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.
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.
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.
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.
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 :)
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?
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.
I was so confused at the beginning... you’re probably the only UA-camr who says “thanks for watching” at the beginning of the video instead of at the end!
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.
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.
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.
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.
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 !
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. :)
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.
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.
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.
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
He splits the x^2+1-1 into two terms, x^2+1 and the number -1, and then splits the integral into two integrals, one for each term. For the first term the x^2+1 cancels with the 1+x^2 in the denominator, and the second term ends up just being the initial integral that defines f(a) just with a factor multiplying it
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.
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.
So the result is not normal, but nearly perfect (Peerfc(1)).
😂😂😂
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.
An example of watering down the curriculum? I'm sure it's an isolated case.😁
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.
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.
it's a theorem usually covered in Measure theory. Sadly physicians don't often have this course.
You're an awesome professor. I laughed a lot and the problem was interesting. Thanks
That french was spot-on and so was the math.
new generation is so lucky to have these videos around. It really helps a lot you cant imagine , thx to prof. :)
So true!
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.
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.
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).
Cool. thanks
Congratulations, you have pleasantly blown my mind again.
Thank you! 🙂
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 :)
nothing tickles my fancy more than math with fun jokes in other languages!
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?
That was beautiful. Thanks for sharing this sir :)
Leibniz: Am I a joke to you?
Lmaoooo
Another wonderful video. Thanks Dr.πm!
13:37 😁
Sie gehen mir NIE auf die Nerven, Dr πm
Hahahaha
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.
Nice video!!! Thank you!! Amazing technique
Gracias por sus vídeos... Manos a la obra
11:16 I didnt know that you speak german ;)
Lots of love From Germany Berlin!.
THanks for showing your Magic to us
I was so confused at the beginning... you’re probably the only UA-camr who says “thanks for watching” at the beginning of the video instead of at the end!
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.
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.
which is incorrect
In Portuguese:
"Por que simplificar,
se se pode complicar?" 😁
Warum einfach, wenn es auch kompliziert geht? - German
Proč dělat něco jednoduše, když to jde složitě? - Czech
@@morgard211 přesně jsem to teď chtěl napsat :D
I love how we all have the same saying 😂
''ⴰⵢⵖⴻⵔ ⴰⴷ ⵜⵏⴻⵔ ⴷ ⵍⵉEⴰ ⵎⴰ ⵏⴻⵣⵎⴻⵔ ⴰⴷ ⵜⵏⴻⵔ ⴷ ⴰⵙⴰⵡⴻⵏ ''In Tamazight
Amazing Dr Peyam. I love Feynman’s technique. Very cool
Thanks. I really enjoy your videos.
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.
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.
1:13 I love this saying ❤️
The title of the video got me lol the equation was amazing as well
i agree this was very beautiful integral and astonishing result.
¡Manos a la obra! , cordiales saludos Dr. U should make videos talking bout' Fourier transform !!! I love your work :)
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.
The most trouble I have with these is: how do I decide where the “a” term goes?
It’s an art, there’s no reason why the a should be at a specific point
beautiful 👏🏻👏🏻👏🏻
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…
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 !
jolie démonstration !
Best and fast...
I remember a long bad proof in university....
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. :)
That x square + 1 in the denominator was looking sooo delicious.
Is the error function complimentary? What is important here? Understanding Euler's equations or perhaps his thoughts on expressing his understanding?
Love from SoCal
🏝🏝🏝
German, French *and* Spanish?! Surely you're joking Dr. Peyam??
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.
If use i^2 as -1
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.
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.
Manos a la obra!! Jejeje... Saludos desde Andalucía (España)
but doesnt a=-1 also get back the original integral, the final evaluation will be different though since erf(-1)=-erf(1)
Your voice feels as if my friend is talking to me😄. Love your presentation
may post a video on numerical calculation og error function
I love how bottom left of the board dances on 720p60
I know, it’s so weird it does that 😂
Hmmm... the gauss'ian integral is so very close to being just a constant.
What criteria should be used to know where to enter the parameter?
Pure guess
11:10 es geht eher an die Perven, oder? :D
we can also solve the 1st order differential equation : f'(a)-2a f(a) = -2square(Pi)
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
As a French Canadian person, I found your accent wonderful !! Did you take any french class ?
I went to a French Lycée :)
@@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 !
There is always beauty in maths!
Brilliant!
What a easy equation! It is a piece of cake.
6:13
you made things complicated. It is a differential equation of order 1 which is very simple to solve.
Ok
Hey everyone! Could somebody please explain the step at 3:28? I dont seem to get how dividing by the 1+x^2 term leads to the following equation.
He splits the x^2+1-1 into two terms, x^2+1 and the number -1, and then splits the integral into two integrals, one for each term.
For the first term the x^2+1 cancels with the 1+x^2 in the denominator, and the second term ends up just being the initial integral that defines f(a) just with a factor multiplying it
Awesome! Great video before going to the job. 😂
Hey what whiteboard do you use?
Very nice presentation.
Can this be solved using complex analysis? If so can you upload a video on it too
Maybe, you come up with a solution :)
when you did f(a) why did you only put 'a' in numerator and not denominator? Denominator also had an 'x' term?
It would make it more complicated! Putting a once is simple yet already solves our problem
If you substitute a as -1 can it still give the same answer like this
Can you please recommend a good reference to study math...?and to understand more please...
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.
Feyman’s idea is very similar to the calculus of variations as discussed in Evan’s PDE textbook
I love this method
Bellísimo
One question, is the number Ei(1) a transcendental number?
Can be solved by using and Gamma or Beta function!!!🤩
mathematics done beautifully
Explain the problem. Y can't you put it integral where numerator is 1?
Belief has no function here
Can all integrals be turned into differential equations like this?
Pourquoi faire ça si on peut faire compliqué ? Vamos a la obra ! Delicious Pi ! Es geht mir auf die Nerven! ....Love u doc.❤
Is that constant, the result of the integral, a transcendental number?
Please make a vedio on design of experiments
Can you provide a proof?
Es geht mir auf die πERF(en).
Haha greetings from Germany 🇩🇪
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.
Interesting!
The answer almost says "perfect"
Saludos, muy interesante el video.
does minus infinity (or infinity for that matter) have any meaning?
Those are the bounds of the integral. If he wanted to he could evaluate the function between those two x values finding the area
The correct thing is to take a finite upper/lower limit, and then take the limit as those grow to +- infinity respectively.
I'm french and I was confused when u started talking french xD
Is Dr. Peyam really doing calculations in the kitchen? I thought the kitchen was meant for cooking, but no, I was wrong.
Set 1 to "t". That would be perfect.
Nice integral
But Erfc(1) is different from Erfc(-1).
Could we replace a=-1?
He said a is positive
@@joao_pedro_c Thanks.
At what point the fact that a>0 was used?
@@nournote i cant rewatch rn but I think he said it when he made the u sub to get the gaussian integral
In Slovak:
Prečo to urobiť jednoducho, keď sa to dá urobiť zložito? :)
You are really a nice guy!
11:20 I think it's actually "Es geht mir auf die Nerven".
What an idea
It is useful. thanks.