The Dirichlet Integral is destroyed by Feynman's Trick

Поделиться
HTML-код
  • Опубликовано: 14 май 2024
  • Get MAPLE LEARN (completely for free) to start making interactive math documents that solve your equations here: ►www.maplesoft.com/products/le.... My thanks to Maple Learn for sponsoring today's video!
    The Dirichlet integral (integral from 0 to infinity of the sin(x)/x also know as the sinc function), is typically not taught in first year calculus courses. But the trick to solve it is actually pretty easy! In this video I show how we can use Feynman's Trick to make it a big messier by including a new exponential factor, but by differentiating under the integral sign the messy x in the denominator gets cleaned right up.
    I mentioned in the video that this was just the Laplace transform, check out my Laplace transform playlist here ►LAPLACE TRANSFORM: • Laplace Transforms and...
    Check out my MATH MERCH line in collaboration with Beautiful Equations
    ►beautifulequations.net/pages/...
    COURSE PLAYLISTS:
    ►DISCRETE MATH: • Discrete Math (Full Co...
    ►LINEAR ALGEBRA: • Linear Algebra (Full C...
    ►CALCULUS I: • Calculus I (Limits, De...
    ► CALCULUS II: • Calculus II (Integrati...
    ►MULTIVARIABLE CALCULUS (Calc III): • Calculus III: Multivar...
    ►VECTOR CALCULUS (Calc IV) • Calculus IV: Vector Ca...
    ►DIFFERENTIAL EQUATIONS: • Ordinary Differential ...
    ►LAPLACE TRANSFORM: • Laplace Transforms and...
    ►GAME THEORY: • Game Theory
    OTHER PLAYLISTS:
    ► Learning Math Series
    • 5 Tips To Make Math Pr...
    ►Cool Math Series:
    • Cool Math Series
    BECOME A MEMBER:
    ►Join: / @drtrefor
    MATH BOOKS I LOVE (affilliate link):
    ► www.amazon.com/shop/treforbazett
    SOCIALS:
    ►Twitter (math based): / treforbazett
    ►Instagram (photography based): / treforphotography

Комментарии • 192

  • @LaughingManRa
    @LaughingManRa 11 месяцев назад +535

    Feynman DESTROYS Dirichlet Integral with FACTS and LOGIC

    • @cristianv2850
      @cristianv2850 11 месяцев назад

      😢

    • @Juan_Carl0s
      @Juan_Carl0s 11 месяцев назад +17

      *Leibniz

    • @cyrillechevallier7835
      @cyrillechevallier7835 11 месяцев назад +16

      Yeah but it relies on the fact that deriving an integral (with respect to a parameter) is the same as integrating the partial derivative :) which is a god level tool (and it’s hard to prove)

    • @orenawaerenyeager
      @orenawaerenyeager 11 месяцев назад

      @@Juan_Carl0s plug the upper limit multiplied by differential of upper limit minus plug the lower limit multiplied by differential if lower limit

    • @kylethompson1379
      @kylethompson1379 11 месяцев назад +6

      @@Juan_Carl0s Thank you. This "Feynman technique" thing is obnoxious. Feynman was great enough on his own merits, he doesn't need any help.

  • @kdmdlo
    @kdmdlo 10 месяцев назад +55

    Trefor: this trick requires that the order of differentiation and integration can be interchanged because your F'(s) is actually d/ds [ int_0^infty e^(-sx) sin(x)/x dx ] . You nonchalantly swapped the order of these two operations. This is thoroughly valid provided the integrand satisfies certain integrability conditions etc. Perhaps it would be worth noting this (and going over these conditions), so if people are looking to use Feynman's trick, they will be on the look out for potential tripping points.

  • @Sugarman96
    @Sugarman96 11 месяцев назад +76

    Since you briefly mentioned the Laplace transform, I feel like it'd be a waste not to mention the super important Fourier transform in this context, because the Fourier transform lets you solve the Dirichlet integral almost immediately. It turns out, the Fourier transform of a window function of from -1 to 1 is sin(w)/w, so using the inverse fourier transform, you get the value of the Dirichlet integral.

    • @donaldmcronald2331
      @donaldmcronald2331 11 месяцев назад +3

      I learnt both the fourier transform and reverse transform and immediately knew sine(x)/x. I wasn't aware of the link between them lol.
      I'd add that if you view sine(x)/x as a spectrum, its energy across actual time is indeed related to the integral of sine(x)/x from 0 to infinity.

    • @MrPoornakumar
      @MrPoornakumar 11 месяцев назад +3

      @@donaldmcronald2331
      Your second sentence restates "Parseval" equation. It physically means, no matter in what domain (frequency or time) you integrate, the energy is same.

  • @Djenzh
    @Djenzh 11 месяцев назад +88

    Beautiful! Funny that the Laplace transform shows up. I only knew how to do this integral by changing sin(x)/x into sin(xt)/x and then taking the Laplace transform of the entire thing, but your solution seems much easier :)

    • @DrTrefor
      @DrTrefor  11 месяцев назад +24

      I suppose this method is mostly equivalent. I'd actually suggest your is more true to the spirit of Feynman's trick and a bit more generalizable, but perhaps the way I showed slightly more efficient for 1st year calc students

    • @leif1075
      @leif1075 9 месяцев назад +2

      @@DrTrefor But why and how did Feynman or whoever else come up with this? I could never admit I couldn't or Im not a math whiz who would come up with this or something similar--why not just have e^x or e^sx where s is just a constant and make it a function of x still--was this tried--Thanks for sharing and hope to hear from you.

  • @NumbToons
    @NumbToons 11 месяцев назад +4

    I just (today) learnt this integral in Fourier Transform, and here you come up with a video to make it permanent my memory.

  • @ketankyadar5228
    @ketankyadar5228 11 месяцев назад +3

    I am really happy to see Differentiation Under Integral sign rule here to calculate integration of Sinx/x.
    Actually today in the class i taught this rule and after that i saw your video.
    I amazed that how you start with combining exponential term in the integral, In real life there are so many situations where you can use this cause there exist always parameter with your function.

  • @emanuellandeholm5657
    @emanuellandeholm5657 11 месяцев назад +12

    I always love to watch different people's take on the Dirichlet integral. It's second only to the Gaussian integral for me. :)
    The interesting thing about the Dirichlet integral is that it's not Lebesgue-integrable. Put some of that stuff in your pipe and smoke it!
    Dr Bazett's take is basically a Laplace transform, and I think it's cute!

  • @speedbird7587
    @speedbird7587 9 месяцев назад +1

    Hello Professor,
    Thanks for all your brilliant videos,
    It was a really nice and technical( a bit similar to laplace technique)
    although I couldn't find a proof or nonexample for this technique from the internet,
    I am very curious to know that could we use this technique by any function other than exponential terms, or is it because of the uniform continuity of the laplace transform that we can use this trick?
    since the problem can also be thought as a differential equation/ a dynamical system biforcating on the parameter s .
    I really would like to know more about it.
    Thank you.
    Br,

  • @moritzberner8402
    @moritzberner8402 11 месяцев назад +5

    I directly tried to solve the integral of cos(x)/x with the same method and found out that this one diverges. Great video!

    • @smiley_1000
      @smiley_1000 11 месяцев назад +8

      The reason is the behavior as x tends to 0, where the function behaves approximately like 1/x. Perhaps you should consider the integral from 1 to infinity in order to get a more interesting result.

  • @purplenanite
    @purplenanite 11 месяцев назад +7

    That's a neat trick!
    i came across this integral yesterday, interestingly enough - although now I know how to do it faster!

    • @DrTrefor
      @DrTrefor  11 месяцев назад +4

      Isn't it cool!?

    • @leif1075
      @leif1075 9 месяцев назад

      @@DrTrefor I don't see why you take the limit at 5:52 instead of say plugging in the value of zero for s to get C? Isn't that more logical and intuitive?

    • @leif1075
      @leif1075 9 месяцев назад

      @@DrTrefor and I don't see why anyone would take the limit as S goes to infinity--isn't there some other way to get C--if you set s to zero you get F(0)=C

  • @johnchessant3012
    @johnchessant3012 11 месяцев назад +17

    Fun fact: the integral of sin(x)^2/x^2 from 0 to infinity is also pi/2

    • @Ninja20704
      @Ninja20704 11 месяцев назад +8

      But go to the third power and then it breaks down. That integral will be 3*pi/8

    • @NumbToons
      @NumbToons 11 месяцев назад +3

      wow

    • @OmegaQuark
      @OmegaQuark 11 месяцев назад

      ​@@Ninja20704 Is there a generalized closed formula for the n-th power of sin(x)/x? Like with the Dirichlet series, or the Zeta function for the even positive integers

    • @violintegral
      @violintegral 11 месяцев назад

      @@OmegaQuarkyes, but it's a bit complicated

    • @Ninja20704
      @Ninja20704 11 месяцев назад

      @@OmegaQuark for the zeta function at even positive integers, its been proven that zeta(2n) will always be some rational multiple of pi^2n. Figuring out that rational multiple is pretty complicated but doable.

  • @pygmalionsrobot1896
    @pygmalionsrobot1896 11 месяцев назад

    Brilliant !! Thanks for making this video :D !!!

  • @ShanBojack
    @ShanBojack 11 месяцев назад +3

    You remind me of my tuition teacher who is also a big mathemagician and you both are my ideals 🙌

    • @DrTrefor
      @DrTrefor  11 месяцев назад +3

      So kind!

  • @dfcastro
    @dfcastro 9 месяцев назад

    What if instead of during the integration by parts you apply again the Feymann technique and than (probably) will get a differential equation and solve it?
    Would it work?

  • @iamreallybadatphysicsbutda8198
    @iamreallybadatphysicsbutda8198 11 месяцев назад +1

    Thank you so much Dr. Trefor

    • @DrTrefor
      @DrTrefor  11 месяцев назад +1

      You're most welcome!

  • @Helibenone
    @Helibenone 11 месяцев назад

    I love your vids could you do some on statistics and probability

  • @scollyer.tuition
    @scollyer.tuition 11 месяцев назад +4

    The terminology related to this method is itself somewhat interesting - when I first came across it (about 40 years ago), I don't think it was given any specific name (except differentiating under the integral sign), then the name "Leibniz rule" seemed to become more popular, and in the last 5 years or so, the "Feynman method" began to reign supreme - a tribute to continuing popularity of Mr Feynman, I guess.
    And it's worth pointing out that there are conditions required for the method to work: continuity of f(x,s) in both x and s and (partial) df/ds over the region of integration, IIRC. (corrections gratefully accepted if I misremember).
    Also, there's a generalisation of the method that takes account of variable limits depending on s.

    • @frenchimp
      @frenchimp 11 месяцев назад +1

      I guess Feynman's contribution to the Leibniz method was "to hell with boring mathematical justifications".

    • @radekvecerka1115
      @radekvecerka1115 5 месяцев назад +1

      there are 4 conditions
      1)continuity (or more generaly the function has to messurable)
      2)continuos partial derivatives
      3)very important is you have to find a mayorant to derivatice of the intagrated function with respect to the parameter, which must have finite value when integrated
      4)you have to find a least 1 parameter for which you can calculate the integral
      Overall the most important part about this ''trick'' is figuring out whether you can even use it!

  • @giovanni1946
    @giovanni1946 11 месяцев назад +6

    You can't plug in s = 0 as the Feynman trick can only be applied with s > 0, due to the absolute convergence requirement, though the limit as s goes to 0 indeed equals pi/2

    • @miloweising9781
      @miloweising9781 11 месяцев назад

      Yeah you need to be careful when limiting to zero here. Probably there’s a nice dominated convergence argument to say that F is continuous at 0 from the right.

    • @xaxuser5033
      @xaxuser5033 10 месяцев назад

      @@miloweising9781 it is actually continus at x=0 and here is a complete proof of this fact: ( let f(s,x) be the function inside the integral )
      -for all x>0 the function s --> f(s,x) is continus.
      -for all s in R : x-->f(s,x) is continus.
      - there exist a continus integrable and positive function g : R+--> R+ such that for all x>0 and for all s in R we have :
      |f(s,x)|

    • @giovanni1946
      @giovanni1946 10 месяцев назад

      @@xaxuser5033 That's not correct, as e^(-sx) -> 1 when s -> 0, which is not integrable on R+

    • @xaxuser5033
      @xaxuser5033 10 месяцев назад

      @@giovanni1946 i didn't understand what u want to say , where did i write e^(-sx) ?

    • @giovanni1946
      @giovanni1946 9 месяцев назад

      @@xaxuser5033 This theorem cannot be applied here, you can't choose g(x) to be e^(-x) as e^(-sx) gets bigger when s -> 0

  • @General12th
    @General12th 11 месяцев назад +3

    Hi Dr. Bazett!
    Feynman's technique is one of my favorite in all of integral math.

    • @frenchimp
      @frenchimp 11 месяцев назад +2

      So long as you are aware it owes nothing to Feynman.

    • @General12th
      @General12th 11 месяцев назад

      @@frenchimp No. I'm not aware of that. Why don't you explain it to me?

    • @kylethompson1379
      @kylethompson1379 11 месяцев назад +1

      @@General12th Feynman is on video on youtube saying how he read it in a book, and anyway it's all just based on work done by Leibniz 100s of years ago. Feynman did re-popularise it though.

  • @fatemekashkouie3662
    @fatemekashkouie3662 11 месяцев назад +2

    Hello Dr.Trefor Bazzet,
    I wanted to take a moment to thank you for all the beautiful content you're creating. They're awesome.
    By the way, I'm supposed to do one of my class projects using maple. But I haven't gotten used to it. Do you suggest any particular tutorial teaching how to use maple?

    • @DrTrefor
      @DrTrefor  11 месяцев назад +3

      Thank you! Sorry I don’t have any particularly great resources at my fingertips, mostly because it can do so much it really depends what you need to use it for!

    • @fatemekashkouie3662
      @fatemekashkouie3662 11 месяцев назад

      @@DrTrefor I am supposed to solve differential equations for a dynamical systems course.
      Anyways, I think if I google every step, it will be somehow manageable.

  •  3 месяца назад

    Muy buen video Genio!

  • @andrewharrison8436
    @andrewharrison8436 11 месяцев назад +1

    Some bits of maths are like p v np, really hard to first find the method (np), but manageable to verify that the solution works (p). Actually it isn't "some bits", it is a lot of the bits and it is cumulative - Newton's "If I have seen further than other men it is by standing on the shoulders of giants".

  • @flamitique7819
    @flamitique7819 11 месяцев назад +6

    Great video ! I just wanted to point out that even though the result is correct, the reasoning here wasn't completely true, or was at least incomplete : by this reasoning, which uses the dominated convergence theorem and it's equivalents to derivate under the integral, you can't directly prove this formula for all s greater or equal to zero, but the formula is only true for s stricly bigger than zero, meaning you can't directly plug in 0 at the end (this is because you can only dominate the first function you want to derive under the integral for all s>0). But the formula still holds for all s>0. So the correct way to prove it is to show that the limit as s goes to zero of F(s) is indeed the integral of sin(x)/x from 0 to infinity, and you can use the fact that the right side of the equation is continuous at zero to give the final result. However, showing that you can interchange the limit and the integral as s goes to zero is not that trivial, since you can't dominate the function properly. To do that, you first need to integrate by parts and only after that you can dominate the function properly and do an interchange of limits and integral that is valid, and get the final result.

    • @aua6330
      @aua6330 11 месяцев назад

      For the continuity at 0, you can also show that if you take the sequence of functions F_n(s) = integral from 0 to n of (...), the sequence converges uniformly on [0, infty(, and each F_n is continuous.

  • @robertcrompton2733
    @robertcrompton2733 11 месяцев назад +1

    Wow! Loved it!

  • @hvok99
    @hvok99 11 месяцев назад +1

    Wow that was so satisfying.

  • @techsolabacademy9980
    @techsolabacademy9980 11 месяцев назад +2

    Dr. You are amazing❤

    • @DrTrefor
      @DrTrefor  11 месяцев назад +1

      Thank you!

  • @David-dvr
    @David-dvr 11 месяцев назад +4

    Although this trick is named after Feynman, I believe he found it in an advanced calculus book his high school physics teacher gave him. It was apparently developed, at least partially, by Leibniz:
    “I had learned to do integrals by various methods shown in a book that my high school physics teacher Mr. Bader had given me. [It] 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. [If] guys at MIT or Princeton had trouble doing a certain integral, [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 their tools on it before giving the problem to me.” (Surely you’re Joking, Mr. Feynman!)
    www.cantorsparadise.com/richard-feynmans-integral-trick-e7afae85e25c

    • @DrTrefor
      @DrTrefor  11 месяцев назад +1

      Cool!

    • @andrewharrison8436
      @andrewharrison8436 11 месяцев назад

      Must reread my Feynman books - nice shout out by Feynman to his teacher.

  • @maroc4747
    @maroc4747 11 месяцев назад +2

    Great explanation. My follow up question would be, why does this work and when should one use this trick? I only knew the Double integral solution for this Problem.

    • @DrTrefor
      @DrTrefor  11 месяцев назад +1

      With a lot of these integration tricks, ultimately it works when it works. With some practice you can gain some intuition for when you can parameterize and integral and do this differentiate with respect to the parameter trick, but there isn't some general rule for when it always works.

  • @cameronspalding9792
    @cameronspalding9792 11 месяцев назад +1

    @ 3:31 Personally I would have used the formula for sine in terms of complex exponentials

  • @anietiethompson5375
    @anietiethompson5375 2 месяца назад

    Please which app are you using to plot those functions?

  • @jeremytimothy3646
    @jeremytimothy3646 11 месяцев назад +2

    This is just spiced up Laplace transforms 😂. Nice video though always happy to learn abit more math.

    • @anuragkr3026
      @anuragkr3026 3 месяца назад

      I dont anymore people to know about it ,
      Sometimes good things are not good to be shared ❤

  • @soufianehrilla5533
    @soufianehrilla5533 11 месяцев назад +1

    C'est juste très élégant

  • @moeberry8226
    @moeberry8226 10 месяцев назад +3

    It’s more common to think that when applying Feynmans trick that you would put the parameter s inside of the sin(x) function aka sin(sx). The only reason that you use this decreasing exponential function is that so it converges on the interval 0 to infinity because the original integral converges as well. If you use the first method then you will obtain cos(sx) after differentiation with respect to s. Which does not converge. You should explain it this way to the students on RUclips.

    • @Ayush-yj5qv
      @Ayush-yj5qv 2 месяца назад

      true when i did this method i end up getting sin of infinite which i dont even know

  • @imad5206
    @imad5206 3 месяца назад

    good prof. thankks

  • @madhavpr
    @madhavpr 20 дней назад

    Pretty cool trick. I'm okay with all the steps except perhaps the interchange of limit and the integral as s->infinity. Is there some theorem in real analysis that justifies this?

  • @utuberaj60
    @utuberaj60 11 месяцев назад +1

    Hi Dr Brazet. Wonderful video.
    But, I have a question here. Why is this method known as Feynman's technique? In fact the idea of Differentiating Under the Integral sign (DIUS) was due to Leibnitz, and also the idea of introducing a parameter 's' in the form e^-sx is the Laplace transform or can even be thought of as the Gamma function?
    Why do you then still call this the Feynman trick? Just wanted to know.

    • @itellyouforfree7238
      @itellyouforfree7238 11 месяцев назад

      Because at least in the physical community in the USA it has been popularized by Feynman

    • @anuragkr3026
      @anuragkr3026 3 месяца назад

      Sometimes good thing are better not to be shared.

  • @ProfeJulianMacias
    @ProfeJulianMacias 11 месяцев назад +1

    Excellent Math problem

  • @md.musaal-kazim1774
    @md.musaal-kazim1774 3 месяца назад

    we can use laplace transformation?

  • @urnoob5528
    @urnoob5528 11 месяцев назад +2

    damn this is basically a laplace transform trick :O
    wat a coincidence that im learning laplace transform right now

    • @DrTrefor
      @DrTrefor  11 месяцев назад +1

      Laplace transform is SO useful!

  • @xiangbocai-ns4sg
    @xiangbocai-ns4sg 8 месяцев назад +1

    What a fantastic method

    • @DrTrefor
      @DrTrefor  8 месяцев назад +1

      Glad you think so!

  • @phenixorbitall3917
    @phenixorbitall3917 11 месяцев назад +1

    Dr. for which class of function can one use the Feynman trick?
    It was satisfying to watch this video 👌

  • @CapAbcv
    @CapAbcv 3 месяца назад

    Can we apply cauchy integral test here??

  • @RobinTester
    @RobinTester 11 месяцев назад

    Complex analysis ON TOP

  • @maxp3141
    @maxp3141 11 месяцев назад +2

    Nice, I think I would had inserted sin(x) = (e^ix - e^-ix) instead of integrating by parts, but I’m not sure if that makes things easier or not..

    • @active285
      @active285 10 месяцев назад

      Yes similarly! Using the Feynman trick here reflects the relation to complex analysis: Choosing the function f(z) = exp(iz)/z and integrate over the (positive) indented semicircle and some appropriate contour avoiding 0.

  • @vpambs1pt
    @vpambs1pt 11 месяцев назад +1

    fuun, so the are under sin(x)/x over R is pi.
    This function is just, the typical sinusoidal sin(x), but as the x increases, it is divided by the factor x, so each period 2pi, it's just the sin wave getting "linearly" smaller.
    On the other hand, this scalling over each x, makes it such that the area over the whole domain is pi, which is the area of a circle of r=1.
    Which clearly has some meaning, the sin is constantly alternating its sign, and x is either x>0 or x

  • @brianneill4376
    @brianneill4376 11 месяцев назад

    Pi devided by 2 amoubts to 1/3 of a Cubical measure.
    All measures are 3D so must be no more or less than Cubical or "Powered to 3", before and or after that are simply sizes that are fractions, Positive or negative to the control body size.

  • @user-sn1lg9js6m
    @user-sn1lg9js6m 6 месяцев назад +1

    Is there a video for when it is allowed to change the order of an integral sign followed by a summation notation? Thank you.

    • @DrTrefor
      @DrTrefor  6 месяцев назад +1

      I really should make this video

  • @airsquid8532
    @airsquid8532 11 месяцев назад +3

    Always a good day when Dr trefor bazette shows up in my RUclips recommended

  • @shivamkardam4608
    @shivamkardam4608 11 месяцев назад +1

    Amazing sir 😮😮😮😮😮

  • @namelastname2244
    @namelastname2244 11 месяцев назад

    3:22 If s is 0 and not a positive, then why do you consider it a negative exponential?

  • @saidfalah4180
    @saidfalah4180 4 месяца назад

    Can we use (sin x/x) / pi/2 in probability ?

  • @edoardoferretti5493
    @edoardoferretti5493 11 месяцев назад

    What is the limit of F(s) as s goes to - infinity? Doesn't the relation we found fail since the integral is not bounded, while arctan is? How do I define the domain of validity of the identity?

    • @reeeeeplease1178
      @reeeeeplease1178 11 месяцев назад

      It may be that the step at 1:47 only works for positive s. What we are doing is interchanging 2 limits (pulling the derivative inside the integral), which we can't *freely* do

  • @padraiggluck2980
    @padraiggluck2980 3 месяца назад

    When I listen to a Feynman lecture I hear a smart version of Ed Norton.

  • @zegrirsaid2855
    @zegrirsaid2855 7 месяцев назад

    thank you

  • @puh8825
    @puh8825 7 месяцев назад +1

    That's insane

  • @peterpan1886
    @peterpan1886 10 месяцев назад

    You can also view the integral as the imaginary part of the integral of e^(izx)/x, evaluated at z=1. Now integrating under the integral sign yields I'(z)=[e^(izx)/z]_0^\infty = -1/z for complex z with positive imaginary part. Hence I(z)=c-log(z). Remember that we are only interested in the imaginary part of I(z), therefore we only need the imaginary part of c. Now let the imaginary part of z approach infinity, while the real part remains constant. Then the imaginary part of log(z) will approach pi/2, while the integral of e^(izx)/x approaches zero. Hence Im(c)=pi/2 and the limit of Im(c-log(z)) as z approaches 1 is pi/2.

  • @sambhavgupta4653
    @sambhavgupta4653 11 месяцев назад +2

    Can't we use exponential definition of sine, (exp(ix)-exp(-ix))/2i to solve it?

    • @DrTrefor
      @DrTrefor  11 месяцев назад +3

      This can replace part of the computation (the double integration by parts) but you still need to use some trick at the beginning before you can use this.

    • @sambhavgupta4653
      @sambhavgupta4653 11 месяцев назад

      Thanks! I'll try.

    • @mokoufujiwara4209
      @mokoufujiwara4209 11 месяцев назад

      And you use the same Laplace Transform / Feynman's Technique at start right?
      If yes, I recommend you use these rules instead:
      sin(x) = Im(e^xi)
      k * Im(z) = Im(kz)
      ∫ Im(z) dx = Im(∫ z dx)

    • @adamc973
      @adamc973 11 месяцев назад

      Easier to note that this is the imaginary part of e^(iz)/z over the interval [-infty,infty] in the complex plane, and work over a semi-circle contour in the upper half plane.

  • @duckyoutube6318
    @duckyoutube6318 7 месяцев назад +1

    What doesnt Feynman destroy? Guy was a beast.

  • @Toxic__rl
    @Toxic__rl 11 месяцев назад

    any smart way to solve int of x^100 sinx dx?

  • @Ron_DeForest
    @Ron_DeForest 11 месяцев назад

    I’m sure this is beneath your notice but I have to ask and hope for the best. I need to know how to find a point in 3 space that’s equidistant from 3 other points. I’ve been looking online and for the life of my I can’t find how to do it. It’s been a very long time since I took any math classes.

  • @EVERYTHING_FACTORY
    @EVERYTHING_FACTORY 10 месяцев назад

    Please make a mathematics books recommendation video

  • @olegzubelewicz3604
    @olegzubelewicz3604 10 месяцев назад

    Mr. Feynman thought that math began with him. 🤣

  • @kaiserali5928
    @kaiserali5928 11 месяцев назад

    Sir I have a question to you. I am an Engineering student from Bangladesh. Sir I derived an alternative procedure or technique to solve a math content which is being solved in complex ways nowadays. Now I want to publish it. Can I use LaTeX to write my paper? Again how and where I can publish it.
    If you give me some piece of advice then it will be very helpful for me.

    • @JuanRomero-re4qz
      @JuanRomero-re4qz 11 месяцев назад

      Perdón!
      Busca una revista científica de tú localidad, y midiendo el terreno puedes pasar a una publicación en idioma ingles.

  • @MushookieMan
    @MushookieMan 11 месяцев назад +1

    I thought the Laplace transform required a complex 's'. Does that affect this method?

    • @DrTrefor
      @DrTrefor  11 месяцев назад

      It can be complex, but we ultimately are only using it to get this differential equation which we only evaluate at infinity and zero so it doesn’t matter if other values could be complex

    • @mokoufujiwara4209
      @mokoufujiwara4209 11 месяцев назад

      Remember real number set is a subset of complex number set.
      If an equation is valid in the whole complex plane, the same equation is valid in the whole real number line, all we need is to limit the imaginary part to 0.

  • @journeymantraveller3338
    @journeymantraveller3338 11 месяцев назад +2

    Isn't Feynman's trick just using Liebniz's Rule on the Laplace transform? Also, the DI method works well for these IBPs .

  • @ulyssesfewl1059
    @ulyssesfewl1059 5 месяцев назад

    Did I miss something? At 1:29 if you set s = 0, you do not get back what you started with, since the entire exponent is zero if s = 0, and e^0 = 1.

  • @mr_angry_kiddo2560
    @mr_angry_kiddo2560 11 месяцев назад +1

    Will you give a proof for Existence of local Maxima b/w two consecutive local minima😢

  • @Grassmpl
    @Grassmpl 10 месяцев назад

    Doesn't feymann trick needs some sort of uniform convergence?

  • @Volk715
    @Volk715 6 месяцев назад

    Well, Feynman was the first "hacker" in history...I'm not surprised about his ability to tackle dirichelet integrals..

  • @tolkienfan1972
    @tolkienfan1972 11 месяцев назад

    Nice!

  • @fordtimelord8673
    @fordtimelord8673 9 месяцев назад +2

    Contour integration works great for this problem too. Using a semicircle contour and taking the imaginary part of the integral of e^iz/z is maybe even an easier method.

    • @revanth36
      @revanth36 7 месяцев назад +1

      Interestingly yes😀

  • @bachirblackers7299
    @bachirblackers7299 11 месяцев назад +1

    ❤❤❤❤❤❤❤❤❤thanks much

  • @radmir_khusnutdinov
    @radmir_khusnutdinov 11 месяцев назад

    I believe using the residue is the fastest way to calc that integral

  • @amauta5
    @amauta5 7 месяцев назад

    Is this the only way to solve the original integral?

  • @hippospudweb
    @hippospudweb 10 месяцев назад +1

    Where can I get that t-shirt?

    • @DrTrefor
      @DrTrefor  10 месяцев назад +1

      Merch link in description!

  • @marciliocarneiro
    @marciliocarneiro 6 месяцев назад

    If we expand sin(x) in Taylor´s series and divide by x we obtain arctg(x) in just one step and the answer much more fast

  • @alphalunamare
    @alphalunamare 11 месяцев назад

    Is 2:06 even legal? I suppose it must work in most common cases but can you prove it generally? I guess you can't and it is indeed strange. Anyway I really enjoyed that 🙂 It's a bit like with platonic solids, you differentiate its Volume formula and get its Surface Area formula ... and most people are convinced ...except that it only works for Platonic Solids not the general mish mash you'll meet in Real Physics.

  • @baptistebermond2082
    @baptistebermond2082 11 месяцев назад

    well it is beautiful, but the real question is can we define the function as continuous and derivable everywhere which makes the trick less evident

  • @maths_505
    @maths_505 11 месяцев назад

    The Dirichlet integral is AWESOME!
    I made a video on solving it using 5 different ways including Feynman's trick:
    ruclips.net/video/MuP6NtGwLTk/видео.html

    • @daddy_myers
      @daddy_myers 11 месяцев назад +1

      Damn. Imagine posting unsolicited ads for your channel.

    • @maths_505
      @maths_505 11 месяцев назад

      @@daddy_myers 😂😂😂
      Who cares
      I'm the one who did a marathon on it.
      You think I wouldn't be hyping it 😂😂😂

  • @samueldeandrade8535
    @samueldeandrade8535 5 месяцев назад

    We live at a weird time. When doctors make videos with clickbaity titles. *Roll eyes*.

  • @akerosgaming7400
    @akerosgaming7400 11 месяцев назад

    I don't think the conditions needed to say that F is differentiable and take the partial differential inside the integral are there.

  • @bobdavisbeta
    @bobdavisbeta 7 месяцев назад

    All the physicists think that this method is Feynman trick but it existed way way before. It is a little abusive to atribute the credit to Feynman when this is actually extremely classical.

    • @dylangabriel2703
      @dylangabriel2703 6 месяцев назад

      Every physicist knows that the integration technique existed before Feynman, it’s just kind of that Feynman was famous for using it.

  • @xl000
    @xl000 10 месяцев назад

    it must be weird to look and point at a green screen when saying "Like this one" at 0:04

  • @domc3743
    @domc3743 11 месяцев назад +3

    brother didnt mention why we can switch order of differentiation and integration, Leibniz rule applies because F(s) bounded and continuous

    • @DrTrefor
      @DrTrefor  11 месяцев назад +3

      Oh thanks, I actually had it in my notes to point this out verbally but somehow forgot to say it out loud during recording time:)

    • @ShanBojack
      @ShanBojack 11 месяцев назад +1

      ​@@DrTrefor can you elaborate it to me please sir

  • @toastyPredicament
    @toastyPredicament 11 месяцев назад

    How

  • @thinkacademy8377
    @thinkacademy8377 11 месяцев назад +1

    I don't know why this is called Feynman's trick? differentiation under the integral sign is known before Feynman. It's just the application of Leibniz theorem for integrals dependent on a parameter!

    • @DrTrefor
      @DrTrefor  11 месяцев назад

      Also, the types of integrals Feynman was considering were nothing like these, but nevertheless the attribution seems to stick

    • @frenchimp
      @frenchimp 11 месяцев назад +1

      @@DrTrefor So why perpetuate the ridiculous notion that this technique is due to Feynman in these videos?

  • @yesiamrussian
    @yesiamrussian 2 месяца назад

    cannot believe you forgot the + c

  • @maksgo498
    @maksgo498 11 месяцев назад

    I would use taylor

  • @shaun18
    @shaun18 5 месяцев назад

    I dumbass though I could just do the uv rule here as (x^-1)(sinx) then realised it would just keep adding x^n in the denominator in the second part of the uv rule 💀

  • @radekvecerka1115
    @radekvecerka1115 5 месяцев назад

    You should at least mention, that certain conditions need to be met in order to be able to use this method

  • @saleemshaya67
    @saleemshaya67 7 месяцев назад

  • @cantkeepitin
    @cantkeepitin 7 месяцев назад

    Imagine getting this problem in your math class: Destroy the integral by Feynmann's technique. .....
    Find the error. Thumb down for such unsuited video title.

  • @tanwarsingh396
    @tanwarsingh396 8 месяцев назад

    Sir you can generate the WhatsApp group for all mathematics candidates

  • @ACh389
    @ACh389 Месяц назад

    It doesn't bother you take the partial inside of the integral. You have to use DCT in order to be able to do that but you completely ignored that. Great!

  • @davidwright5719
    @davidwright5719 11 месяцев назад

    Argh, don’t use double integration by parts here. Just use sin(x) = (e^{ix} - e^{-ix})/(2i) to get an integral involving only exponentials. This is almost always the best way to do almost any calculus problem involving trig functions. (Also non-calculus problems. Never worry about remembering a double-angle formula again.)

    • @piyusharora5327
      @piyusharora5327 11 месяцев назад

      Our professor taught like what you are saying when he was teaching the same thing in video in the name of Laplace transform.

  • @kevinnielsen1356
    @kevinnielsen1356 11 месяцев назад

    When seeking realization of reality, obsession with math is the cosmic equivalent of going thru life staring at a cell phone.

  • @warhurst1968
    @warhurst1968 11 месяцев назад

    An important caveat: this is true of the Riemann integral but not the Lebesgue integral.

    • @frenchimp
      @frenchimp 11 месяцев назад

      Why do you try to confuse Americans with all those European names? Feynmann trick, no caveats, that's the beauty of it!

    • @warhurst1968
      @warhurst1968 11 месяцев назад

      @@frenchimp sin(x)/x is Riemannian integrable but not Lebesgue integrable over R.

    • @frenchimp
      @frenchimp 11 месяцев назад

      @@warhurst1968 Well it's not Riemann integrable, it's an improper integral.

    • @warhurst1968
      @warhurst1968 11 месяцев назад

      @@frenchimp Fair enough, but that's the point, you have be careful about what you mean by integrable with this example.

  • @walter274
    @walter274 Месяц назад

    If you're a stat person, this is a lot like an MGF.

  • @0sKiDo
    @0sKiDo 11 месяцев назад

    Did you justify why we can take the derivative and switch the ds with the integral ?…

  • @welldonehuang9133
    @welldonehuang9133 День назад

    ✌yes,good vedio