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.
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 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!
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...
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 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!
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
Just saw the Gaussian integral=sqrt(pi)/2 half an hour ago in lecture hall. I didn’t know where it came from while my prof was explaining Laplace Transform of t^(-1/2). And now here… What a small world of Mathematics !
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!)
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!
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
Thanks for remind me of a feeling in Maths again. It is 20+ years ago since my college course. Actually, I love Maths, in the past It is my recreation. But living in real life, I have no time to solve the challenge Maths problem. Life is hard.
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
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
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
hlo there ... i hd seen yewr question last night nd started solving right after .. well i wanna tell yew that i have expand COS( AX) to taylor series form :- summation(n€ (0 to infinity)) (-1)^n (A)^2n X ^(2n) / (2n)! then substitute the above into the integral and then bring out the summation [along with the constant term (-1) ^n A^(2n) / (2n) ! ] nd move the integral inside such that we will get integral (0 to infinity) x^(2n) e^((-x* x) i.e x square) dx now using gamma function definetion this integrand would be written like this 1/2 times gamma function (n+ 1/2 ) times gamma function (1/2) which is (pi)^(1/2) now (n+ 1/2 ) ! which can further be written like (n+1/2) ! = [(2n)! / 4^(n) (n) ! ] now substituting this result back into the summation we will see that the result we get is in the form 0.5 times gamma function(0.5) times summation (n €( 0 to infinity)) C^(n) / (n) ! where C = -A^(2) / 4 finally e^C = summation (n€ (0 to infinity) ) C^(n) / n! so the net result for this kinda integral will be {0.5 times gamma function(0.5) times e^ (- A*A /4 ) }
OH MAH GAWD! I tried to integrate e^x*cosx using this method when I was 15! Holy shit! So this is differentiating under the integral? I did it differently - I simply took the derivative of e^x*sinx to get e^x + e^x*cosx and by integrating we get integral of e^x*cosx = e^x*sinx minus integral of e^x which is just e^x*(sinx - 1)... So for this problem (integral of e^-x^2*cos5x) we take the derivative of e^-x^2*sin5x and rearrange the expression so that e^-x^2*cos5x is on the LHS and we can simplify to get the RHS.
Using the property that: sinh(x) = -i*sin(ix) We get that: sin(ix) = i*sinh(x) Putting that into the formula we now have: i*sinh(x) = 2 Shuffling everything to the right side we get: X = arcsinh(-2*i) If we now use the logarithm form of the hyperbolic sine we get: X = ln(-2*i + i*sqrt(3)) Taking out the i out of the ln we get: X = ln(sqrt(3) - 2) + ln(i) Using the formula: e^(i*pi) = -1 We can square both sides to get: e^(i*pi/2) = i Now take the natural logarithm on both side and you have something in function of ln(i), and the original solution is simplified to: X = Ln(sqrt(3) - 2) + (i*pi)/2
@@ferhatakbulut6572 I did it way differently. sin(ix)=2 (e^(ix)(i)-e^(-ix)(i))/2i=2 e^-x-e^x=4i 1-(e^x)^2=4i(e^x) -(e^x)^2-4i(e^x)+1=0 e=^x=(4i+sqrt(-16+4))/2i e=^x=(4i-sqrt(-16+4))/2i e^x=2+sqrt(3) e^x=2-sqrt(3) x=ln(2+sqrt(3)) x=ln(2-sqrt(3))
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?
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
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 🤤
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 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
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.
As a teacher, I loved you saying "negative fish" and will use that in future. Cheers, always good to watch your videos too.
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!
Very nice use of Feynman’s technique. I’m getting the book rn!
Very nice!! Thanks.
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...
When you set alpha equal to sqrt(2 - 4ln(2)), you get sqrt(pi / e) for the answer. Pure beauty indeed.
Maths with you are wounderfull, thanks
This is the hardest integral I’ve gotten right on my own! So proud of myself
Integrate (e^x)(x^x)(2+logx) wrt to x
Please someone do this
x^x oh no
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 great! Thank you! Richard Feynman really was a genius!
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...
All the feynman's techniques are UNIQUE 👍👍
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 such an elegant proof. Really impressive.
Congrats on 400K subscribers!!!
JZ Animates thank you!!
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?
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!
Excellent explanation. So brilliantly explained. Thanks a million.
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! : )
Just saw the Gaussian integral=sqrt(pi)/2 half an hour ago in lecture hall. I didn’t know where it came from while my prof was explaining Laplace Transform of t^(-1/2). And now here… What a small world of Mathematics !
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?
π and e in a same expression is always beautiful
Sometimes, a lot of integral practices makes me to say Instagram as Integram
lol!
Instagram is the culprit
If there is anything I do not want to forget from my school days is it calculus. Such a beautiful form of math.
Feymann's Technique + Differential Equation
Mokou Fujiwara yes. And Chen lu!
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
11:20 ...............the sentence is veryyy TRUE indeed !!!!
The channel name is blackpenredpen but you also use blue pen
🫤🤡
AND at no extra charge! 😊
A beautiful technique explained beautifully!!
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!
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 ?
You didn't got views but all you got is alots of love from the lover of mathematics
I wish I understood. Someday, maybe. Man that's orders of magnitude beyond what I can comprehend at the moment.
This was a unique derivation technique. Thank you for sharing.
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.
Amazing¡¡¡¡ you must record more videos about this topic¡¡¡¡¡¡¡
Even more beautiful: Since e^(-x^2) cos(2x) is an even function, the integral from -inf to inf just becomes sqrt(pi)/e.
10:54 I was thinking he would let it equal i
I love this channel!
Thanks!
Beautiful. Thank you so much.
I like it. I love your enthusiasm too.
I very much enjoy watching the derivations, even though I know I'd probably never be able to figure it out myself.
A fun trick would also be using the fourier tramsform of the bell curve
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
“This is very very nice ! “
5:48 fish😂😂
I love all your videos, they are hearwarming. Thank you so much !
11:38
I love how satisfied he looked after all that he did.
University teacher: ur exam has integrals
The intégral during the exam:
One of the best crossover episodes ever
That was beautiful man just phenomenal
This is an amazing question for Calc 2.
Thanks for remind me of a feeling in Maths again. It is 20+ years ago since my college course. Actually, I love Maths, in the past It is my recreation. But living in real life, I have no time to solve the challenge Maths problem. Life is hard.
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!!!
Nice and clean trick ! Thank you.
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
One can also observe that f(\alpha) is (up to a constant factor) just the Fourier transform of e^{-x^2}.
Can't say I understand, but I do agree: it's very nice!
These integrals show up quite often in quantum mechanics.
Absolutely elegant
This is one of the most beautiful videos I have seen. ¡Very complete and engaging explanation!
you can also use the taylor series for cos(5x) and use the gamma function
Very nicely done! Thanks!
Nice don't remember running across the Feynman technique before.
you can also use a Fourier transform
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
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
That was an experience. What a crazy and amazing technique
I don't understand it, but I can tell from his excitement that this is some pretty profound shit right here.
lovely video, it's this that makes me love calculus
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.
Wow, nice solution !!!
Beautiful explanation😀
Haven't seen a video for long time wich made me so happy :)
Great video man!
Pure black art; I had to watch twice.
this made my day
Graet integral! Feynman is a genius
Awesome vid, really enjoyed it!!!!!!
very good proof amazing use of differential equations
Just watch this impressive Math channel ruclips.net/channel/UCZDkxpcvd-T1uR65Feuj5Yg
hlo there ...
i hd seen yewr question last night nd started solving right after ..
well
i wanna tell yew that i have expand COS( AX) to taylor series form :- summation(n€ (0 to infinity)) (-1)^n (A)^2n X ^(2n) / (2n)!
then substitute the above into the integral and then bring out the summation [along with the constant term (-1) ^n A^(2n) / (2n) ! ] nd move the integral inside such that we will get
integral (0 to infinity) x^(2n) e^((-x* x) i.e x square) dx
now using gamma function definetion
this integrand would be written like this 1/2 times gamma function (n+ 1/2 ) times gamma function (1/2) which is (pi)^(1/2)
now (n+ 1/2 ) !
which can further be written like
(n+1/2) ! = [(2n)! / 4^(n) (n) ! ]
now substituting this result back into the summation we will see that the result we get is
in the form 0.5 times gamma function(0.5) times
summation (n €( 0 to infinity)) C^(n) / (n) !
where C = -A^(2) / 4
finally e^C = summation (n€ (0 to infinity) ) C^(n) / n!
so the net result for this kinda integral will be {0.5 times gamma function(0.5) times e^ (- A*A /4 ) }
OH MAH GAWD! I tried to integrate e^x*cosx using this method when I was 15! Holy shit! So this is differentiating under the integral? I did it differently -
I simply took the derivative of e^x*sinx to get e^x + e^x*cosx and by integrating we get integral of e^x*cosx = e^x*sinx minus integral of e^x which is just e^x*(sinx - 1)...
So for this problem (integral of e^-x^2*cos5x) we take the derivative of e^-x^2*sin5x and rearrange the expression so that e^-x^2*cos5x is on the LHS and we can simplify to get the RHS.
Feynman is so fucking genius
That was a wild ride!
Can be done directly using chain rule of differentiation
blackpenbluepenredpen
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
sin(ix)=2
x=?
Interesting question, I thought of it recently and want to see if I solved it.
Keep up with the amazing videos!
Using the property that:
sinh(x) = -i*sin(ix)
We get that:
sin(ix) = i*sinh(x)
Putting that into the formula we now have:
i*sinh(x) = 2
Shuffling everything to the right side we get:
X = arcsinh(-2*i)
If we now use the logarithm form of the hyperbolic sine we get:
X = ln(-2*i + i*sqrt(3))
Taking out the i out of the ln we get:
X = ln(sqrt(3) - 2) + ln(i)
Using the formula:
e^(i*pi) = -1
We can square both sides to get:
e^(i*pi/2) = i
Now take the natural logarithm on both side and you have something in function of ln(i), and the original solution is simplified to:
X = Ln(sqrt(3) - 2) + (i*pi)/2
@@ferhatakbulut6572 I did it way differently.
sin(ix)=2
(e^(ix)(i)-e^(-ix)(i))/2i=2
e^-x-e^x=4i
1-(e^x)^2=4i(e^x)
-(e^x)^2-4i(e^x)+1=0
e=^x=(4i+sqrt(-16+4))/2i e=^x=(4i-sqrt(-16+4))/2i
e^x=2+sqrt(3) e^x=2-sqrt(3)
x=ln(2+sqrt(3)) x=ln(2-sqrt(3))
@@georgeklimov3464 I did it that way too! :)