Only FEYNMAN'S TRICKS can help solve this TERRIFYING INTEGRAL
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Nothing makes my morning more than the words: TERRIFYING INTEGRAL. But then, when that pesky π shows up, my day is complete.
Thank you, professor.
1:17 Yeah, I spotted the mistake right away when you wrote "I(3,4) = goal". So I paused the video and looked through all the comments. Since nobody had mentioned it, I assumed-correctly-that you'd fix it at some later time. Which was around 6:03. But then, while you correctly wrote t = b/a, when reading it, you said t = a/b. Now, after many years of catching such errors, I've become somewhat more confident in my own math skills. So thanks, I guess?
As a math student I watched all of your videos and last night i saw you in my dream that you has taught me math 😅
Distance learning
😅
I think he should charge you for subconscious learning.
@@pradumnyadav3950 Epic comment 😁💯
@@sanjaysurya6840 Yeah - the family of YT comments gaining likes "i've been working over it with Feynmann at Alamos", "My aunt was calculating this for the military ballistic project untill feynmann came" or "i was the one who was working on it with Ramanujan... - thanks for reviving the memories" :-D
If you see arctan(z), just use the integral representation
arctan(z) = z * int_0^1 dt/(1 + z^2 * t^2)
Here z = 3/sqrt(16 + x^2).
Integral over x is then trivial:
int_0^infty dx/(16 + x^2 + 9 t^2) = 0.5 * pi/sqrt(16 + 9 t^2).
The integral over t is elementary:
int_0^1 dt/sqrt(16 + 9 t^2) = (1/3) * asinh(3/4).
Combining all together, we get
0.5 * pi * asinh(3/4).
No need for the Feynman trick 😜
Feynman trick is always neet and creative.
But yeah, there's not only one way evaluationg this.
i did the exact same thing
Your approach is not so far from Leibniz rule
My approach involves only basic integration techniques
such as substitution,integration by parts , linearity of integral , additive property with respect to interval of integration
and some log properties
There are always many methods. You could even use Fubini on this one.
@@SimsHacks but my approach is available also for beginner , for calculus II student
I find it interesting that this sort of integral often results in solutions containing ln(2) and Pi
Amazing video! Thanks Prof. Penn!
I understood nothing and I loved it. That means I need to watch and read more on the subject.
always thankfull
Exciting problem and an amazing solution
Nice clickbait but it is possible without Leibniz rule
1. Euler substitution (scaled if necessary)
2. Integration by parts with arctan as part to be differentiated
3. Partial fraction of rational factor
(This can be tricky byt fortunately we have convergent integral)
Now we use substitution to get interval of integration to be [0;1]
(We may also need to use additive property with respect to interval of integration)
then we use geometric series expansion
To use series expansion interval of integration must be subinterval of interval [0,1]
and we must be careful to get convergent integrals in each step (to be sure that none of this situations like cancelations of infinity happens etc)
There is even no need for series expansion in approach wich i gave
because problematic integrals will cancel out and finally i have got
ln(2)*(arctan(1/2)+arctan(2))
Use substitution sqrt(16+x^2) = t + x
Scale interval of integration by substitution t = 4y
Do integration by parts with u = arctan(3/2*y/(y^2+1)) and dv = 1/y dy
We receive integral 6Int((y^2-1)/(4y^4+17y^2+4)ln(y),y=0..1)
Now we can use partial fraction of this rational factor (y^2-1)/(4y^4+17y^2+4)
but once we factor denominator we can guess this decomposition without undetermined coefficients because
4y^4+17y^2+4 = (4y^2+1)(y^2+4)
and (4y^2+1) - (y^2+4) = 3y^2 - 3
so we will have two integrals to evaluate
2(Int(ln(y)/(y^2+4),y=0..1) - Int(ln(y)/(4y^2+1),y=0..1))
Fortunatelly both integrals are convergent so we will not get cancellation of infinity or other indeterminated forms
Now we can play with simple substitutions , log properties, and additive property of integral with respect to interval of integration
Can you say something about which types of integrals are (potentially at least) solveable, and which ones aren't no matter what advanced methods and clever tricks you pull out of your sleeve?
What are the limits?
The length of curves will often lead to some evil integrals - for instance just simple ellipses ( but for the special case of circles that bit is masked in the various infinite series for PI ).
And some classes of functions are "simply" defined by integrals of other functions, so you can only approximate them.
To sharpen your query, maybe you're looking for criteria for indefinite integrals to be expressed in terms of elementary functions? Nice question!
@@abbeleon Yes, thank you - I think (?) that was what I was trying to say 😂
the usual mandatory comment that the last integral can also and probably more cleanly be solved by substituting t=sinh(α) to get I(t)=π/2*arsinh(t)
I did it with a direct approach of feynmann technique, for some reason got pi/8 arctan(3/4).
edit: oh no! I did a fatal mistake, I forgot the square root! yeah I think another hyperbolic trig sub should do the trick, aprox same result.
astounding work as always Doctor penn.
I seriously love these thumbnails :D
Feynman‘s technique is so brilliant. The integral looks so difficult but it’s doable- great Michael 👍
Great vid michael, as always!
I want to see chalk’s monster truck garage!
Nice video.... I always like seeing new opportunities to use Feynman's trick. Two comments: I think your initial 2-parameter approach really obscured what was going on. I just replaced the 3 with A (so that we're looking for I(3)) and left the 16 in as a constant. Secondly, you used two more u-subs than I think was necessary (one to go to y and one to go back) which also felt unnecessarily confusing.
Yeah, I think he only needed one parameter for this integral.
It is easier to put b=3 and differentiate by b. We will have dI/db=Int(dx/(x^2+16+b^2))(from x=0 to inf, which is equal (pi/2)/sqrt(16+b^2). The Integrating of this formula leads to I(b)=pi/2*Asinh(b/4) and so on
11:17 why not make it a definite intergral instead of later fiddling with the integration constant?
Just to avoid introducing a new integration variable (i.e. for didactic reasons)?
Thumbnail 10/10
1:42 Hyperbolic substitution "x = 4 * sinh(t)" works just as well, and leads to (slightly) less terms.
I wish people would stop calling it the “Feynman trick”. Feynman didn’t invent it, he read it in a calculus book - it was already an established method of solving definite integrals. Feynman just used it famously to solve many integrals (eg at Los Alamos) so it became popularised by him.
It should be called “Integrating by differentiating under the integral” instead.
In a different lane I'm dealing a lot with Feynman (kac) "tricks" aswell this semester on stochastic differential equations in continous time arbitrage theory.
Nice video thanks professor i learned a lot from your videos
Is this "Feynman trick" something developed by particle physicist/quantum mechanics expert, Richard Feynman?
Yes!
This trick had been developed by Leibniz more than 200 years before Feynman was born
Leibniz created the Rule but Feynman made it popular
@@zeravam and this is closer to the truth
@@zeravam but,un my opinion, the proof needs Lebesgue dominated's convergence theorem ofLebesgue 's integral theory 40years before Feynmann!
Curious that it's now called "Feynman's trick." Feynman said it came from studying a copy of "Advanced Calculus" by Fredrick S. Woods, which his high school math teacher gave him, telling him to go sit in the back of the room and to say nothing more in class until he understood the entire text. My dad also had a copy of that text, and sure enough, Woods covers both differentiating and integrating under the integral. Too bad I became acquainted with it in my 30's rather than in high school, but then I'm certainly no Feynman!
Must have been used for a long time as many larger tables of integrals list the results of differentiation and integration under the integral of certain functions contiguously!
Elegant ✨
After all that it simplifies to pi/2*ln(2) 😂
Illusions, Michael
Love and respect from Bangladesh ❤
Good video, you could have used the leibniz rule right at the original integral to get the same equations but with less steps
Nice video!
smol math man lol.
The solution is too long. Mr. Feynman would be disappointed. The solution is in two lines!
I(a,b) is an integral on the board. I(a,0)=0.
dI(a,b)/db= ∫(from 0 to ∞) dx/[x^2+(a^2+b^2)]= ( π/2)/√(a^2 +b^2).
I(a,b)= ( π/2)*∫(0 to b) db/√(a^2 +b^2)= (π/2)*{ln[b+ √(a^2 +b^2)] - ln a}= (π/2)*ln[b/a+ √(1 +(b/a)^2)] .
👏👏👏👏👏
So when you say "only" Feynman's trick can solve the integral, is that literally true? Maybe there are other techniques, but Feynman's trick happens to be the handiest? I don't know; I'm just asking. As an engineer, I almost always end up using numerical methods. Of course, before doing that, I do try to look it up online (earlier I used to use dead tree handbooks, references, and encyclopedias). I also try to solve it using symbolic algebra software. If those fail, there's no point in trying to solve it by hand.
Thanks for, really, it's an imaginary method.
Nice video, but man were you filming this on the surface of the sun?
cool
👏👏👏👏👏👏👏
I(4,3)
What would have happend if the two parameters a and b could not be substituted by a unic parameter t?
Feynmann :)
Feynman's trick as in guessing the solution?
This integral couldn’t solve Mathematica but numerically
The solution is in front of you
Anyone notice Mr Penn kinda looks like an older Jesse Pinkman?
Clickbaited math video. You love to see it LMAO
My calculator solved this with only one parameter.
Dude, please don’t jump on the “idiotic facial expression in the thumbnail” bandwagon. Unless this was a onetime joke 😂
it's always a joke when I do it. I do what plays nice with the algorithm even if I don't really like it myself. You'll never see one like this that doesn't have the real problem on it :)
-Stephanie
MP Producer, Editor
Hmm..., I'm not a fan of this Feynman tricks. It seems to work just for definite integrals.
The simplification works, because you "intermediately" evaluate the integral, by applying the limits. I've seen better tricks than this.
Every integration trick has its uses. This one is pretty fun though perhaps it should be called the Leibniz rule (less catchy name)
I'm a fan if we verify that it may be used. Putting the derivative inside the integral isn't obvious and doesn't always work.
@@SimsHacks I think putting the derivitive inside the integral may require the dominated convergence theorem.
@@sujitsivadanam You're absolutely right :)
Here indefinite integral involves complex numbers and dilog function
PLEASE don't go down the embarrassing pull-a-stupid-face thumbnail route. You're one of the last grown-up channels on CZcams.
It’s funny lol
I thought it was satire LOL
@@ethanlipson1637 You do things ironiclly enough, they become part of you
Sorry but your are very far from being rigorous in the most important steps such as :
-You don't even mention that the conditions to take derivative under the integral sign are verified .