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Feynman's Calculus

  1. Jun 30, 2005 #1
    Feynman wrote in his book ‘Surely You’re Joking, Mr. Feynman!’:

    “That book [Advanced Calculus, by Woods] also showed how to differentiate parameters under the integral sign – it’s a certain operation. It turns out that’s not taught very much in the universities; they don’t emphasize it. But I caught on how to use that method, and I used that one damn tool again and again. So because I was self-taught using that book, I had peculiar methods of doing integrals.”

    What is this ‘differentiating under the integral sign’? Does anyone know this? Can anyone please help me?

    And another thing Feynman wrote about is contour integration. What is this contour integration? Can anyone help me with that too?

    In another chapter, Feynman wrote about computing the cube root of a number. The number was 1729.03. He wrote:

    “I happened to know that a cubic foot contains 1728 cubic inches, so the answer is a tiny bit more than 12. The excess, 1.03, is only one part in nearly 2000, and I have learned in calculus that for small fractions, the cube root’s excess is one-third of the number’s excess. So all I had to do is find the fraction 1/1728 and multiply by 4 (divide by 3 and multiply by 12). So I was able to pull out a whole lot of digits that way.”

    Can anyone explain that to me? What is the ‘cube root’s excess’? What is the ‘number’s excess’?

    thanks in advance to anyone who can help.
  2. jcsd
  3. Jun 30, 2005 #2


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    Differentiating under the integral: Leibniz's rule- [tex]\frac{d}{dx}\int_{a(x)}^{b(x)} f(x,t)dt= \int_{a(x)}^{b(x)}\frac{\partial f(x,t)}{\partial x}dt+ \frac{da(x)}{dx}f(x,a(x))- \frac{db(x)}{dx}f(x,b(x))[/tex].

    In the second problem, "excess" is not referring to a number's excess but simply how much larger one number is than another. He happened to know that 123= 1728 (that is, that 12 is the cube root of 1728) and he wanted to find the cube root of 1729.03. The "excess", in this problem, is 1729.03- 1728= 1.03.

    The derivative of x1/3 is (1/3)x-2/3 so d(x1/3 (the "change" in x2 if x changes slightly- what Feynman is calling the "excess") is (1/3)x-2/3 dx. Here x= 1728 so x-2/3= 1/122= 1/144. (1/3)(1/144)= 1/432 and dx= 1.03. d(x1/3)= 1.03/144= 0.007222... The third root of 1729.03= 12.007222...
  4. Jun 30, 2005 #3


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    What is this ‘differentiating under the integral sign’?
    When an integrant is nice the integral of the derivative equals the derivative of the integral. So if you have an integral with a parameter (a varible other than the variable of integration) or you insert a parameter you can integrate the derivative of the integrant to find the derivative of the integral. Using the derivative of the integral one can write often find the original integral.
    Here is an example
    find [tex]\int_0^\infty\exp(-x^2-1/x^2) dx[/tex]
    let [tex]F(t)= \int_0^\infty\exp(-x^2-t^2/x^2) dx[/tex]
    now we want F(1)
    F'(t)=-2F(t) so
    F(t)=F(0) exp(-2 t)
    Clearly F(0)=(2^-1) sqrt(pi)
    F(1)=(2^-1) sqrt(pi)exp(-2)~.1194
    What is this contour integration?
    Contour integration what happens when you integrate a function of several along a contour. In 1-D you integrate from a to b you must follow a line in n-D you have n-1 degrees of freedom. In 2-D if you wanted to integrate from (-1,0) to (1,0) you could integrate along a stait line or along the top half of a circle centered at the origin with radius 1, or any number of other paths.
    What is the ‘cube root’s excess’? What is the ‘number’s excess’?
    let f(x)=x^(1/3)
    Cube roots excess=f(x)-f(y)
    number’s excess=x-y
    in calculus one learns
    number’s excess~(1/3)Cube roots excess
    or f(x)-f(y)~(1/3)(x-y)
    where ~ means "is about" and x~y
    use 1.03~1
    since 1+1/1728~1
    So one can get pretty close easily this way.
    Last edited: Jun 30, 2005
  5. Jun 30, 2005 #4


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    in michael sopivaks little book, calculus on manifolds, he explains path ingration very well, and also shoiws that the three basic theorems on interchange of limits are essentially equivalent:

    1) interchange of derivative and integral (differentiating under the integral sign)
    2) interchange of order of (partial) derivatives
    3) interchange of order of integration (fubini)
  6. Jun 30, 2005 #5
    Here is an example of differentiating under the integral sign:

    [tex] \int x Cos (a x) dx = \int \frac{\partial}{\partial a} -Sin (a x) dx =\frac{\partial}{\partial a}\int -Sin (a x) dx [/tex]

    Contour integration is a method of evaluating bucketloads of indefinite integrals that cannot be done using the fundamental theorem of calculus, such as:

    [tex] \int_{-\infty} ^\infty \frac{Sin (x)}{x} dx = \pi [/tex]

    In fact, any time you see a hard definite integral that has an answer containing pi, odds are that it can be evaluated elegantly using contour integration (which requires complex variables).

    The 'number's excess' is the difference between the number he recognizes and the number he is given. Similarly, the cube roots excess is the difference between the cube root he recognizes and the one he must compute. In this case Feynman most likely used the binomial series ( a special case of taylor series, which you may have studied).
  7. Jun 30, 2005 #6


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    That is a bit optimistic. Contour integration in the complex plane can be a helpfull tool, but it is not always simple or elegant.
  8. Jul 1, 2005 #7
    For the most part agree with this, but I qualified my statement with "odds are..." and it was meant to be funny (because it is slightly true) more then anything.
  9. Jul 1, 2005 #8
    Optimistic or not, contour integration on the complex plane can be very beautiful mathematically.
  10. Jul 2, 2005 #9
    how would you solve this one using the Feynman integration method? I tried using Euler's formula to convert it into an integral involving


    then i got stuck. i noted that [tex]\int_{-\infty}^{+\infty} \frac {e^{ix}}{x} dx = \int_{-\infty}^{+\infty} \frac{\partial}{\partial i} \frac {e^{ix}}{x^2} dx[/tex]

    but don't know where to go from there...am i on the right track?
  11. Jul 2, 2005 #10


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    If you mean using differentiation with respect to a parameter (ie under the integral sign it can be done line this.
    [tex] \int_{-\infty} ^\infty \frac{Sin (x)}{x} dx =2\int_{0} ^\infty \frac{Sin (x)}{x} dx [/tex]
    so we want 2f(0) where
    [tex] f(s)=\int_{0} ^\infty \frac{Sin (x)}{x}{exp(-s x)} dx [/tex]
    [tex] f'(s)=-\int_{0} ^\infty {Sin (x)}{exp(-s x)} dx [/tex]
    [tex] f'(s)=-\frac{1}{1+s^2}[/tex]
    [tex] f(s)=C-Arctan(s)[/tex]
    for some constant C but f(infinity)=0
    [tex]\int_{-\infty} ^\infty \frac{Sin (x)}{x} dx=2f(0)={\pi} [/tex]
    This can also be done with contour integration in the complex plane. It is in fact a famous example. Take f=exp(i z)/z and the contour be lines connecting (-H,0) to (-h,0) and (H,0) to (h,0) and upper semicircles centered at (0,0) conecting (-H,0) to (H,0) and (-h,0) to (h,0) where 0<h<H and take limits h->0 and H-> infinity. The upper arc is 0 the lines add to the integral of 2i sin(x)/x on (0,infinity) and the small arc is pi i since that sum of all these is 0 we know the integral. Also you can't differentiate with respect to i in this context as it is constant it would be like differentiating with respect to 11.
    Last edited: Jul 2, 2005
  12. Jul 3, 2005 #11
    very cool, thanks lurflurf!
  13. Jul 5, 2005 #12
    Does anyone knows some webpage with a proof of this Leibniz rule?
  14. Jul 5, 2005 #13


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    It's rather straightforward to work out: it doesn't really require knowing much more than the definition of a derivative and the basic properties of integrals. You have help, since you know what the rule looks like. :smile: You should give it a try.

    (Note that HallsOfIvy's statement assumed that f was continuous. a and b, of course, must be differentiable)
    Last edited: Jul 5, 2005
  15. Jul 5, 2005 #14
    [tex]\frac{d}{dx} f(x)g(x) = \lim_{\Delta x \rightarrow 0} \frac{f(x + \Delta x)g(x + \Delta x) - f(x)g(x)}{\Delta x}[/tex]

    take the expression in the numerator and "add and subtract" [itex]g(x)f(x + \Delta x)[/itex]

    [tex] = \lim_{\Delta x \rightarrow 0} \frac{g(x)f(x + \Delta x) - g(x)f(x) + f(x + \Delta x)g(x + \Delta x) - g(x)f(x + \Delta x)}{\Delta x} = \lim_{\Delta x \rightarrow 0} \frac{g(x)(f(x + \Delta x) - f(x)) + f(x + \Delta x)(g(x + \Delta x) - g(x))}{\Delta x}[/tex]

    since f and g are continuous at x, you can take the limits of each term and then you will have the familiar product rule.
  16. Jul 6, 2005 #15
    OK, Quetalcoatl, but I was asking about a proof of the differentiation under the integral sign...
  17. Jul 7, 2005 #16
    Feynman's calculus: Question to lurflurf

    how did you get F'(t) = -2F(t) in the first place?
    i understood the part after that.
    Last edited: Jul 7, 2005
  18. Jul 7, 2005 #17


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    find F'(t) by differentiating under the integral sign, then simplify the integral with the substitution u=t/x. Then you can see F'(t)=-2F'(t)
    [tex]F'(t)=\frac{d}{dt} \int_0^\infty\exp(-x^2-t^2/x^2) dx[/tex]
    differentiate under the integral sign
    [tex]F'(t)= \int_0^\infty\frac{d}{dt}\exp(-x^2-t^2/x^2) dx[/tex]
    perform the differentiation
    [tex]F'(t)= \int_0^\infty\frac{-2t}{x^2}\exp(-x^2-t^2/x^2) dx[/tex]
    let u=t/x
    [tex]F'(t)= -2\int_0^\infty\exp(-t^2/u^2-u^2) du[/tex]
    now see that
    [tex]F'(t)= -2F(t)[/tex]
    Last edited: Jul 7, 2005
  19. Jul 7, 2005 #18
    not to nitpick, but shouldn't it be [tex]\frac{d}{dt}[/tex] ?
  20. Jul 7, 2005 #19


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    Yes it was a typo.
  21. Jul 8, 2005 #20


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    You know what, I just don't see that. Can someone help me? When I take the derivative I get:

    [tex]\frac{d}{dt}\int_0^{\infty}\exp(-x^2-t^2/x^2) dx=-2t\int_0^{\infty}\frac{1}{x^2}e^{-(x^2+\frac{t^2}{x^2})}dx[/tex]

    and thus I don't see how the derivative is -2F(t).
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