Restate Feynman’s “Great Identity”

In a letter dated July 7, 1948, Richard Feynman, a Nobel laureate in physics (1965) made a claim:

I am the possessor of a swanky new scheme to do each problem in terms of one with one less energy denominator. It is based on the great identity \frac{1}{ab} = \int \limits_{0}^{1} \frac{1}{(ax+b(1-x))^2}\;dx.

(The Beat of a Different Drum: The Life and Science of Richard Feynman, by Jagdish Mehra, page 262)

Assuming non-zero constants a, b are both real but otherwise arbitrary, let’s check the validity of Feynman’s “great identity”.

If a \ne b,

\int \frac{1}{(ax+b(1-x))^2}\;dx

= \int \frac{1}{((a-b)x+b)^2}\;dx

=  \int \frac{1}{a-b}\cdot \frac{a-b}{((a-b)x+b)^2}\;dx

= \frac{1}{a-b}\int \frac{((a-b)x+b)'}{((a-b)x+b)^2}\;dx

= \frac{1}{a-b}\cdot \frac{-1}{(a-b)x+b}.

Using Leibniz’s rule,

\int\limits_{0}^{1}\frac{1}{(ax+b(1-x))^2}\;dx = \frac{1}{a-b}\cdot\frac{-1}{(a-b)x+b}\bigg| _{0}^{1} =\frac{1}{a-b}(\frac{-1}{a}-\frac{-1}{b})=\frac{1}{ab}.

When a=b,

\int \frac{1}{(ax+b(1-x))^2}\;dx = \int \frac{1}{b^2}\;dx =\frac{1}{b^2}x


\int \limits_{0}^{1}\frac{1}{(ax+b(1-x))^2}\;dx = \frac{1}{b^2}x\bigg|_{0}^{1} = \frac{1}{b^2}=\frac{1}{ab}.

All is as Feynman claimed:

\int\limits_{0}^{1}\frac{1}{(ax+b(1-x))^2}\;dx = \frac{1}{ab}\quad\quad\quad(\star)

There is something amiss:

If a and b have opposite signs i.e., ab<0 then the right hand side of (\star) is negative. But the integrand is squared so the integral on the left hand side of (\star) is never negative, no matter what a and b may be.

Let’s figure it out !

In its full glory, Leibniz’s rule we used to obtain (\star) is

If the real-valued function F on an open interval I in R has the continuous derivative f and a, b \in I then \int\limits_{a}^{b} f(x)\; dx = F(b)-F(a).

Essentially, the rule requires the integrand f to be a continuous function on an open interval that contains a and b.

Solving ax+b(1-x)=0 for x yields

x = \frac{b}{b-a},

the singularity of integrand \frac{1}{(ax+(1-x))^2} in (\star).

For ab<0, we consider the following two cases:

Case (1-1) (a>0, b<0) \implies (a>0, b<0, b-a<0)\implies (\frac{b}{b-a}>0,

\frac{b}{b-a} = \frac{b-a+a}{b-a} = 1+\frac{a}{b-a}<1)\implies 0<\frac{b}{b-a}<1

Case (1-2) (a<0, b>0) \implies (a<0, b>0, b-a>0) \implies (\frac{b}{b-a}>0,

\frac{b}{b-a} = \frac{b-a+a}{b-a} = 1+\frac{a}{b-a}<1) \implies  0<\frac{b}{b-a}<1

From both cases, we see that

when ab<0, \frac{1}{(ax+b(1-x))^2} has a singularity in (0, 1) \implies \frac{1}{(ax+b(1-x))^2} is not continuous in (0, 1).

Applying Leibniz’s rule to \int\limits_{0}^{1} \frac{1}{(ax+b(1-x))^2}\;dx regardless of integrand’s singularity thus ensured an outcome of absurdity.

However, ab>0 paints a different picture.

We have

Case (2-0) a=b \implies \frac{1}{(ax+b(1-x))^2}=\frac{1}{b^2} \implies no singularity

Case (2-1) (a>b, a>0, b>0) \implies (b-a<0, a>0, b>0) \implies \frac{b}{b-a}<0

Case (2-2) (a<b, a>0, b>0) \implies (b-a>0, a>0, b>0)

\implies \frac{b}{b-a} = \frac{b-a+a}{b-a}=1+\frac{a}{b-a}>1

Case (2-3) (a>b, a<0, b<0) \implies (b-a<0, a<0, b<0) \implies \frac{b}{b-a}=1+\frac{a}{b-a}>1

Case (2-4) (a<b, a<0, b<0) \implies (b-a>0, a<0, b<0) \implies \frac{b}{b-a} <0

All cases show that when ab>0, the integrand has no singularity in (0,1).

It means that \frac{1}{(ax+b(1-x))^2} is continuous in (0, 1) and therefore, Leibniz’s rule applies.

So let’s restate Feynman’s “great identity”:

a\cdot b > 0 \iff \frac{1}{ab} = \int\limits_{0}^{1}\frac{1}{(ax+b(1-x))^2}\;dx

Exercise-1 Evaluate \int\limits_{0}^{1}\frac{1}{(ax+b(1-x))^2}\;dx using Omega CAS Explorer. For example,

(hint : for ab>0, specify a > b or b>a)

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