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”.
Using Leibniz’s rule,
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.
Case (2-0) no singularity
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 )