Exploring Feynman's Calculus: Differentiating & Contour Integration

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In summary, Feynman wrote in his book ‘Surely You’re Joking, Mr. Feynman!’ that he used the ‘differentiating under the integral sign’ and the ‘contour integration’ to solve problems in calculus. He also wrote about computing the cube root of a number and about contour integration.
  • #1
murshid_islam
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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.
 
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  • #2
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...
 
  • #3
murshid_islam said:
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.
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
f(1729.03)=12*f(1729.03/1728)=12*f(1+1.03/1728)
use 1.03~1
f(1729.03)~12*f(1+1/1728)=12(1+(f(1+1/1728)-f(1))
since 1+1/1728~1
12(1+(f(1+1/1728)-f(1))]~12(1+1/3/1728)=12+4/172812=12+1/432~12.0023
f(1729.03)~12.00238378569171812
So one can get pretty close easily this way.
 
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  • #4
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)
 
  • #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).

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

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).
 
  • #6
Crosson said:
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).
That is a bit optimistic. Contour integration in the complex plane can be a helpfull tool, but it is not always simple or elegant.
 
  • #7
That is a bit optimistic. Contour integration in the complex plane can be a helpfull tool, but it is not always simple or elegant.

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.
 
  • #8
Optimistic or not, contour integration on the complex plane can be very beautiful mathematically.
 
  • #9
Crosson said:
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]

how would you solve this one using the Feynman integration method? I tried using Euler's formula to convert it into an integral involving

[tex]\frac{e^{ix}}{x}[/tex]

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?
 
  • #10
quetzalcoatl9 said:
how would you solve this one using the Feynman integration method? I tried using Euler's formula to convert it into an integral involving

[tex]\frac{e^{ix}}{x}[/tex]

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?
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
f(s)=pi/2-Arctan(s)
[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.
 
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  • #11
very cool, thanks lurflurf!
 
  • #12
Does anyone knows some webpage with a proof of this Leibniz rule?
 
  • #13
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)
 
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  • #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.
 
  • #15
OK, Quetalcoatl, but I was asking about a proof of the differentiation under the integral sign...
 
  • #16
Feynman's calculus: Question to lurflurf

lurflurf said:
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

how did you get F'(t) = -2F(t) in the first place?
i understood the part after that.
 
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  • #17
murshid_islam said:
how did you get F'(t) = -2F(t) in the first place?
i understood the part after that.
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]
 
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  • #18
not to nitpick, but shouldn't it be [tex]\frac{d}{dt}[/tex] ?
 
  • #19
quetzalcoatl9 said:
not to nitpick, but shouldn't it be [tex]\frac{d}{dt}[/tex] ?
Yes it was a typo.
[tex]\frac{d}{dt}F(t)=F'(t)=-2F(t)[/tex]
while
[tex]\frac{d}{dx}F(t)=0[/tex]
 
  • #20
lurflurf said:
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

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).
 
  • #21
saltydog said:
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).
That step (F'(t)=-2F(t)) was a little sudden so upon request I added a few intermidiate steps a few posts up. You can see the equality more easily if a substitution like u=t/x is made.
 
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  • #22
lurflurf said:
First thing you have a sign error in your derivative +t^2/x^2 should have a minus sign.
That step (F'(t)=-2F(t)) was a little sudden so upon request I added a few intermidiate steps a few posts up. You can see that equality more easily if a substitution like u=t/x is made.

Very good Lurflurf. I see that now. Also, I took the minus sign out of parenthesis so I think I was Ok with that. :smile:
 
  • #23
For the change of variable I am using these functions:
u = u(x) = t/x (t fixed)
<´¨¨´¨¨´¨k¨´¨´´´´j´´¨¨¨¨z¨´s¨s¨s´s´s´´´<<sp(u) = exp(-t¨Z¨Z´´ä¨zñS´¨pa´zo¨¨¨Qq´qqá´´AááÁÁAÍ´SD´S
and I obtained the result, but with 2 instead of -2.
 
  • #24
Sorry for that post.

Castilla.
 
  • #25
So if one integrates from -infinity to infinity, can you always change the range (or whatever it's called) of integration to 0 to infinity, and multiply the remaining integral by 2? Or is that a property of the sin that makes it symmetrical to where the negative side doesn't affect it? In other words, why doesn't a -2 come out?

lurflurf said:
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]

How does this exp(i z) work? Is it like i to the z power?

I may have seen some exp functions with three variables like exp (x y z) or so. Does that mean anything?

lurflurf said:
Take f=exp(i z)/z
 
  • #26
nanoWatt said:
So if one integrates from -infinity to infinity, can you always change the range (or whatever it's called) of integration to 0 to infinity, and multiply the remaining integral by 2? Or is that a property of the sin that makes it symmetrical to where the negative side doesn't affect it? In other words, why doesn't a -2 come out?

The property you described depends on whether a function is even or odd. An even function is a function such that f(-x) = f(x), while an odd function is one such that g(-x) = -g(x).

So, for example, cosine is an even function, since cos(-x) = cos(x), while sine is an odd function, since sin(-x) = -sin(x). The exponential function is neither odd nor even, as e^(-x) does not equal either e^(x) (for any arbitrary value of x) or -e^(x).

Because an even function looks the same in the x > 0 half plane as it does in the x < 0 half plane, if you have symmetric limits about x = 0, then integrating from -L to L of an even function is just like integrating from 0 to L twice. This is only true for even functions. For odd functions, the contribution from the negative half plane will cancel out that from the positive half plane, so the result will be zero.

To summarize:

[tex]\int_{-L}^{L} dx f(x) = 2 \int_{0}^{L} dx f(x)~\mbox{if f(x) is even}[/tex]
[tex]\int_{-L}^{L} dx f(x) = 0~\mbox{if f(x) is odd}[/tex]



How does this exp(i z) work? Is it like i to the z power?

I may have seen some exp functions with three variables like exp (x y z) or so. Does that mean anything?


Have you ever heard of the imaginary unit i? It is the number defined such that [itex]i^2 = -1[/itex]. With this number one can define the complex exponential function, which has the property that

[tex]e^{ix} = \cos x + i \sin x[/tex]

Since exponentials typically have nicer properties than sines or cosines, by considering the integral of exp{ix}/x you might be able to get the integral for sin(x)/x in an easier fashion (and, you'll probably also get the integral of cos(x)/x out of it too, if that happens to be finite, which I don't think it is).
 
  • #27
Thanks for the breakdown.

So how would you expand exp (i z)?
Is that the same as [tex] e^{iz} [/tex] ?
 
  • #28
nanoWatt said:
Thanks for the breakdown.

So how would you expand exp (i z)?
Is that the same as [tex] e^{iz} [/tex] ?

exp{z} is just another notation for e^z.

If z is a complex number z = x + iy (where x and y are real), then

e^{z} = e^{x}e^{iy} = e^{x}(cos(y) + i sin(y))

And e^{i z} = e^{-y + ix} = e^{-y}(cos(x) + i sin(x))
 
  • #29
HallsofIvy said:
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].
...

It seems to me that the signs on the last two terms of the right-hand side are reversed; should the formula be
[tex]\frac{d}{dx}\int_{a(x)}^{b(x)} f(x,t)dt= \left(\int_{a(x)}^{b(x)}\frac{\partial f(x,t)}{\partial x}dt\right)+ \frac{db(x)}{dx}f(x,b(x))- \frac{da(x)}{dx}f(x,a(x))[/tex]
?

I guess so, but I haven't figured out how to prove this formula yet...
 
  • #30
swensonj said:
I guess so, but I haven't figured out how to prove this formula yet...
It's actually pretty straightforward; e.g. you might simply apply the limit formula for the derivative, and then splitting and rearranging the resulting expression into pieces that can be approximated well.

It's one of those things that, if you understand the ideas behind using limits and approximations, will be very straightforward. And if you don't find it straightforward, then it's really worth studying as an exercise.
 
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1. What is Feynman's Calculus?

Feynman's Calculus, also known as the path integral formulation, is a mathematical framework used to describe the behavior of quantum systems. It was developed by physicist Richard Feynman in the 1940s and is based on the idea that a particle can take all possible paths simultaneously, with each path having a certain probability amplitude.

2. How is Differentiation used in Feynman's Calculus?

Differentiation is used in Feynman's Calculus to calculate the probability amplitude for a particle to travel from one point to another. The derivative of the action (a mathematical quantity that describes the behavior of a system) with respect to the position of the particle is used to determine the probability amplitude.

3. What is Contour Integration in Feynman's Calculus?

Contour integration, also known as complex integration, is a mathematical technique used to evaluate integrals over complex numbers. In Feynman's Calculus, contour integration is used to calculate the probability amplitude for a particle to travel from one point to another in a certain amount of time.

4. How is Feynman's Calculus used in quantum mechanics?

Feynman's Calculus is a fundamental tool in quantum mechanics. It is used to calculate the probability of outcomes for quantum systems, such as the position or momentum of a particle. It is also used to calculate transition amplitudes, which describe the probability of a system transitioning from one state to another.

5. What are some real-world applications of Feynman's Calculus?

Feynman's Calculus has many real-world applications, including in quantum computing, quantum field theory, and particle physics. It is also used in various fields such as chemistry, biology, and finance to model complex systems and calculate probabilities. Additionally, Feynman's Calculus has been used in the development of technologies such as lasers and transistors.

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