Feynman's Integral Trick with Math With Bad Drawings

hello maths fans i am incredibly excited to bring you my latest video with ben allen from maths with bad drawings ben has written two books they are absolutely brilliant and you should definitely go and read them as soon as you finish watching this video they're all about the maths of everything and i mean everything but today we're of course not talking about everything i'm here to talk about richard feynman and in particular his favorite maths trick differentiating under the integral sign if you've done some calculus you'll know that derivatives aren't too bad there's a few general rules sure and you have to crunch through some calculation but usually you'll get to where you need to be ben compares derivatives to an office building where everything is predictable and easy to find if a little boring integrals on the other hand are more like a fun house with each room throwing up its own surprise and requiring its own map or technique in order to get through let's take the example of 1 over 1 plus x squared now this you may recognize is a known integral the answer is a known function and is equal to arctan x plus c now if i change the integral ever so slightly by just increasing the power of x on the denominator to x cubed so basically the same thing right as i said before integrals much less straightforward so what can we do about it borrowing from feynman let's try combining the two differentiation plus integration and to see how this works let's take an example from the apparently most difficult undergraduate math class in the us at least according to wikipedia harvard math 55. so the integral we're going to look at is from naught to infinity x to the power n e to the minus x integrated with respect to x one approach might be to try repeated integration by parts we have a polynomial power of x here which we can reduce by continued differentiation and the exponential function will integrate to basically give itself so if we do this enough times we will hopefully get down to a zero power on the x and then some kind of exponential function that hopefully we can integrate the first step in any integration by parts is to figure out what am i differentiating what is my u and what am i integrating what represents dv so here x to the n will be u and dv will be given by e to the minus x now we want to keep differentiating the power of x so d u equals n x to the n minus 1 and we want to integrate the exponential so v is going to be minus e to the minus x now using my integration by parts formula this is equal to u dv is equal to the integrated part u v so that's minus e to the minus x x to the power n evaluated at zero and infinity and then it's minus the integral of d u times v so minus minus gives me a plus and then i've got the integral from naught to infinity of n x to the n minus 1 e to the minus x so after one step of integration by parts it looks as follows now if we evaluate these limits if i plug in infinity the exponential is e to the minus infinity so that goes to zero that kills the polynomial exponents beat polynomials so that's zero and then if i plug in the limit of zero the exponential is one the polynomial is zero zero times one is zero so this whole term will actually vanish and this will happen at every step in fact so what we've done after one step of integration by parts is bring down an n x to a reduced power and we have the same exponential function so this is good this is what we were hoping was going to happen so now we want to repeat the same process on our new integral so if i take n x to the n minus 1 that will be my u u is n x to the n minus 1. my dv is unchanged so dv is e to the minus x which means v is unchanged so minus e to the minus x and now d u differentiating i'll get n minus 1 times n x to the n minus 2. now it's basically the same as what we had before so when i substitute all of these in and do integration by parts the boundary terms the u and v will again vanish because u has got this power of x so that will go to zero when x is zero and v is the exponential which will go to zero as x goes to infinity so we again lose boundary terms and now i'm left with the integral from naught to infinity of n n minus 1 x to the n minus 2 e to the minus x dx so after two steps of integration by parts we have reduced the power of x down by two and what we've got in fact are these two powers in front of the otherwise basically the same form of the integral now we could carry on and repeat the exact same process do another set of integration by parts and continue on basically until this goes all the way to zero or we can try and spot the pattern so what's happening here is that each time we differentiate the x to reduce its power in each step of integration by parts the boundary terms will always vanish because you'll always have a power of x positive power of x which goes to zero at x equals naught and the exponential to the minus x which goes to zero as x goes to infinity so the boundary terms will always vanish at each step and so in fact what we're going to get here is after n steps so after n times integration by parts and n of course here could be very very large it could be a thousand it could be 4 billion 796 million 385 212 who knows after n lots of integration by parts we're just going to have all of these terms in front of x and here we've done it twice and it was x to the n minus 2. so if i do it n times it becomes x to the n minus n which is x to the naught which is 1. and that's what we wanted here so we're going to end up with n times n minus 1 times m minus 2 times m minus 3 dot dot dot dot times 2 times 1 also known as n factorial so we have n factorial then we've got x to the n minus n x to the naught one and then we have the integral from naught to infinity and we've seen at each step that v is always uh minus e to the minus x and dv is e to the minus x so dv here e to the minus x e to the minus x it will be the same here so that term is unchanged e to the minus x dx so we're almost done we've got all the way down to something which i think we can now hopefully integrate so the n factorial just stays out the front if i integrate this we did this before minus e to the minus x between 0 to infinity plugging in at infinity e to the minus infinity is zero and then we've got minus minus one because e to the naught is one the two minuses cancel so i just get one so the final answer is n factorial that that was uh that was a lot of work so we got there sure this is the right answer n factorial but that was i think you'll agree one hell of a slog so this is where the magic of differentiation under the integral sign comes in so over to you mr feynman our starting point is actually the last step of my very involved integration by parts calculation that i just went through so we begin with the integral from naught to infinity of e to the minus x dx is equal to one and we're going to make a very slight modification to this expression so if i take my purple chalk and what we're going to do is actually add a here so this is now e to the minus a x so that when we integrate this we now have to divide by a and so the answer now will be 1 over a so this is still a true statement and we could in fact have started here but i just thought it was quite nice to begin with the the last point of our previous calculation now this is where the magic happens this is where we're going to differentiate under the integral sine and we're going to in fact differentiate both sides of our expression with respect to a so if i differentiate both sides with respect to a what's going to happen here is i get a minus and i still have the integral from naught to infinity i have differentiated with respect to a so i've brought down an x from here then i've got e to the minus a x dx so that's the left-hand side differentiate with respect to a and then i do the same on the right 1 over a or a to the minus 1 so it's minus a to the minus 2 so it's minus 1 over a squared now so far we've differentiated with respect to a once and we've ended up with x to the power 1. seeing as we want x to the power n we in fact want to differentiate both sides of the expression n times with respect to a so we're going to differentiate under our integral sign a total of n times now we've differentiated once and we got a minus and an x so if we differentiate n times we will get minus one n times so to the power n we then have the integral from zero to infinity every time we differentiate with respect to a we get a power of x so this will be x to the n e to the minus a x dx now for the right hand side same idea with the minuses the sign changes every time we differentiate the power of a so we still get a minus 1 to the n the power of a itself increases every time so that's 1 over a to the n plus one because we started with a to the one on the denominator and every time we differentiate we bring down the power so if i differentiate here i get a two then i'd have an a to the minus three then i'd get a three next time so you in fact get an m factorial term on the top from all of these continued differentiations now the final step is just to simplify this expression so first of all we can cancel the minus 1 to the n term from both sides as it's the same and now what we want to do very clever is to actually now set a equal to one so if we set a equals one as our final step this becomes a one and over here we've got one over one to the power n plus 1 so that's just 1. so just to neaten it all up we're just left with n factorial so setting a equal to 1 turns this into just e to the minus x this is now our original integral and the right-hand side all those powers of a disappear because a is one and we get n factorial this result while it's looking pretty neat is also really powerful because what we have here is an integral definition of the factorial function with the old definition we said that n factorial was n times n minus 1 times n minus 2 all the way down to 1. and so that would only work for positive whole numbers because otherwise we didn't know when to stop but what we've shown here is that the new definition using this integral formula is actually equivalent to the old definition and therefore this can be used for any value of n because plugging in any value of n here makes complete sense in the integral context this same trick of differentiating under the integral sign can be used in countless other situations and indeed was readily employed by feynman a famous example comes from his time at the los alamos labs during world war ii he struggled to find his feet in several departments when he first arrived until one day a researcher presented him with a problem that had stumped the researchers for three months three entire months working on the same integral can you imagine trying to do a homework question for that length of time when feynman saw the integral he suggested trying differentiation under the integral sign and supposedly the problem was solved in under 30 minutes thank you everyone for watching a huge thank you to ben from maths with bad drawings for all of his brilliant cartoons and also for contributing the story which comes from his latest book about calculus called change is the only constant remember to check out all of ben's books and also remember to subscribe to my channel if you've enjoyed this video and i'll see you soon you

Channel: Tom Rocks Maths

Views: 88,247

Keywords: math with bad drawings, feynman, richard feynman, feynman integral technique, feynman integral trick, feynman integration, MWBD, differentiation under the integral sign, integral differentiation, integral derivative, leibniz rule, leibniz rule differentiation, los alamos lab, ben orlin, tom crawford, oxford maths, oxford mathematician, tom rocks maths, tomrocksmaths, tom crawford maths, oxford calculus, math cartoon, tom rocks oxford, differentiation inside integral, integral

Id: 4RIHTHYD2SQ

Length: 15min 35sec (935 seconds)

Published: Wed Oct 28 2020

Reddit Comments

[u/mr_ryh]

This is discussed in Surely You're Joking, Mr Feynman! ("A Different Box of Tools", pp86-7 in the Norton paperback):

So every physics class, I paid no attention to what was going on with Pascal's Law, or whatever they were doing. I was up in the back with this book: Advanced Calculus, by Woods. Bader knew I had studied Calculus for the Practical Man a little bit, so he gave me the real works - it was for a junior or senior in college. It had Fourier series, Bessel functions, determinants, elliptic functions -- all kinds of wonderful stuff I didn't know anything about.

That book also showed me 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.

The result was, when guys at MIT or Princeton had trouble doing a certain integral, it was because they couldn't do it with the standard methods they had learned in school. If it was contour integration, they would have found it; if it was a simple series expansion, they would have found it. Then I come along and try differentiating under the integral sign, and often it worked. So I got a great reputation for doing integrals, only because my box of tools was different from everybody else's, and they had tried all of their tools on it before giving it to me.

Apparently this has piqued some interest over the years and googling the phrase "feynman differentiating under integral" gives some nice links: https://www3.nd.edu/~math/restricted/CourseArchive/100Level/166/1662000S/Misc/DiffInt.pdf https://kconrad.math.uconn.edu/blurbs/analysis/diffunderint.pdf https://web.williams.edu/Mathematics/lg5/Feynman.pdf

[u/Kyru_G]

The usage is fine but the reasoning behind why this is an acceptable tactic is lacking. I do feel that if any students want to do this they proceed with caution and take the time to really learn where this process is acceptable.

[u/DaveBeleren02]

Bruh it's called the Leibniz rule... Physicists always appropriating of stuff

[u/[deleted]]

this is a new frontier for machine gun kelly

[u/aaa_azidoazideazide]

Subtleties but I think in the integration by parts: dv = e^-x. dx and not just e^-x. I have seen a lot of people write it this way. Edit : correction. Also didn’t know reddit interpreted superscripts by just using the up symbol. Neat.

[u/tomrocksmaths]

Richard Feynman famously used differentiation under the integral sign to solve many difficult problems, including one during his time at Los Alamos Laboratory during World War II that had stumped researchers for 3 months.

Learn how Feynman's Integral Technique works from Oxford Mathematician Dr Tom Crawford with the help of Ben Orlin from Math With Bad Drawings and his brilliant cartoons.

You can find out more about Ben at https://mathwithbaddrawings.com/

Produced by Dr Tom Crawford at the University of Oxford. Tom is an Early-Career Teaching and Outreach Fellow at St Edmund Hall: https://www.seh.ox.ac.uk/people/tom-crawford

[u/DAT1729]

When I learned this for the Putnam exam it was called Parameter Integration

[u/swni]

I would be interested to see an example where using this technique makes solving the problem easier rather than more complicated than standard techniques.

[u/[deleted]]

Gamma function no? Gamma(n+1)