Using Feynman's technique to solve this really cool Berkeley Math Tournament integral
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- Опубликовано: 7 фев 2025
- Last time around we solved this integral using some elegant properties of the definite integral:
• Is this hard integral ...
Those elegant properties are stated and explained in the link below
• Some really nice prope...
This time, however, I couldn't resist solving it using Feynman's technique and really enjoyed solving this particular problem. I hope you enjoy the video too and remember to like and subscribe for more maths and physics content
If you are grading your hand writing with an E, I'm giving you an A for presentation.
And yes, I learned that technique in some lesson at the university, but it was only mentioned with some simple example. Years later I met it again when reading Feynman's "Surely you're joking", but you make a big show out of it. Fascinating and many thanks, your videos are always great bed time stories.
Everytime I see this channel I can only see thumbnail of feymann
😂😂😂
It's my favourite technique so I can't help it. Although I have solved this integral using just the basics in a previous video
And you might even see some DEs and thumbnails of Newton and Leibniz too 😂
I am falling in love with Feynman's Technique to be honest. Why didn't they teach me that in High School!
Every thing was very pritty but in last step I coyd't understand how to swich the integrals borders from 0-infinity to 0-1
thanks for all nice videos
He wants I(1). He knows I(0). By the Fundamental Theorem of Calculus, I(1) - I(0) = integral from 0 to 1 of I'(a) da. Use this to solve for I(1).
Heaviside cover up method works here with complex and makes way easier the partial fraction decomposition
Indeed it does
@@maths_505 love your videos, btw. Awsome content.
How do you know that the Leibinz rule can be applied to a given integral with a parameter? To do this, you first need to determine, if it converges uniformly with respect to the alpha parameter, isn’t it? Please, sorry for my possibly incorrectly formulated sentences in English. I don't know it very well, and I composed these sentences with the help of a translator.
Can you integrate the Bernoulli integral using the Feynman method?
Hello! Is there a "simple" function whose integral has a very "complex" closed form? By a simple function, I mean a function similar to the one in the video, consisting of a few elementary functions, at most rational. By a very complex closed form, I mean expressions that involve various constants, raficals, logarithms, sums and multiplications etc. (ellipitc integrals or hylergeometric functions are not allowed, solution must be close). Thanks for any insights!
Gamma and zeta function are not allowed as solutions. Elementary functions only
That's actually an interesting question
That's a nice one to solve. But your answer here is + Pi/2 ^ ln(2), but in your other video that you solved using elementary methods (no exotic Feynman technique), the same answer is negative of the above answer. Can you please explain why?
Good evening. What is the app you used for solution here?
U could have just used "The king's Property" and add both the integrals making it a simpler integration of just cotx.
Wouldn't work...try it and see for yourself
I thought perhaps you were to going to write the x in the numerator as atan(a*tan(x)) 🤔
Nah....not elegant enough
FAROUT! Sir,your handwriting is not bad; you should see mine. Even my chickens can not read it!
i don't think that's your fault though, chickens can't read much tbh
I prefer previous version
another method-
I= integral(xcotx)
apply by parts to get:
I=integral(-ln(sinx)) from x=0 to x=pi/2
Using symmetry:
I=integral(-ln(cosx)) from x=0 to x=pi/2
Adding:
2I=Integral( -ln(sin2x) + ln2) from x=0 to x=pi/2
=integral(-ln(sin2x)) + pi/2 ln2
Interestingly: integral(-ln(sin2x)) from x=0 to x=pi/2 is equal to I (can be proved by subsitution u=2x, followed by use of symmetry of sinx about x=pi/2)
2I=I+pi/2 ln2
I=pi/2 ln2
Done
I have an easy solution to this that was pointed out to me by my friend, who solved it in less than a minute after i spent 10 minutes trying to show him the power of Feynman’s technique (🫣)
Rewrite the integrand as x * cosx/sinx
Now integrate by parts, differentiate x, integrate cosx/sinx = ln(sinx)
The integral then becomes,
I = [ xln(sinx) ]_{0,π/2} - int_{0,π/2} [ ln(sinx) ] dx
The first term is obviously zero, and the second term (the integral) is just the famous Eular’s integral whose value is -π/2 ln2. So simplyfying,
I = π/2 * ln2
Sometimes we get so used to the advanced techniques that our mind misses the simplest of things 🫠🫠🫠
The first video I made on this integral solved it the way you pointed out....the video also involved the proof of Eular's log trig integral using symmetry arguments
@@maths_505 can you share the link maybe ?
ruclips.net/video/iS0rf7SchlM/видео.html