Feynman's trick for indefinite integrals

user434058

Jul 31, 2020 17:50

Hi! Could anybody explain me the comments by zwim in this comment thread?

Ted Shifrin

@FakeMod: The problem is that the indefinite integral is not a function, so it makes no sense to differentiate it with respect to the parameter. The constant of integration could very well be a function of the parameter as well. I think it's hopeless. You have to fix limits of integration (one of which will be your independent variable, if you like), and then you have a well-defined antiderivative.

user434058

@TedShifrin Hmmm... I understand your second sentence, however, I don't see why the final expression of the indefinite integral won't be a function of the parameter we introduced in the integrand?

Ted Shifrin

It's not a well-defined function. The indefinite integral is not a function. It's a symbol for "any antiderivative."

user434058

@TedShifrin Oh, yeah. So except the arbitrary integration constant, the rest is a well characterised function in the parameter that we introduced, and the integration variable, right?

Ted Shifrin

But the integration constant is not actually a constant, as I already said. That's the fallacy. It may vary and be an arbitrary function of your parameter $b$.

user434058

Jul 31, 2020 18:00

@TedShifrin "You have to fix limits of integration (one of which will be your independent variable, if you like), and then you have a well-defined antiderivative." But how do I get any closer to using Feynman's technique after doing that?

user434058

@TedShifrin Oh! I see.

Ted Shifrin

I don't know why Feynman's name gets attached to this, incidentally. That's totally bizarre.

If it were a Feynman path integral, OK.

user434058

@TedShifrin Prolly, 'cause he thought of it first (I may be terribly wrong)

Ted Shifrin

That's totally not true.

It was around in mathematics and in physics long before him.

user434058

@TedShifrin Then the only reason I can think of is that he popularized it.

Ted Shifrin

Jul 31, 2020 18:04

Nah. I've never heard of this before.

user434058

@TedShifrin Anyways, I still don't know why you said ths \/

user434058

5 mins ago, by FakeMod

@TedShifrin "You have to fix limits of integration (one of which will be your independent variable, if you like), and then you have a well-defined antiderivative." But how do I get any closer to using Feynman's technique after doing that?

Ted Shifrin

If you consider $F(x,b) = \int_a^x g(t,b)\,dt$, then $\partial F/\partial b = \int_a^x \partial g/\partial b\,dt$ is valid.

user434058

@TedShifrin That's alright, but does it get me any closer to evaluating the integral this way?

Balarka Sen

I think the story is Feynman liked to say to people that no matter how complicated an integral a mathematician can give to him he can always do it very fast, and differentiation under the integral sign with respect to a different parameter thing was a trick upon his sleeve

Ted Shifrin

Jul 31, 2020 18:09

I have no idea if it gets you closer to evaluating the integral. The differentiation under the integral trick works sometimes and doesn't work others.

user434058

@TedShifrin So any chances, whatsoever, of it working with indefinite integrals transformed to indefinite integrals?

Ted Shifrin

No.

Charlie Shuffler

Hi all

Ted Shifrin

I explained why it's nonsensical.

$Mathematics$ Mathematics

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