Question

What is the Feynman integral trick?

Answer 1

See below

Explanation:

It's a way of solving otherwise tricky definite integrals using a lot of creativity and, as the common theme, the Liebnitz rule for differentiating under the integral sign.

It is not a process or method in the sense that you don't just follow some rules and find the answer. This example might make this clear.

`I = int_0^oo (sin x)/x dx qquad circ`

It's hardly intuitive to me, this is one I knew beforehand, but let's consider this integral instead:

`bar I(alpha) = int_0^oo e^(-alpha x) (sin x)/x dx qquad triangle`, "kinda like" a Laplace Transform, I guess

and then

`(d bar I)/(d color(red)(alpha)) = int_0^oo - color(blue)(x) e^(-alpha x) (sin x)/(color(blue)(x)) dx`

`=- int_0^oo e^(-alpha x) sin x dx`

that's now très do-able as an indefinite integral, using, say, IBP:

`- int e^(-alpha x) sin(x) dx = (e^ (- alpha x) (alpha sin(x)+cos(x)))/(alpha^2+1)+C`

So

`(d bar I)/(d alpha) = ( (e^ (- alpha x) (alpha sin(x)+cos(x)))/(alpha^2+1) )_0^oo`

the expression `e^(- alpha x)` nullifies the `oo` limit and so we are left with

`(d bar I)/(d alpha) = - 1/(alpha^2 +1)` and by tan sub

`bar I(alpha) = - arctan alpha + C qquad square`

We then steal an IV from `triangle` by noting that `bar I(alpha to oo) = 0`. [Finding this IV is often the sticking point, when you have simplified an integral in this way.]

so `square` becomes

`bar I(alpha to oo) = - arctan (alpha to oo) + C`

`0 = - pi/2 + C, C = pi/2`

`implies bar I(alpha) = - arctan alpha + pi/2`

The original integral in `circ` has `alpha = 0` so we now know that

`I = bar I(0) = - arctan 0 + pi/2 = pi/2`

It's a pretty neat trick with Laplace and Fourier Transforms, in my own experience, but it's still very hit and miss. I reckon the best thing to do, if you wanna learn it, is to find a book or website and work a load of examples, with hints where necessary.

After thought: It can also be used on indefinite integrals but I have no experience of those.