Question

What is the Feynman integral trick?

Answer 1

Hint : Feynman integral trick is a trick which was discovered by Richard Feynman. It’s an easy way of solving tricky definite integrals using a common theme, the Leibniz rule for differentiating under the integral sign and with creativity.

Complete step-by-step answer :
Feynman integral trick is not a process or method that just follows some rules and finds the answer. It’s much more than that.
Let’s see by an example:
We have an integral function $I = \int\limits_0^\infty {\dfrac{{\sin x}}{x}dx}$ :
Adding a constant $\alpha$ , we get the terms as:
$I\left( \alpha \right) = \int\limits_0^\infty {{e^{ - \alpha x}}\dfrac{{\sin x}}{x}dx}$ ……..(i)
This is a kind of a Laplace Transform.
The Feynman integral trick we know, uses the Leibniz rule for differentiating under the integral sign and applying that, we get:
$\dfrac{{dI}}{{d\alpha }} = \int\limits_0^\infty { - x{e^{ - \alpha x}}\dfrac{{\sin x}}{x}dx} \\\ = - \int\limits_0^\infty {{e^{ - \alpha x}}\sin xdx} \;$
Using some rules of indefinite integrals, like IBP, we get: $- \int\limits_0^\infty {{e^{ - \alpha x}}\sin xdx} = \dfrac{{{e^{ - ax}}\left( {\alpha \sin \left( x \right) + \cos \left( x \right)} \right)}}{{{\alpha ^2} + 1}} + C$
On further solving with limits, we get:
$\dfrac{{dI}}{{d\alpha }} = \left( {\dfrac{{{e^{ - ax}}\left( {\alpha \sin \left( x \right) + \cos \left( x \right)} \right)}}{{{\alpha ^2} + 1}}} \right)_0^\infty$
As, we know that ${e^{ - ax}}$ will nullify the $\infty$ limit, so we get:
$\dfrac{{dI}}{{d\alpha }} = \dfrac{{ - 1}}{{{\alpha ^2} + 1}} + C$
From tan sub, we can write:
$I\left( \alpha \right) = - \arctan \alpha + C$ …….(ii)
When, we simplify the equation with respect to (i), in get:
$I\left( {\alpha \to \infty } \right) = - \arctan \left( {\alpha \to \infty } \right) + C \\\ 0 = - \dfrac{\pi }{2} + C,C = \dfrac{\pi }{2} \\\ = > I\left( \alpha \right) = - \arctan \left( \alpha \right) + \dfrac{\pi }{2} \;$
The original integral which we started with had $\alpha = 0$ , so
$= > I = I\left( 0 \right) = - \arctan \left( 0 \right) + \dfrac{\pi }{2} = \dfrac{\pi }{2}$

Therefore, the integral of $I = \int\limits_0^\infty {\dfrac{{\sin x}}{x}dx}$ gives the value $\dfrac{\pi }{2}$
Overall, this trick is a messy trick with Laplace and Fourier Transforms.

Note : It requires a detailed study of Laplace and Fourier Transforms, to solve the values of integral using Feynman integral trick.
Do not rush in solving the question instead always go step by step otherwise it would lead to various confusions.
It can also be used for indefinite integrals.
Always cross check the answer for once.