Feynman's Integration Technique — Differentiation Under the Integral Sign (4)

Taking advantage of some free time this morning, I’ll get this article written. The Gaokao (National College Entrance Examination) results will be announced this afternoon, and after that, I might be busy for a while. This should be the final article in the “Feynman’s Integration Technique” series. It mainly discusses another example of Feynman’s integration method. What’s different is that this is the first time BoJone has independently solved a problem using Feynman’s integration method. The examples mentioned previously were provided by books along with hints, which allowed BoJone to calculate them. Therefore, this problem holds a bit of commemorative significance.

Wayne on the Mathematics R&D Forum once sought proof for the following proposition:

\int_0^{\infty}\frac{f(x,2m-1)-\sin x}{x^{2m+1}}dx where f(x,2m-1) represents the (2m-1)-th order Taylor expansion of \sin x.

For example, when m=1: \int_0^{\infty}\frac{x-\sin x}{x^3}dx

When m=2: \int_0^{\infty}\frac{x-\frac{x^3}{6}-\sin x}{x^5}dx

With the help of software, I found the result to be: \frac{\pi(-1)^{m-1}}{2(2m)!}

Someone provided a proof using the Laplace transform, but I only have a conceptual understanding of the Laplace transform and do not know the specific methods. I used Feynman’s integration method to generalize the problem. Suppose we have a function y=f(x), and we want to find the integral:

F(t)=\int_a^b \frac{f(0)+f'(0)(tx)+f''(0)\frac{(tx)^2}{2}+\dots+f^{(n)}(0)\frac{(tx)^n}{n!}-f(tx)}{x^{n+1}} dx

We have: \begin{aligned}\frac{d F(t)}{dt}=\int_a^b \frac{f'(0)+f''(0)(tx)+\dots+f^{(n)}(0)\frac{(tx)^{n-1}}{(n-1)!}-f'(tx)}{x^n} dx\end{aligned}

Differentiating n times consecutively yields: \begin{aligned}\frac{d^n F(t)}{dt^n}=\int_a^b \frac{f^{(n)}(0)-f^{(n)}(tx)}{x} dx\end{aligned}

In some cases, we can differentiate n+1 times to get: \begin{aligned}\frac{d^{n+1} F(t)}{dt^{n+1}}=\int_a^b -f^{(n+1)}(tx) dx = \left. -\frac{1}{t} f^{(n)}(tx) \right|_a^b\end{aligned}

Then, integrating with respect to the variable t for n or n+1 times will yield several integration constants. We know that when t=0, the original integral value is 0, which allows us to determine that each integration constant is 0.

For Wayne’s problem, f(x)=\sin(x), n=2m, and f^{(2m)}(x)=(-1)^m \sin x. \begin{aligned}\frac{d^{2m} F(t)}{dt^{2m}}=\int_0^{\infty} \frac{-(-1)^m \sin (tx)}{x} dx\end{aligned}

(Here we cannot differentiate further, as differentiating again would yield -(-1)^m \sin (tx)|_0^{\infty}, which is meaningless.)

According to \int_0^{\infty} \frac{\sin x}{x}dx=\frac{\pi}{2}, we have: \frac{d^{2m} F(t)}{dt^{2m}}=-(-1)^m\frac{\pi}{2}=(-1)^{m-1} \frac{\pi}{2}

After integrating 2m times: F(t)=\frac{(-1)^{m-1} \pi \cdot t^{2m}}{2(2m)!}.

Simply taking t=1 gives the result.

Thus, an interesting definite integral is beautifully solved using Feynman’s integration method!

Appendix: Proof via Laplace Transform