“Differentiation under the integral sign”, a.k.a., “Feynman’s trick” is a clever application of Leibniz’s rule (LR-1):
Let be continuous and have a continuous derivative in a domain of plane that includes the rectangle
“Feynman’s trick” is known to be an effective technique for evaluating difficult definite integral such as
Is Feynman’s “trick” applicable to indefinite integrals too?
In other words, is it also true that
Let’s apply to indefinite integral
Since , the right-hand side of (1) is
For , we have
It checks out:
Let’s also evaluate
Let , (2) yields
It checks out too:
Now that we have gained confidence in the validity of let’s prove it.
where is a function of , we have,
i.e., for , a function of both and
It follows that
Exercise-1 Evaluate .