex numbers, they can be transformed into each other. For example, the Schrödinger equation is the differential form of quantum mechanics, while path integrals can essentially be considered the integral form of quantum mechanics. This gives me some thoughts: does every mathematical theory in differential form have a corresponding version in integral form? If so, is the differential version superior or the integral version?
In mathematical analysis, we often feel that differentiation is much easier than integration, as differentiation has ready-made formulas and so on. However, differentiation has a major drawback: it is multi-component. For instance, a potential function is a scalar, but after differentiation (taking the gradient), it becomes a three-component vector (i.e., force). Multi-component quantities are actually harder to handle, and to deal with such problems, a large number of operators are introduced. The characteristic of integration lies in its scalar nature; although the calculation may be complex, the underlying idea is easy to grasp. I prefer theories in integral form (such as the principle of least action, path integrals, etc.).
When discussing common and famous definite integrals in mathematical analysis, one must mention the following trigonometric integrals: \int_0^{\pi/2} \sin^{2n} \theta d\theta It is not difficult to prove that it also equals \int_0^{\pi/2} \cos^{2n} \theta d\theta
There are many ways to evaluate this definite integral. The method given in textbooks is based on recurrence relations via mathematical induction. However, I find a simpler and more memorable method (meaning that even if forgotten, it can be quickly re-derived) is to use the exponential form of trigonometric functions. Simply use \cos\theta = \frac{e^{i\theta} + e^{-i\theta}}{2} and write the integral as: \begin{aligned} \int_0^{\pi/2} \cos^{2n} \theta d\theta &= \int_0^{\pi/2} \left(\frac{e^{i\theta} + e^{-i\theta}}{2}\right)^{2n} d\theta \\ &= \frac{1}{2^{2n}} \int_0^{\pi/2} \left(e^{i\theta} + e^{-i\theta}\right)^{2n} d\theta \end{aligned}
Consider (e^{i\theta} + e^{-i\theta})^{2n}; it must take the form: \frac{(2n)!}{(n!)^2} + \sum_{k=-n, k \neq 0}^{n} a_k e^{2ik\theta} (One could write out the specific form of the coefficients a_k, but it is not important; it is sufficient to know they are real numbers). Furthermore: \int_0^{\pi/2} e^{2ik\theta} d\theta = \left. -i \frac{1}{2k} e^{2ik\theta} \right|_0^{\pi/2} = \frac{[(-1)^{k+1} + 1]i}{2k} In fact, there is no need to calculate the exact value of this integral; one only needs to determine that it is a purely imaginary number. Then: \begin{aligned} \int_0^{\pi/2} \cos^{2n} \theta d\theta &= \frac{1}{2^{2n}} \int_0^{\pi/2} \left[ \frac{(2n)!}{(n!)^2} + \sum_{k=-n, k \neq 0}^{n} a_k e^{2ik\theta} \right] d\theta \\ &= \frac{1}{2^{2n}} \frac{(2n)!}{(n!)^2} \frac{\pi}{2} + i \sum_{k=-n, k \neq 0}^{n} \frac{[(-1)^{k+1} + 1]a_k}{2k} \end{aligned}
Regardless of the form of a_k, after summation and multiplication by i, it must result in a purely imaginary number. However, the left side is a real number, so the imaginary terms must be zero. Therefore: \begin{aligned} \int_0^{\pi/2} \cos^{2n} \theta d\theta &= \frac{1}{2^{2n}} \frac{(2n)!}{(n!)^2} \frac{\pi}{2} \\ &= \frac{(2n)!}{((2n)!!)^2} \frac{\pi}{2} \end{aligned}
However, if one wants to use the same technique to evaluate \int_0^{\pi/2} \sin^{2n+1} \theta d\theta and \int_0^{\pi/2} \cos^{2n+1} \theta d\theta, it is no longer simple. In fact, for odd powers, there seems to be no relatively simple and direct method.
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