So the result is not normal, but nearly perfect (Peerfc(1)).
@drpeyam3 жыл бұрын
😂😂😂
@puerulus3 жыл бұрын
The differentiation under the integral sign technique is covered in many calculus textbooks from the 1800s and early 1900s. Richard Feynman himself pointed that out. He probably wouldn't have wanted the technique named after him.
@euva2092 жыл бұрын
An example of watering down the curriculum? I'm sure it's an isolated case.😁
@UnforsakenXII3 жыл бұрын
I remember being obessesed with these kind of integrals back in sophomore year of undergrad, needless to say, it paid off tremendously and I wish it was taught more standardly at least in physics curricula.
@jacobharris58943 жыл бұрын
I’m still obsessed with them now but I learned it pretty recently. Whenever I see a hard looking integral I try to find any excuse I can to use it.
@SimsHacks Жыл бұрын
it's a theorem usually covered in Measure theory. Sadly physicians don't often have this course.
@thelink99593 жыл бұрын
That french was spot-on and so was the math.
@ChariotduNord3 жыл бұрын
nothing tickles my fancy more than math with fun jokes in other languages!
@AnkitSharma-ef7md3 жыл бұрын
While others get to read early morning text from their sweet and loved ones, I get a notification from youtube that Dr. Peyam has uploaded another astonishing problem from mathematics. So good to see you professor, Amazing approach. I was puzzled at the step where you nicely adjusted the error function. Loved it. Seasonal greetings to you.
@noonesperfect3 жыл бұрын
new generation is so lucky to have these videos around. It really helps a lot you cant imagine , thx to prof. :)
@constantinototis48593 жыл бұрын
So true!
@77Chester773 жыл бұрын
13:37 😁 Sie gehen mir NIE auf die Nerven, Dr πm
@drpeyam3 жыл бұрын
Hahahaha
@thomasborgsmidt98013 жыл бұрын
Ich glaube Du hast im Schwarzen getroffen! πm does not need to make the student feel like an idiot - but he points out and lets the student decide for himself, that maybe he has to reevaluate. Math is - like all sciences - not about finding answers, but finding the relevant question. It is like Kepler that just asked himself: Does planetary orbits HAVE to be circular? Well, they don't - the concept of circular orbits was made to make calculations simpler - and elipses do not have a rough and ready formula for the length of the circumfence. But then again the quest for making a simple circular equations valid led to a religious conviction that was much more complicated as an end result.
@GRBtutorials3 жыл бұрын
I was so confused at the beginning... you’re probably the only KZbinr who says “thanks for watching” at the beginning of the video instead of at the end!
@hOREP2453 жыл бұрын
One of my favourite integrals! This can be easily extended using the Gamma and Incomplete function for any n>0 (where this is the special case n = 2).
@robertl.crawford43692 жыл бұрын
Cool. thanks
@mudkip_btw3 жыл бұрын
Living dangerously pays off, that's why I like it :} Though I feel like it's getting time for me to learn the proper theorems at this point, like dominated convergence as you talked about. I did quite badly in my calculus 1, 2 classes but much better later, once we got to vector calculus etc. since I'm more of a visual learner. Glad to have people like you around to keep me interested in proper maths as a physics student :)
@naveensagar24083 жыл бұрын
Integration under diff followed by lebini t' z rule feynman technique is very awsome ! being a masters in physics i can tell only 1 thing this technique is very useful in solving the famous dirichlet integrals . & Many problems in physics.
@muratkaradag37033 жыл бұрын
11:16 I didnt know that you speak german ;) Lots of love From Germany Berlin!. THanks for showing your Magic to us
@alexdemoura99723 жыл бұрын
In Portuguese: "Por que simplificar, se se pode complicar?" 😁
@TheTKPizza3 жыл бұрын
Warum einfach, wenn es auch kompliziert geht? - German
@morgard2113 жыл бұрын
Proč dělat něco jednoduše, když to jde složitě? - Czech
You went way over my head but one thing I learned is that you are afflicted with the same disease that I am, you are left handed.
@integralboi29003 жыл бұрын
Leibniz: Am I a joke to you?
@marinmaths38263 жыл бұрын
Lmaoooo
@colonelburak29063 жыл бұрын
I'm a PhD student in applied maths, and I still learn so much from your videos, Dr. Peyam! The French saying we have in Sweden as well: Varför göra något enkelt när man kan göra det svårt?
@66127703 жыл бұрын
Congratulations, you have pleasantly blown my mind again. Thank you! 🙂
@channalbert3 жыл бұрын
By instinct, maybe biased by "Feynman" in the title, I tried to find the most direct way to apply Feynman's trick, and I came up with multiplying outside the integral by e and inside by e^-1. The latter, together with e^(-x^2) in the integrand form e^[-(1+x^2)] and you can now see an easy parametrization to do the trick: e^[-a(1+x^2)] with the limits from 1 to infty.
@vai_-cn9br3 жыл бұрын
That was beautiful. Thanks for sharing this sir :)
@adritobiswas19822 жыл бұрын
That x square + 1 in the denominator was looking sooo delicious.
@chriswinchell15703 жыл бұрын
Hi Dr. Peyam, I accept your challenge to make things worse: 1. Expand 1/(x^2) in a series and integrate against guassian function. You get an infinite sum of alternating even moments. (A sum of gamma functions, I think.) 2. Expand the guassian and integrate against the Cauchy distribution so you get a sum of moments of Cauchy. 3. Find the integral of the inverse Fourier transform of the convolution of guass and Fourier transform of Cauchy (It’s something like e^-abs(x)). I wonder if this works out better than it sounds. 4. Things aren’t always difficult, remember: toujours aimer le vent qui leve les joupons.
@spintwohalves3 жыл бұрын
Another wonderful video. Thanks Dr.πm!
@CoffeeTroll3 жыл бұрын
1:13 I love this saying ❤️
@umerfarooq48313 жыл бұрын
The title of the video got me lol the equation was amazing as well
@Lupercus-ht1xt11 ай бұрын
You're an awesome professor. I laughed a lot and the problem was interesting. Thanks
@alphaglucopyranose69283 жыл бұрын
Or just use residue integral and integrate on a semicircle on the upper half plane. The function has a simple pole at z=i. Plug in z=i into 2πi*e^(-z^2)/(z+i). We get πe.
@theelk801 Жыл бұрын
which is incorrect
@joaquingutierrez30723 жыл бұрын
Nice video!!! Thank you!! Amazing technique
@vivelesport81973 жыл бұрын
Best and fast... I remember a long bad proof in university....
@itswakke3 жыл бұрын
The most trouble I have with these is: how do I decide where the “a” term goes?
@drpeyam3 жыл бұрын
It’s an art, there’s no reason why the a should be at a specific point
@triton626743 жыл бұрын
German, French *and* Spanish?! Surely you're joking Dr. Peyam??
@mertaliyigit32883 жыл бұрын
I love how bottom left of the board dances on 720p60
@drpeyam3 жыл бұрын
I know, it’s so weird it does that 😂
@marinmaths38263 жыл бұрын
Amazing Dr Peyam. I love Feynman’s technique. Very cool
@maxsch.65553 жыл бұрын
I tried to solve this integral by myself. I introduced the factor e^(-a(1+x^2)) in the integrand and used feynmans trick. I defined f(a) to be the integral from 0 to inf of e^(-x^2)/(1+x^2) * e^(-a(1+x^2)) dx. Notice that the integral we want is 2f(0). For f'(a) i got -sqrt(pi)/2 * e^(-a)/sqrt(1+a). To get the original function f(a) I integrated from inf to a. I could express f(a) using the incomplete gamma function: f(a) = sqrt(pi)/2 * e * gamma(1/2, 1+a) So the original integral is 2f(a) = sqrt(pi)*e*gamma(1/2, 1) If we use the relation between the incomplete gamma function and the error function we get the same answer as in the video. :)
@ny6u3 жыл бұрын
beautiful 👏🏻👏🏻👏🏻
@MrRyanroberson13 жыл бұрын
Erf(1) is such an underrated constant and really puts things into perspective when working wuth statistics. Erf(1)*2 many people lie withing just one standard deviation of the average, and so on, as a fundamental definition.
@bprpfast3 жыл бұрын
Love from SoCal
@drpeyam3 жыл бұрын
🏝🏝🏝
@shreyashegde40322 жыл бұрын
Your voice feels as if my friend is talking to me😄. Love your presentation
@joeremus90392 жыл бұрын
Thanks. I really enjoy your videos.
@KarlMarX_933 жыл бұрын
Awesome! Great video before going to the job. 😂
@yaskynemma9220 Жыл бұрын
The answer almost says "perfect"
@peterdecupis82962 жыл бұрын
This brilliant example paradoxically highlights that the success of Feynman method is critically dependent on a "fortunate" guess of the parametric generalization of the integrand function.
@bernardlemaitre47013 жыл бұрын
jolie démonstration !
@ahmedgaafar53693 жыл бұрын
i agree this was very beautiful integral and astonishing result.
@uxueanderealdazabal1282 жыл бұрын
Gracias por sus vídeos... Manos a la obra
@quinktap3 жыл бұрын
Belief has no function here
@LucPatry3 жыл бұрын
mathematics done beautifully
@luna92003 жыл бұрын
Nice application of this that I thought of: If you do the same thing but with sin(x^2)/(1+x^2), you differentiate twice and use the Fresnel Integrals, and you get a second order ODE! A nice answer comes out if you're willing to use the Fresnel S and C functions.
@thomasborgsmidt98013 жыл бұрын
Hmmm... the gauss'ian integral is so very close to being just a constant.
@emiliomontes20433 жыл бұрын
¡Manos a la obra! , cordiales saludos Dr. U should make videos talking bout' Fourier transform !!! I love your work :)
@jesusalej13 жыл бұрын
There is always beauty in maths!
@YWanek-ft2fs3 жыл бұрын
Es geht mir auf die πERF(en). Haha greetings from Germany 🇩🇪
@acac73533 жыл бұрын
Manos a la obra!! Jejeje... Saludos desde Andalucía (España)
@AstroB73 жыл бұрын
As a French Canadian person, I found your accent wonderful !! Did you take any french class ?
@drpeyam3 жыл бұрын
I went to a French Lycée :)
@AstroB73 жыл бұрын
@@drpeyam oh that’s fantastic, french is a very hard language ! It’s always nice to see your great content ! Keep up the good work !
@h4z4rd283 жыл бұрын
In Slovak: Prečo to urobiť jednoducho, keď sa to dá urobiť zložito? :)
@chemistrychamp3369 Жыл бұрын
Thanks amazing integration. I Never really know using this method, when u integrate for da how to put the limits of the integrale and why…
@paulg4443 жыл бұрын
The first thing I would do is bound that integral by putting x=0 in the denominator term (1+x^2)=1. .. that way I have a quick relation with the ERF function, proceed then to solve and then I double check that my solution does not exceed the upper bound. Else Ive got so many moving parts that I dont know if Im right at the end or if I made a few teen mistakes !
@superkarnal133 жыл бұрын
Bellísimo
@peterchindove71462 жыл бұрын
Brilliant!
@stochasticxalid9853 Жыл бұрын
Pourquoi faire ça si on peut faire compliqué ? Vamos a la obra ! Delicious Pi ! Es geht mir auf die Nerven! ....Love u doc.❤
@mlgswagman6002 Жыл бұрын
Nice, took a similar approach but it doesn't go through the differential equation route. Let I(a) = integral from -inf to inf of (e^(a(-x^2-1)) / (x^2 + 1)) Note that e*I(1) is the desired integral. Differentiating under the integral now completely destroys the denominator. I'(a) = -sqrt(pi)/(e^a * sqrt(a)) Now, just directly solve for I(a) using that lim as a -> inf I(a) = 0. You can do some trickery with integration by parts and a substitution to get the same result you did.
@bruwayn21582 жыл бұрын
we can also solve the 1st order differential equation : f'(a)-2a f(a) = -2square(Pi)
@justjacqueline20043 жыл бұрын
Ballet in a math tutu.
@petervinella55452 жыл бұрын
Feyman’s idea is very similar to the calculus of variations as discussed in Evan’s PDE textbook
@PhilesArt3 жыл бұрын
I'm french and I was confused when u started talking french xD
@Idtelos3 жыл бұрын
And don't call me Shirley.
@alexcaesar53773 жыл бұрын
What a easy equation! It is a piece of cake.
@quinktap3 жыл бұрын
Is the error function complimentary? What is important here? Understanding Euler's equations or perhaps his thoughts on expressing his understanding?
@ericpham82053 жыл бұрын
If use i^2 as -1
@brendanlawlor2214 Жыл бұрын
fantastic integral wish Peyam was my college teacher 🔥😃
@alecapin3 жыл бұрын
Dicho por el Negro Olmedo: “¡Si lo vamo’ a hacer, lo vamo’ a hacer bien!” :D
@DeanCalhoun2 жыл бұрын
this problem just screams residue theorem to me
@yashagrahari3 жыл бұрын
Nice integral
@joniiithan2 жыл бұрын
your jokes are hilarious
@ilovephysics3503 жыл бұрын
Set 1 to "t". That would be perfect.
@AndDiracisHisProphet3 жыл бұрын
11:10 es geht eher an die Perven, oder? :D
@shanmugasundaram96883 жыл бұрын
Very nice presentation.
@karabodibakoane32023 жыл бұрын
Is Dr. Peyam really doing calculations in the kitchen? I thought the kitchen was meant for cooking, but no, I was wrong.
@andresvivas28143 жыл бұрын
Me, whatching this when I can Hardy Solve a simple integral using U substitution...
@g-smith44663 жыл бұрын
Actually what is the meaning of " why do something simply, when it can be done in a more complicated way?" It hits home somehow, but I am not sure I get it right.
@alireza-vq6ul3 жыл бұрын
It is useful. thanks.
@ismailaitabdelkarim7164 Жыл бұрын
6:13 you made things complicated. It is a differential equation of order 1 which is very simple to solve.
@drpeyam Жыл бұрын
Ok
@shibhanlalpandita69753 жыл бұрын
Explain the problem. Y can't you put it integral where numerator is 1?
@arupabinash22633 жыл бұрын
I love this method
@physicsboy31083 жыл бұрын
Can be solved by using and Gamma or Beta function!!!🤩
@camilocastrojimenez86122 жыл бұрын
Saludos, muy interesante el video.
@dihinamarasinghe9278 Жыл бұрын
Can this be solved using complex analysis? If so can you upload a video on it too
@drpeyam Жыл бұрын
Maybe, you come up with a solution :)
@thomasfritz81742 жыл бұрын
And the function f(a) actually has a name: it is the Faddeeva function w(iz) , well exactly it is f(a) = w(ia). I deal with the Faddeeva fct. very often.
@drpeyam2 жыл бұрын
Interesting!
@jamesbentonticer47062 жыл бұрын
It's cool when you say things in French or other languages you speak. You should do that more often.
@drpeyam2 жыл бұрын
Will do :)
@aneeshsrinivas90887 ай бұрын
I’m surprised nobody brings up int_{-infty}^{infty} e^(ix)/(x^2+1)dx=pi/e as beautiful identities involving fundamental cosntants.
@psjt92303 жыл бұрын
What an idea
@meiwinspoi50803 жыл бұрын
may post a video on numerical calculation og error function
@the_nuwarrior3 жыл бұрын
What criteria should be used to know where to enter the parameter?
@drpeyam3 жыл бұрын
Pure guess
@AndDiracisHisProphet3 жыл бұрын
13:34 pretty peerfct
@tgx35293 жыл бұрын
Dr Peyam, please , what is the result for this integral.We are not able get primitiv function for integral exp-x^2 from -infinity till a. I am not sure, mabye I can take origin integral as e* integral from (exp -(1+x^2))/(1+x^2), and then calculate integral f(a) =exp (-a(1+x^2))/(1+x^2), there will be cancel (1+x^2) for f'(a).... , but I am not sure, if this way is OK
@michaelbaum6796 Жыл бұрын
Genius solution, I love ist👍
@dr.mohamedfawzy9633 жыл бұрын
Wonderful
@MrZeno0013 жыл бұрын
You are really a nice guy!
@alpcanakaydn69863 жыл бұрын
Very good
@DynestiGTI3 жыл бұрын
Can all integrals be turned into differential equations like this?