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 *.

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

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

If ,

.

Using Leibniz’s rule,

.

When ,

and,

.

All is as Feynman claimed:

There is something amiss:

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

Let’s figure it out !

In its full glory, Leibniz’s rule we used to obtain is

If the real-valued function on an open interval in has the continuous derivative and then .

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

Solving for yields

,

the singularity of integrand in .

For , we consider the following two cases:

Case (1-1) ,

Case (1-2) ,

From both cases, we see that

when has a singularity in is *not* continuous in .

Applying Leibniz’s rule to regardless of integrand’s singularity thus ensured an outcome of absurdity.

However, paints a different picture.

We have

Case (2-0) no singularity

Case (2-1)

Case (2-2)

Case (2-3)

Case (2-4)

All cases show that when , the integrand has no singularity in .

It means that is continuous in and therefore, Leibniz’s rule applies.

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

Exercise-1 Evaluate using Omega CAS Explorer. For example,

*(hint *: for , specify or )