Wow... an integral question solved by partial derivatives, integration by parts, differential equations and the Gaussian Integral to top it all off. Amazing! More Feymann technique questions, please!!
You could also write the cos term as the real part of e^i5x, and then complete the square in the exponential to get the final answer. Physicists use that trick a lot in quantum field theory.
f(a) := integral from 0 to oo of exp(-x^2) cos(ax) dx g(a) := integral from 0 to oo of exp(-x^2) sin(ax) dx H(a) := integral from 0 to oo of exp(-x^2) exp(iax) dx H(a) = f(a) + ig(a) ∴ f(a) = Re(H(a)) && g(a) = Im(H(a)) ------------------------------------------------------------------------------------- exp(-x^2) * exp(iax) = exp( -x^2 + iax ) = exp(-( x^2 - iax )) = exp(-( x^2 - 2(ia/2)x + (ia/2)^2 - (ia/2)^2 )) = = exp(-( (x - ia/2)^2 + a^2/4 )) = exp( -(x - ia/2)^2 - a^2/4 ) = exp(-(x - ia/2)^2) exp(-a^2/4) ------------------------------------------------------------------------------------- H(a) = integral from 0 to oo of exp(-(x - ia/2)^2) exp(-a^2/4) dx = = exp(-a^2/4) integral from 0 to oo of exp(-(x - ia/2)^2) dx ------------------------------------------------------------------------------------- i am stuck at this moment. i tried the transformation u := x - ia/2 but i don't know what to do with the integral: integral from -ia/2 to (oo - ia/2) of exp(-u^2) du that has complex limits (i don't know if that is how i was supposed to set the limits of u variable either) and I am not able to split it into two integrals of real variable either. can you give me a hint how can i proceed from here?
@@michalbotor you did all that before understanding the basic concept of substitution :) Exp(-x^2) if multiplied by the euler's theorem would lead to addition of i in the expression whose integral in forward solving is a pain in butt (from past experiences) So moral is to find a logical concept and think on it before just scribbling this is pro tip in competitive level prep. Be well my friend.
I was going to point it out as my way. But, I guess, the hosts wants to teach the Feynman's method. By the way, Feynman was a physicist if I remember correctly.
@@cpotisch oh thanks this brings back memories from when I was trying to learn calculus by youtube (self learnt) and didn't know the terms thanks for explaining it now since now I have more broad understanding than what I did 3 months ago
AugustoDRA : ))) I actually didn’t learn this when I was in school too. Thanks to my viewers who have suggested me this in the past. I haven a video on integral of sin(x)/x and that’s the first time I did Feynman’s technique.
It's covered in measure theory (math majors only) as one of the conditions to use the theorem is to find a L¹ function such that |d/da f(x,a)| ≤g(x) for almost all x. L¹ = set of functions with finite Lebesgue integral (not ±∞)
If you’re sad about that, you don’t belong in engineering. arcane mathematical techniques are nothing but a tool to an engineer, the primary of objective of an engineer is the creative process of ideating new machine designs, and this on its own is a massively difficult issue that takes enormous creative power. If you’re focusing on learning esoteric integration techniques, you aren’t focusing on engineering. I bet you aren’t an engineer now.
I really wish youtube existed when I was studying mathematics. The potential to be educated in advanced topics without paying a hefty fee for university tuition will hopefully change this world for the better.
I remember this method, because in the video contest I did the integral of (e^-(x^2))*cos(2x) from 0 to infinity. BTW whenever I see e^(-x^2), I always think about feynman technique.
My initial intuition was to use Feynman to get rid of the exponential term, because if you can get rid of that, trig functions are easy. The thing I didn't think through was the limits of integration: a trig function has no limit at infinity. So quite counterintuitively, it was the cosine that was going to be the troublesome element in all this, while the exponential term was what made the thing solvable.
I really enjoy your enthusiasm while explaining things :) Thank you for the videos and please, never lose the energy, liveliness, and passion that you have now. Very nice!
It's a bit crazy to call that the Feynmann technique. It goes back to Leibniz and it"s just deriving an integral depending on a parameter. Which by the way demands justification (either uniform convergence or dominated convergence). And in order to make this work you have to be extremely lucky and have a good intuition because you need 1) to find the right parametrization (here it's pretty obvious) ; 2) to be able to integrate the partial derivative for each value of the parameter (which is most of the time not possible) 3) to end up with a differential equation which you can solve (which is most of the time impossible), 4) to be able to compute a special value (here you need to know the value of the Gaussian integral, which is in itself tricky). So, I'd say it's a nice trick when it works but doesn"t qualify as a method...
It looks like the this problem was purposely designed to arrive at an aesthetically pleasing solution. (Given all the justifications/special circumstances/restrictions you mentioned)
@@Hmmmmmm487Feynman learnt this method in a random book during his undergrad and he famously showed off to basically everyone that he could solve otherwise very hard integrals.
Makes me a bit mad when people call it Feynman's technique. The guy did a lot of good things, but this one has nothing to do with him. They're basically saying that only an American in the middle of 20th century could come up with such idea... What did people all over the world do before that, when calculus was already so advanced, and things like FT and others were well known...
I am agreeing that Feynman's technique is having a good strong hold in solving exponential integrals...but rather than complicating we could have solved it by manipulating "cos(5x)" as (e^5ix + e^-5ix)..it also saves the time...
I'm halfway through algebra 1, and yet somehow I understand and enjoy most of these videos. You and other channels like you (e.g. Mathologer) make this stuff really accessible, and importantly, fun. (Not to say I don't enjoy my algebra 1 class!)
Well I must say ty to you Mr. @blackpenredpen . Thanks to your videos I finished Differential Equations with a B. It was on of my last 2 math classes for my mathematics BS
I don't have an advanced level of English but that's one of a lot of thing that I love Maths, it's an universal language and your passion in every video is the thing because of I'm still here. Imagine! If I can understand you and I don't speak English fluently, you're MORE THAN AMAZING. Lots of love from Mexicoooo ꒰⑅ᵕ༚ᵕ꒱˖♡
I notice the *Feynman' technique* (aka. _Leibniz Integral Rule_ ) depends basically upon parameterizing the parts expansion here; its the _by-parts_ part that gives it the power in my opinion for what its worth!
Because the function is even, you can take the integral from -infinity to infinity and then that would double your answer so the final answer (given alpha = 2) would just be sqrt(pi)/e which i think is even cooler
convert cosine to sum of exponentials, complete square, Gaussian integral is root pi, can do it in your head in a few minutes, under a minute if you're confident, and almost instantly if you've seen a few of these. Cool to see feynman's technique at work though, great video!
let's make it e^-sx² cos(5x) x²=t, dt=2√t dt e^-st cos(5√t)2√t dt This is just the Laplace transform of 2cos(5√t)√t Find the Laplace transform and put s=1 Use cosx expansion
the sinx over exp x^2 when x goes to infinity needs a bit more rigor when calculating, you can't just say it's a finite number on the nominator (max +1 or min -1) because the lim of the sin function when x goes to infinity doesn't exist. I believe one way to alleviate this, is by using the "sandwitch" theorem; wikipedia -> Squeeze_theorem
Sure but I think it's safe to assume that if the viewer understands Feynman integration, they also know (or intuitively understand, at the very least) why the expression evaluates to zero at inf
You can also notice that the function is even and replace the integral with half the integral from -inf to inf. Then you break up the cosine into two complex exponentials, separate into two integrals. For each one you can complete the square in the exponent and reduce to the integral of exp(-x^2) by shifting the variable.
I am leaving the constant to the reader f(x) = e^{-x^2) => f''(x) = xf(x) using Fourier F_w(w) = wF'(w) => F(w) = e^{-w^2} => F(5) = 1/2\int_0^infinity e^{-x^2} cos(5x) and we are done
Радмир Хуснутдинов No, not quite. It means you have to integrate two exponential functions with quadratic arguments, and to complicate it further, those arguments have complex coefficients.
@@angelmendez-rivera351 you right. But it's no poles of this functions and residue will be 0, so integral should be same as the integral along the real axis
Is it fish or alpha?
Use something like Ž than you cant mess up
alfish
Maybe *alpha fish* 😅
Fish of course
It is *a* fish
Wow... an integral question solved by partial derivatives, integration by parts, differential equations and the Gaussian Integral to top it all off. Amazing! More Feymann technique questions, please!!
You could also write the cos term as the real part of e^i5x, and then complete the square in the exponential to get the final answer. Physicists use that trick a lot in quantum field theory.
f(a) := integral from 0 to oo of exp(-x^2) cos(ax) dx
g(a) := integral from 0 to oo of exp(-x^2) sin(ax) dx
H(a) := integral from 0 to oo of exp(-x^2) exp(iax) dx
H(a) = f(a) + ig(a)
∴ f(a) = Re(H(a)) && g(a) = Im(H(a))
-------------------------------------------------------------------------------------
exp(-x^2) * exp(iax) = exp( -x^2 + iax ) = exp(-( x^2 - iax )) = exp(-( x^2 - 2(ia/2)x + (ia/2)^2 - (ia/2)^2 )) =
= exp(-( (x - ia/2)^2 + a^2/4 )) = exp( -(x - ia/2)^2 - a^2/4 ) = exp(-(x - ia/2)^2) exp(-a^2/4)
-------------------------------------------------------------------------------------
H(a) = integral from 0 to oo of exp(-(x - ia/2)^2) exp(-a^2/4) dx =
= exp(-a^2/4) integral from 0 to oo of exp(-(x - ia/2)^2) dx
-------------------------------------------------------------------------------------
i am stuck at this moment.
i tried the transformation u := x - ia/2 but i don't know what to do with the integral:
integral from -ia/2 to (oo - ia/2) of exp(-u^2) du
that has complex limits (i don't know if that is how i was supposed to set the limits of u variable either) and I am not able to split it into two integrals of real variable either.
can you give me a hint how can i proceed from here?
@@michalbotor you did all that before understanding the basic concept of substitution :)
Exp(-x^2) if multiplied by the euler's theorem would lead to addition of i in the expression whose integral in forward solving is a pain in butt (from past experiences)
So moral is to find a logical concept and think on it before just scribbling this is pro tip in competitive level prep.
Be well my friend.
Trueee
I was going to point it out as my way.
But, I guess, the hosts wants to teach the Feynman's method.
By the way, Feynman was a physicist if I remember correctly.
try y =x+-alpha*x/2
Wow! Feyman’s technique, DI method, Gaussian, ODE all in one. What else can top this? Adding a bit of FTC perhaps
It inherently involves FTC because it involves indefinite integrals.
What's the full form of ftc?
AND the chen lu
@@executorarktanis2323 Fundamental Theorem of Calculus. Which there already was plenty of, so I don’t see how OP thinks it was missing.
@@cpotisch oh thanks this brings back memories from when I was trying to learn calculus by youtube (self learnt) and didn't know the terms thanks for explaining it now since now I have more broad understanding than what I did 3 months ago
This is the coolest thing I watched today
The coolest thing so far
That's insane!!!!!!!!!!!!!!!!!!!! I love it.
It makes me sad they don't teach this in my engineering courses :(
AugustoDRA : )))
I actually didn’t learn this when I was in school too. Thanks to my viewers who have suggested me this in the past. I haven a video on integral of sin(x)/x and that’s the first time I did Feynman’s technique.
It's covered in measure theory (math majors only) as one of the conditions to use the theorem is to find a L¹ function such that |d/da f(x,a)| ≤g(x) for almost all x.
L¹ = set of functions with finite Lebesgue integral (not ±∞)
If you’re sad about that, you don’t belong in engineering.
arcane mathematical techniques are nothing but a tool to an engineer, the primary of objective of an engineer is the creative process of ideating new machine designs, and this on its own is a massively difficult issue that takes enormous creative power.
If you’re focusing on learning esoteric integration techniques, you aren’t focusing on engineering.
I bet you aren’t an engineer now.
@@maalikserebryakov hahaha, you hit the nail on the head.
@@GusTheWolfgang Was he right?
If nothing works to solve a integral
Then *feynman technique* would work😉
BTW in the description of book, your name was also there 😁
Chirayu Jain yup! I gave a review of the book : )))
Not that much... Sometimes we need to use complex analysis which includes residue theorem or Cauchy's Theorem
I really wish youtube existed when I was studying mathematics. The potential to be educated in advanced topics without paying a hefty fee for university tuition will hopefully change this world for the better.
I found the book in college that Feynman learned this trick from, it's Advanced Calculus By Frederick Shenstone Woods · 1926.
Wow, that's nice
The channel name is blackpenredpen but you also use blue pen
🫤🤡
Sometimes, a lot of integral practices makes me to say Instagram as Integram
lol!
Instagram is the culprit
I remember this method, because in the video contest I did the integral of (e^-(x^2))*cos(2x) from 0 to infinity. BTW whenever I see e^(-x^2), I always think about feynman technique.
Chirayu Jain
Oh yea you did. And you did a great job on that. : )
I changed my profile picture recently
@@mariomario-ih6mn Me too
@@jumpman3773 Hi
Very nice use of Feynman’s technique. I’m getting the book rn!
Very nice!! Thanks.
My initial intuition was to use Feynman to get rid of the exponential term, because if you can get rid of that, trig functions are easy. The thing I didn't think through was the limits of integration: a trig function has no limit at infinity. So quite counterintuitively, it was the cosine that was going to be the troublesome element in all this, while the exponential term was what made the thing solvable.
I really enjoy your enthusiasm while explaining things :)
Thank you for the videos and please, never lose the energy, liveliness, and passion that you have now. Very nice!
Maths with you are wounderfull, thanks
As a teacher, I loved you saying "negative fish" and will use that in future. Cheers, always good to watch your videos too.
It's a bit crazy to call that the Feynmann technique. It goes back to Leibniz and it"s just deriving an integral depending on a parameter. Which by the way demands justification (either uniform convergence or dominated convergence). And in order to make this work you have to be extremely lucky and have a good intuition because you need 1) to find the right parametrization (here it's pretty obvious) ; 2) to be able to integrate the partial derivative for each value of the parameter (which is most of the time not possible) 3) to end up with a differential equation which you can solve (which is most of the time impossible), 4) to be able to compute a special value (here you need to know the value of the Gaussian integral, which is in itself tricky). So, I'd say it's a nice trick when it works but doesn"t qualify as a method...
It looks like the this problem was purposely designed to arrive at an aesthetically pleasing solution. (Given all the justifications/special circumstances/restrictions you mentioned)
@@Hmmmmmm487Feynman learnt this method in a random book during his undergrad and he famously showed off to basically everyone that he could solve otherwise very hard integrals.
Makes me a bit mad when people call it Feynman's technique. The guy did a lot of good things, but this one has nothing to do with him. They're basically saying that only an American in the middle of 20th century could come up with such idea... What did people all over the world do before that, when calculus was already so advanced, and things like FT and others were well known...
Congrats on 400K subscribers!!!
JZ Animates thank you!!
When you set alpha equal to sqrt(2 - 4ln(2)), you get sqrt(pi / e) for the answer. Pure beauty indeed.
All the feynman's techniques are UNIQUE 👍👍
I am agreeing that Feynman's technique is having a good strong hold in solving exponential integrals...but rather than complicating we could have solved it by manipulating "cos(5x)" as (e^5ix + e^-5ix)..it also saves the time...
I'm halfway through algebra 1, and yet somehow I understand and enjoy most of these videos. You and other channels like you (e.g. Mathologer) make this stuff really accessible, and importantly, fun.
(Not to say I don't enjoy my algebra 1 class!)
A great teacher is everything, right?
Well I must say ty to you Mr. @blackpenredpen . Thanks to your videos I finished Differential Equations with a B. It was on of my last 2 math classes for my mathematics BS
Nice! I am very glad to hear! : )
Feymann's Technique + Differential Equation
Mokou Fujiwara yes. And Chen lu!
I don't have an advanced level of English but that's one of a lot of thing that I love Maths, it's an universal language and your passion in every video is the thing because of I'm still here.
Imagine! If I can understand you and I don't speak English fluently, you're MORE THAN AMAZING.
Lots of love from Mexicoooo ꒰⑅ᵕ༚ᵕ꒱˖♡
This is great! Thank you! Richard Feynman really was a genius!
I've read some of the book's reviews and it looks awesome. I might pick one soon, the applications and integration techniques look interesting
This is such an elegant proof. Really impressive.
DAYUM, that's one of the most elegant solutions I've ever seen! Why none of my professors was teaching this when I was studying?
POV: you can't sleep now, there are monsters nearby 7:36
Excellent explanation. So brilliantly explained. Thanks a million.
π and e in a same expression is always beautiful
That was a wild ride!
I wish I understood. Someday, maybe. Man that's orders of magnitude beyond what I can comprehend at the moment.
These integrals show up quite often in quantum mechanics.
10:54 I was thinking he would let it equal i
If there is anything I do not want to forget from my school days is it calculus. Such a beautiful form of math.
Beautiful. Thank you so much.
11:20 ...............the sentence is veryyy TRUE indeed !!!!
Haven't seen a video for long time wich made me so happy :)
you can also use a Fourier transform
I notice the *Feynman' technique* (aka. _Leibniz Integral Rule_ ) depends basically upon parameterizing the parts expansion here; its the _by-parts_ part that gives it the power in my opinion for what its worth!
This was a unique derivation technique. Thank you for sharing.
A beautiful technique explained beautifully!!
You didn't got views but all you got is alots of love from the lover of mathematics
Because the function is even, you can take the integral from -infinity to infinity and then that would double your answer so the final answer (given alpha = 2) would just be sqrt(pi)/e which i think is even cooler
I love this channel!
Thanks!
I love all your videos, they are hearwarming. Thank you so much !
“This is very very nice ! “
I like it. I love your enthusiasm too.
this made my day
I don't understand it, but I can tell from his excitement that this is some pretty profound shit right here.
A fun trick would also be using the fourier tramsform of the bell curve
Such an elegant and clever integration technique. Bravo to Feynman and to you, of course. Very cool indeed.
Bravo to Feynman? For appropriating an integration technique known to Leibniz around 300 years earlier?
@@epicmarschmallow5049 ?
"fish over 2" - said like a true mathematician...
11:38
I love how satisfied he looked after all that he did.
int lnx/(1+e^x) from 0 to infinity
convert cosine to sum of exponentials, complete square, Gaussian integral is root pi, can do it in your head in a few minutes, under a minute if you're confident, and almost instantly if you've seen a few of these. Cool to see feynman's technique at work though, great video!
Can be done directly using chain rule of differentiation
Apparently that’s why it’s “Feynman’s”.
Universities didn’t teach it. He learned it from an obscure textbook.
Hopefully more are teaching it today.
Beautiful!
One of the best crossover episodes ever
8:32 Casually growls at whiteboard.
Even more beautiful: Since e^(-x^2) cos(2x) is an even function, the integral from -inf to inf just becomes sqrt(pi)/e.
Thank you guy.
This is one of the most beautiful videos I have seen. ¡Very complete and engaging explanation!
No fish were harmed in the making of this video
Feynman is so fucking genius
Absolutely elegant
200 integrals in 1 take world record is going to be broken by me this week.
KingArthur nice!! Send me the link once you have it done!
@@blackpenredpen I'll send it on Twitter and I'll see if I can livestream it
Graet integral! Feynman is a genius
Very nicely done! Thanks!
This is an amazing question for Calc 2.
let's make it
e^-sx² cos(5x)
x²=t, dt=2√t dt
e^-st cos(5√t)2√t dt
This is just the Laplace transform of 2cos(5√t)√t
Find the Laplace transform and put s=1
Use cosx expansion
lovely video, it's this that makes me love calculus
Chen Lu is cool, Feynman's technique is also cool, and when they are combined by you, it is cool cubed 😎
Cool^4 due to the Gaussian integral
Hahhaha thanks!!!
the sinx over exp x^2 when x goes to infinity needs a bit more rigor when calculating, you can't just say it's a finite number on the nominator (max +1 or min -1) because the lim of the sin function when x goes to infinity doesn't exist. I believe one way to alleviate this, is by using the "sandwitch" theorem; wikipedia -> Squeeze_theorem
Sure but I think it's safe to assume that if the viewer understands Feynman integration, they also know (or intuitively understand, at the very least) why the expression evaluates to zero at inf
You can also notice that the function is even and replace the integral with half the integral from -inf to inf.
Then you break up the cosine into two complex exponentials, separate into two integrals. For each one you can complete the square in the exponent and reduce to the integral of exp(-x^2) by shifting the variable.
Niceee 🤤
That was an experience. What a crazy and amazing technique
This is a direct application of the Leibniz integral rule. Feynman may have rediscovered it by himself, but it is more a trick since it is a theorem.
Can't say I understand, but I do agree: it's very nice!
👀great work sir
blackpenbluepenredpen
very good proof amazing use of differential equations
Just watch this impressive Math channel th-cam.com/channels/ZDkxpcvd-T1uR65Feuj5Yg.html
Feynman was just a genius!
Awesome vid, really enjoyed it!!!!!!
One can also observe that f(\alpha) is (up to a constant factor) just the Fourier transform of e^{-x^2}.
I am leaving the constant to the reader f(x) = e^{-x^2) => f''(x) = xf(x) using Fourier F_w(w) = wF'(w) => F(w) = e^{-w^2} => F(5) = 1/2\int_0^infinity e^{-x^2} cos(5x) and we are done
Nice don't remember running across the Feynman technique before.
Pure black art; I had to watch twice.
Very nice!
Thanks. I just ordered amazon unlimited to gain unlimited access to the advanced calculus book.
He forgot to negate the second integral... integration by parts is uv MINUS the integral of vdu
Gg blackpenredpen
This is Leibnitz' rule for differentiating integrals
beautiful, thank you
5:48 fish😂😂
such a beautiful video thanks
I would like to solve this another way. Cos(5x)=Re(e^(i5x)). Then it's really easy to take this integral.
Радмир Хуснутдинов No, not quite. It means you have to integrate two exponential functions with quadratic arguments, and to complicate it further, those arguments have complex coefficients.
@@angelmendez-rivera351 you right. But it's no poles of this functions and residue will be 0, so integral should be same as the integral along the real axis