Introducing Feynman’s Integral Method (MAA Meeting, Spring 2024)

/ Michael Xue

Introducing Feynman’s Integral Method

Michael Xue

Indiana Section MAA Spring 2024 MEETING
Marian University – Indianapolis
April 5-6, 2024

“The first principle is that you must not fool yourself and you are the easiest person to fool.”

— Richard Feynman

This is one of my favorite quotes because it speaks to the importance of humility when approaching mathematics and science. To me, it means not being constrained by my existing knowledge or beliefs, nor by previous learning or observations of others’ methods. It’s about freeing my mind to view things from a entirely new perspective.

Feynman’s Integral Method

Leibniz : “\displaystyle \frac{d}{d\beta}\int\limits_{a}^{b}f(x, \beta) \;dx = \int\limits_{a}^{b}\frac{\partial}{\partial\beta}f(x, \beta)\;dx”

Feynman: “To evaluate hard \displaystyle \int\limits_{a}^{b} f(x)\;dx, differentiate under the integral sign!”

Feynman’s integral method is a potent mathematical technique developed to simplify complex integrals, often transforming them into more manageable forms. This method, pioneered by Richard Feynman, effectively brought Leibniz’s theorem into practical use, showcasing its real-world applicability in solving intricate mathematical problems.

Two algorithms implementing Feynman’s Integral method

Algorithm #1 Reducing \displaystyle \int\limits_{a}^{b} f(x)\;dx

Algorithm #2 Generating \displaystyle \int\limits_{a}^{b}f(x)\;dx from a known definite integral

Like many other mathematicians and physicists, I’ve extensively experimented with Feynman’s method in solving numerous problems. Through a gradual, inductive approach, I’ve discerned patterns that allowed me to synthesize these into two efficient algorithms. Today’s talk will focus on these algorithms, demonstrating their effectiveness in implementing Feynman’s method. I’ll provide multiple examples to illustrate how these algorithms can be used to solve a wide range of mathematical problems.

Example-1

Evaluate \displaystyle I=\int\limits_{0}^{1}\frac{x-1}{\log(x)}\;dx

\displaystyle J(\beta) = \int\limits_{0}^{1}\frac{x^\beta-1}{\log(x)}\;dx \implies J(1)=I, J(0)=0

\displaystyle \frac{d}{d\beta}J(\beta) = \int\limits_{0}^{1}\frac{\partial}{\partial \beta}\left(\frac{x^\beta-1}{\log(x)}\right)\;dx = \int\limits_{0}^{1}x^\beta\;dx=\frac{x^{\beta+1}}{\beta+1}\bigg|_{0}^{1}=\frac{1}{\beta+1}

\displaystyle \int\limits_{0}^{1}\frac{d}{d\beta}J(\beta)\;d\beta = \int\limits_{0}^{1}\frac{1}{\beta+1}\;d\beta \implies J(1)-J(0) = \log(\beta+1)\bigg|_{0}^{1} = \log(2)

\displaystyle \overset{J(1)=I, J(0)=0}{\implies} I = \log(2)

Example-2

Evaluate \displaystyle I = \int\limits_{0}^{\infty} e^{-x^2}\;dx

\displaystyle J(\beta) = \int\limits_{0}^{\infty}\frac{e^{-\beta^2(x^2+1)}}{1+x^2}\;dx \implies J(0)=\arctan(x)\bigg|_{0}^{\infty}=\frac{\pi}{2}, \quad \underline{J(\infty)=0}

\displaystyle \frac{d}{d\beta}J(\beta) = \int\limits_{0}^{\infty}\frac{\partial}{\partial \beta}\left(\frac{e^{-\beta^2(1+x^2)}}{1+x^2}\right)\;dx = -2\beta e^{-\beta^2}\int\limits_{0}^{\infty}e^{-\beta^2x^2}\;dx

\displaystyle \overset{\phi(t)=\frac{t}{\beta}, \phi'(t)\frac{1}{\beta}}{=}-2e^{-\beta^2}\int\limits_{0}^{\infty}e^{-t^2}\;dt = -2e^{-\beta^2}I

\displaystyle \int\limits_{0}^{\infty}\frac{d}{d\beta}J(\beta)\;d\beta=\int\limits_{0}^{\infty}-2e^{-\beta^2}I\; d\beta = -2I\int\limits_{0}^{\infty}e^{-\beta^2}\;d\beta = -2 I \cdot I = -2I^2

\displaystyle J(\infty) - J(0) = -2I^2 \overset{J(0)=\frac{\pi}{2}, J(\infty)=0}{\implies} 0 - \frac{\pi}{2}=-2I^2 \implies I=\frac{\sqrt{\pi}}{2}

Example-3

Evaluate \displaystyle \int\limits_{0}^{1}\log(x)\;dx

For \displaystyle \beta \ge 0, \int\limits_{0}^{1}x^\beta\;dx = \frac{1}{\beta+1} \implies \frac{d}{d\beta}\int\limits_{0}^{1}x^\beta\;dx = \frac{d}{d\beta}\left(\frac{1}{\beta+1}\right)

\displaystyle \implies \int\limits_{0}^{1}\frac{\partial}{\partial \beta}x^\beta\;dx = \frac{-1}{(\beta+1)^2}\implies \int\limits_{0}^{1}x^\beta\log(x)\;dx = \frac{-1}{(\beta+1)^2}

\displaystyle \overset{\beta=0}{\implies} \int\limits_{0}^{1}\log(x)\;dx = -1

\displaystyle \int\limits_{0}^{1}\log(x)\;dx = \int\limits_{0}^{1}\frac{d}{dx}x\cdot\log(x)\;dx = \underbrace{x\log(x)\bigg|_{0}^{1}}_{\blacksquare}-\int\limits_{0}^{1}\;dx

Using traditional methods, you often encounter problematic forms like 0 multiplied by infinity. However, Feynman’s method effectively circumvents this issue without the need to invoke L’Hôpital’s rule.

Example-4

Given \displaystyle \int\limits_{-\infty}^{\infty}\frac{e^{2x}}{a e^{3x}+b}\;dx = \frac{2\pi}{3\sqrt{3}a^\frac{2}{3} b^\frac{1}{3}}, a, b >0, find \displaystyle \int\limits_{-\infty}^{\infty}\frac{e^{2x}}{(e^{3x}+1)^2}\;dx.

\displaystyle \frac{d}{db}\int\limits_{-\infty}^{\infty}\frac{e^{2x}}{a e^{3x}+b}\;dx = \frac{d}{db}\left(\frac{2\pi}{3\sqrt{3}a^\frac{2}{3} b^\frac{1}{3}}\right) \implies \int\limits_{-\infty}^{\infty}-\frac{e^{2x}}{(a e^{3x}+b)^2}\;dx = -\frac{2\pi}{3^\frac{5}{2}a^\frac{2}{3}b^\frac{4}{3}}

\displaystyle \overset{a=1, b=1}{\implies}\int\limits_{-\infty}^{\infty}\frac{e^{2x}}{(e^{3x}+1)^2}\;dx=\frac{2\pi}{9\sqrt{3}}

While preparing for this talk last weekend, I developed an example and here is an abbreviated version of the proof to highlight the key steps.

Example-5

Prove \displaystyle 1 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + ... = \frac{\pi^2}{6}\quad\quad(\star)

Since \displaystyle\sum\limits_{i=1}\frac{1}{i^2}=\sum\limits_{i=1}\frac{1}{(2i)^2} + \sum\limits_{i=0}\frac{1}{(2i+1)^2}\implies \sum\limits_{i=0}^{\infty}\frac{1}{(2i+1)^2} = \frac{3}{4}\sum\limits_{i=1}^{\infty}\frac{1}{i^2},

to prove (\star), it suffices to demonstrate that

\displaystyle \sum\limits_{i=0}^{\infty}\frac{1}{(2i+1)^2} = \frac{\pi^2}{8}

\displaystyle I = \int\limits_{0}^{\infty}\frac{\arctan(x)}{1+x^2}\;dx, \quad J(\beta) =\int\limits_{0} ^{\infty}\frac{\arctan(\beta x)}{1+x^2}\;dx

\displaystyle \implies J(1) = I =\frac{1}{2}\arctan^2(x)\bigg|_{0}^{\infty} = \frac{\pi^2}{8}, \quad J(0)=0

\displaystyle \frac{d}{d\beta}J(\beta) = ... = \int\limits_{0}^{\infty}\frac{x}{(1+x^2)(\beta^2 x^2+1)}\;dx \implies \int\limits_{0}^{1}\frac{d}{d\beta}J(\beta)d\beta=\int\limits_{0}^{1}\int\limits_{0}^{\infty}\frac{x}{(1+x^2)(\beta^2 x^2+1)}\;dx d\beta

\displaystyle \implies J(1)-J(0) = \int\limits_{0}^{1}\int\limits_{0}^{\infty}\frac{x}{(1+x^2)(\beta^2 x^2+1)}\;dx\;d\beta=...=\int\limits_{0}^{1}\frac{\log(\beta)}{\beta^2-1}\;d\beta = ...

\displaystyle =\int\limits_{0}^{1}\sum\limits_{i=0}^{\infty}\frac{\beta^{2i}}{2i+1}\;d\beta =\sum\limits_{i=0}^{\infty}\frac{1}{(2i+1)^2} \quad \overset{J(1)=\frac{\pi^2}{8}, J(0)=0}{\implies} \quad \sum\limits_{i=0}^{\infty}\frac{1}{(2i+1)^2}= \frac{\pi^2}{8}

For full-length proof, read my posting “An Analytic Proof of the Extraordinary Euler Sum” on

https://nuomegamath.wordpress.com/2024/04/02

Many of you may know Dr. Zeilberger, “a champion of using computers and algorithms to perform mathematics quickly and efficiently.” (Wikipedia) You can find more about his work on his website. After completing the proof a few days ago, I reached out to him for a review. He graciously agreed to do so at short notice. Dr. Zeilberger is known for his straightforward style, always telling it like it is.

Dr. Zeilberger, Professor at Rutgers University.

https://sites.math.rutgers.edu/~zeilberg/

I would like to revisit example 3 and solve it using a different approach than Feynman’s. This aligns with Feynman’s fundamental principle of not fooling oneself and emphasizes the importance of viewing problems from entirely new perspectives.

Question: WWFD?

Answer: I see A = \displaystyle\int\limits_{-\infty}^{0}e^y\; dy= e^y\bigg|_{-\infty}^{0} = 1-0 =1 and \displaystyle \int\limits_{0}^{1}|\log(x)|\;dx = A. Since for 0 < x \le 1, \log(x) \le 0, I have \displaystyle\int\limits_{0}^{1}-\log(x)\;dx=A so \displaystyle\int_{0}^{1}\log(x)\;dx = -A = -1

Leave a comment