Learning Feynman's Trick for Integrals

tacitusarc an hour ago | parent | next [-]

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.

	▲	chrisweekly 41 minutes ago \| parent [-]
		That's one of the things I like best about https://betterexplained.com -- it focuses on ways to gain intuition about a given math concept, using visuals and metaphors as appropriate. If only math education were always presented like that....

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lordnacho an hour ago | parent | prev | next [-]

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.

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zeroonetwothree 27 minutes ago | parent [-]

That’s how most of math works past high school. It requires a lot of practice and intuition.

	▲	lordnacho 23 minutes ago \| parent [-]
		I don't know about this particular case though, I get the feeling there's a system to it that can be exploited by eg Wolfram. It's just that you're in the dark for a long time before you find the switch.

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zeroonetwothree 25 minutes ago | parent | prev | next [-]

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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impossiblefork 2 hours ago | parent | prev [-]

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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owalt an hour ago | parent [-]

No, it is correct. The integral is with respect to x, and the ordinary/partial derivatives are with respect to t. Written out fully, the derivative computation is

d/dt (x^t - 1)/ln(x) = d/dt [exp(ln(x)t) - 1]/ln(x) = ln(x)exp(ln(x)t)/ln(x) = exp(ln(x)t) = x^t.

Edit: d/dt exp(ln(x)t) = ln(x)exp(ln(x)t) by the chain rule, while d/dt (1/ln(x)) = 0 since the expression is constant with respect to t.

There are convergence considerations that were not discussed in the blog post, but the computations seem to be correct.

	▲	impossiblefork 18 minutes ago \| parent [-]
		Ah, yes. I don't understand how I differentiate with respect to x instead of t, but...