# 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 $a=b$,

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

and,

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

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) $(a0, b>0) \implies (b-a>0, a>0, b>0)$

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

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) $(a0, a<0, b<0) \implies \frac{b}{b-a} <0$

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 $b>a$)