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.
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.
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.
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!
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?
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.
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.
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.
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.
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.
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. :)
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.
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 !
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.
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.
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.
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.
That french was spot-on and so was the math.
nothing tickles my fancy more than math with fun jokes in other languages!
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.
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.
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! 🙂
You're an awesome professor. I laughed a lot and the problem was interesting. Thanks
Leibniz: Am I a joke to you?
Lmaoooo
new generation is so lucky to have these videos around. It really helps a lot you cant imagine , thx to prof. :)
So true!
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!
That x square + 1 in the denominator was looking sooo delicious.
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 :)
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
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.
Nice video!!! Thank you!! Amazing technique
That was beautiful. Thanks for sharing this sir :)
The title of the video got me lol the equation was amazing as well
Another wonderful video. Thanks Dr.πm!
Best and fast...
I remember a long bad proof in university....
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?
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.
jolie démonstration !
Amazing Dr Peyam. I love Feynman’s technique. Very cool
1:13 I love this saying ❤️
Thanks. I really enjoy your videos.
beautiful 👏🏻👏🏻👏🏻
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.
i agree this was very beautiful integral and astonishing result.
I love how bottom left of the board dances on 720p60
I know, it’s so weird it does that 😂
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
Gracias por sus vídeos... Manos a la obra
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.
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.
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. :)
Your voice feels as if my friend is talking to me😄. Love your presentation
Awesome! Great video before going to the job. 😂
German, French *and* Spanish?! Surely you're joking Dr. Peyam??
¡Manos a la obra! , cordiales saludos Dr. U should make videos talking bout' Fourier transform !!! I love your work :)
mathematics done beautifully
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
There is always beauty in maths!
What a easy equation! It is a piece of cake.
Love from SoCal
🏝🏝🏝
I love this method
Hmmm... the gauss'ian integral is so very close to being just a constant.
Feyman’s idea is very similar to the calculus of variations as discussed in Evan’s PDE textbook
The answer almost says "perfect"
Manos a la obra!! Jejeje... Saludos desde Andalucía (España)
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.
Belief has no function here
Es geht mir auf die πERF(en).
Haha greetings from Germany 🇩🇪
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 !
Very nice presentation.
Brilliant!
Nice integral
Pourquoi faire ça si on peut faire compliqué ? Vamos a la obra ! Delicious Pi ! Es geht mir auf die Nerven! ....Love u doc.❤
fantastic integral wish Peyam was my college teacher 🔥😃
If use i^2 as -1
Is the error function complimentary? What is important here? Understanding Euler's equations or perhaps his thoughts on expressing his understanding?
6:13
you made things complicated. It is a differential equation of order 1 which is very simple to solve.
Ok
Bellísimo
In Slovak:
Prečo to urobiť jednoducho, keď sa to dá urobiť zložito? :)
we can also solve the 1st order differential equation : f'(a)-2a f(a) = -2square(Pi)
What an idea
I'm french and I was confused when u started talking french xD
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.
Ballet in a math tutu.
Dicho por el Negro Olmedo: “¡Si lo vamo’ a hacer, lo vamo’ a hacer bien!” :D
And don't call me Shirley.
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 !
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…
Can be solved by using and Gamma or Beta function!!!🤩
Me, whatching this when I can Hardy Solve a simple integral using U substitution...
I didn't expect that spanish quote xd, nice prob.
You are really a nice guy!
Genius solution, I love ist👍
but doesnt a=-1 also get back the original integral, the final evaluation will be different though since erf(-1)=-erf(1)
this problem just screams residue theorem to me
11:10 es geht eher an die Perven, oder? :D
It is useful. thanks.
your jokes are hilarious
11:20 I think it's actually "Es geht mir auf die Nerven".
Set 1 to "t". That would be perfect.
Saludos, muy interesante el video.
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!
Is Dr. Peyam really doing calculations in the kitchen? I thought the kitchen was meant for cooking, but no, I was wrong.
Wonderful
Very good
may post a video on numerical calculation og error function
What criteria should be used to know where to enter the parameter?
Pure guess
13:34 pretty peerfct
Très content de vous entre parler français
very good accents!
I’m surprised nobody brings up int_{-infty}^{infty} e^(ix)/(x^2+1)dx=pi/e as beautiful identities involving fundamental cosntants.
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.