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Learning Feynman's Trick for Integrals

62 pointsby Zen1thtoday at 7:55 PM9 commentsview on HN

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tacitusarctoday at 10:18 PM

I just finished Mathematica by David Bessis and I wish this information was presented in the way he talks about math: using words and imagery to explain what is happening, and only using the equations to prove the words are true.

I just haven’t had to use integral calculus in so many years, I don’t recall what the symbols mean and I certainly don’t care about them. That doesn’t mean I wouldn’t find the problem domain interesting, if it was expressed as such. Instead, though, I get a strong dose of mathematical formalism disconnected from anything I can meaningfully reason about. Too bad.

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lordnachotoday at 10:09 PM

My issue with both this and u-substitution is that you don't know what expression to use. There are a LOT of expressions that plausibly simplify the integral. But you have to do a bunch of algebra for each one (and not screw it up!), without really knowing whether it actually helps.

OTOH, if I'm given the expression, it's just mechanical and unrewarding.

zeroonetwothreetoday at 10:44 PM

It’s interesting he mentions he doesn’t like contour integration since many integrals can be done either way.

Feynman’s trick is equivalent to extending it into a double integral and then switching the order of integration.

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impossibleforktoday at 9:37 PM

It starts off with a pretty major error.

I'(t)=\int_0^1 \partial/(\partial t)((x^t - 1)/(ln x))dx = \int_0^1 x^t dx=1/(t+1), when it is actually equal to \int_0^1 x^{t-1}/ln(x)dx.

These two are definitely not always equal to each other.

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