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)


Oh! Matryoshka!

Given polynomial f(x) = a_0 + a_1 x+a_2 x^2 + ... + a_{n-1}x^{n-1}+a_n x^n, we wish to evaluate integral

\int \frac{f(x)}{(x-a)^p}\;dx, \quad p \in N^+\quad\quad\quad(1)

When p = 1,

\int \frac{f(x)}{x-a} \;dx= \int \frac{f(x)-f(a)+f(a)}{x-a}\;dx

= \int \frac{f(x)-f(a)}{x-a}\;dx + \int \frac{f(a)}{x-a}\;dx

=\int \frac{f(x)-f(a)}{x-a}\;dx + f(a)\cdot \log(x-a).


f(x) = a_0 + a_1x + a_2x^2 + ... + a_{n-1}x^{n-1} + a_n x^n


f(a) = a_0 + a_1 a + a_2 a^2 + ... + a_{n-1}a^{n-1} + a_n a^n

It follows that

f(x)-f(a) = a_1(x-a) + a_2(x^2-a^2) + ... + a_{n-1}(x^{n-1}-a^{n-1}) + a_n (x^n-a^n).

That is

f(x)-f(a) = \sum\limits_{k=1}^{n}a_k(x^k-a^k)

By the fact (see “Every dog has its day“) that

x^k-a^k =(x-a)\sum\limits_{i=1}^{k}x^{k-i}a^{i-1},

we have

f(x)-f(a) = \sum\limits_{k=1}^{n}a_k(x-a)\sum\limits_{i=1}^{k}x^{k-i}a^{i-1}=(x-a)\sum\limits_{k=1}^{n}(a_k\sum\limits_{i=1}^{k}x^{k-i}a^{i-1})


\frac{f(x)-f(a)}{x-a}=  \sum\limits_{k=1}^{n}(a_k\sum\limits_{i=1}^{k}x^{k-i}a^{i-1})\quad\quad\quad(2)


\int\frac{f(x)}{x-a}\;dx = \int \sum\limits_{k=1}^{n}(a_k \sum\limits_{i=1}^{k}x^{k-i}a^{i-1})\;dx + f(a)\log(x-a)

=\sum\limits_{k=1}^{n}(a_k \sum\limits_{i=1}^{k}\int x^{k-i}a^{i-1}\; dx)+ f(a)\log(x-a)


\int \frac{f(x)}{x-a} = \sum\limits_{k=1}^{n}(a_k\sum\limits_{i=1}^{k}\frac{x^{k-i+1}}{k-i+1}a^{i-1})+ f(a)\log(x-a)

Let us now consider the case when p>1:

\int \frac{f(x)}{(x-a)^p}\; dx

=\int \frac{f(x)-f(a)+f(a)}{(x-a)^p}\;dx

=\int  \frac{f(x)-f(a)}{(x-a)^p} + \frac{f(a)}{(x-a)^p}\;dx

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

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

\overset{(2)}{=}\int \frac{g(x)}{(x-a)^{p-1}}\;dx + \frac{f(a)(x-a)^{1-p}}{1-p}


g(x) = \frac{f(x)-f(a)}{x-a}=\sum\limits_{k=1}^{n}(a_k\sum\limits_{i=1}^{k}x^{k-i}a^{i-1}), a polynomial of order n-1.

What emerges from the two cases of p is a recursive algorithm for evaluating (1):

Given polynomial f(x) = \sum\limits_{k=0}^{n} a_k x^k,

\int \frac{f(x)}{(x-a)^p} \;dx, \; p \in N^+= \begin{cases}p=1: \sum\limits_{k=1}^{n}(a_k\sum\limits_{i=1}^{k}\frac{x^{k-i+1}}{k-i+1}a^{i-1})+ f(a)\log(x-a) \\p>1:  \int \frac{g(x)}{(x-a)^{p-1}}\;dx + \frac{f(a)(x-a)^{1-p}}{1-p}, \\ \quad\quad\quad g(x) = \frac{f(x)-f(a)}{x-a}=\sum\limits_{k=1}^{n}(a_k\sum\limits_{i=1}^{k}x^{k-i}a^{i-1}). \end{cases}

Exercise-1 Optimize the above recursive algorithm (hint: examine how it handles the case when f(x)=0)

Integration of Trigonometric Expressions

We will introduce an algorithm for obtaining indefinite integrals such as

\int \frac{(1+\sin(x))}{\sin(x)(1+\cos(x))}\;dx

or, in general, integral of the form

\int R(\sin(x), \cos(x))\;dx\quad\quad\quad(1)

where R is any rational function R(p, q), with p=\sin(x), q=\cos(x).


t = \tan(\frac{x}{2})\quad\quad(2)

Solving (2) for x, we have

x = 2\cdot\arctan(t)\quad\quad\quad(3)

which provides

\frac{dx}{dt} = \frac{2}{1+t^2}\quad\quad\quad(4)


\sin(x) =2\sin(\frac{x}{2})\cos(\frac{x}{2})\overset{\cos^(\frac{x}{2})+\sin^2(\frac{x}{2})=1}{=}\frac{2\sin(\frac{x}{2})\cos(\frac{x}{2})}{\cos^2(\frac{x}{2})+\sin^2(\frac{x}{2})}=\frac{2\frac{\sin(\frac{x}{2})}{\cos(\frac{x}{2})}}{1+\frac{\sin^2(\frac{x}{2})}{\cos^2(\frac{x}{2})}}=\frac{2\tan(\frac{x}{2})}{1+\tan^2(\frac{x}{2})}


\sin(x) = \frac{2 t}{1+t^2}\quad\quad\quad(5)


\cos(x) = \cos^2(\frac{x}{2})-\sin^2(\frac{x}{2})=\frac{\cos^2(\frac{x}{2})-\sin^2(\frac{x}{2})}{\cos^2(\frac{x}{2})+\sin^2(\frac{x}{2})}=\frac{1+\frac{\sin^2(\frac{x}{2})}{\cos^2(\frac{x}{2})}}{1+\frac{\sin^2(\frac{x}{2})}{\cos^2(\frac{x}{2})}}=\frac{1-\tan^2(\frac{x}{2})}{1+\tan^2(\frac{x}{2})}



We also have (see “Finding Indefinite Integrals” )

\int f(x)\;dx \overset{x=\phi(t)}{=} \int f(\phi(t))\cdot\frac{d\phi(t)}{dt}\;dt.


\int R(\cos(x), \sin(x))\;dx \overset{(2), (4), (5), (6)}{=} \int R(\frac{1-t^2}{1+t^2}, \frac{2t}{1+t^2})\cdot\frac{2}{1+t^2}\;dt,

and (1) is reduced to an integral of rational functions in t.

Example-1 Evaluate \int \csc(x)\;dx.

Solution: \csc(x) = \frac{1}{\sin(x)}\implies \int \csc(x)\;dx = \int \frac{1}{\sin(x)}\;dx

= \int \frac{1}{\frac{2t}{1+t^2}}\cdot\frac{2}{1+t^2}\;dt=\int\frac{1}{t}\;dt = \log(t) = \log(\tan(\frac{x}{2})).

Example-2 Evaluate \int \sec(x)\;dx.

Solution: \sec(x) = \frac{1}{\cos(x)}\implies \int \sec(x)\; dx =\int \frac{1}{\cos(x)}\;dx

=  \int \frac{1}{\frac{1-t^2}{1+t^2}}\cdot \frac{2}{1+t^2}\; dt=\int \frac{2}{1-t^2}\;dt=\int \frac{2}{(1+t)(1-t)}\;dt=\int \frac{1}{1+t} + \frac{1}{1-t}\;dt

=\int \frac{1}{1+t}\;dt - \int \frac{-1}{1-t}\;dt

=\log(1+t) -\log(1-t) =\log\frac{1+t}{1-t}=\log(\frac{1+\tan(\frac{x}{2})}{1-\tan(\frac{x}{2})}).

According to CAS (see Fig. 1),

Fig. 1

However, the two results are equivalent as a CAS-aided verification (see Fig. 2) confirms their difference is a constant (see Corollary 2 in “Sprint to FTC“).

Fig. 2

Exercise-1 According to CAS,

Show that it is equivalent to the result obtained in Example-1

Exercise-2 Try

\int \frac{1}{\sin(x)+1}\;dx

\int \frac{1}{\sin(x)+\cos(x)}\;dx

\int \frac{1}{(2+\cos(x))\sin(x)}\;dx

\int \frac{1}{5+4\sin(x)}\;dx

\int \frac{1}{2\sin(x)-\cos(x)+5}\;dx

and of course,

\int \frac{1+\sin(x)}{\sin(x)(1+\cos(x))}\;dx