Feynman’s Technique is Unstoppable!!! | Fresnel Integrals

Today I solve the Fresnel integrals using Feynman’s integration technique. This video teaches you how to apply Feynman’s method to the fresnel integrals, leading to an amazing result.

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#maths #advancedmath #calculus #feynmantechnique #JEE #feynman #fresnel

[mohandoshi]

Awesome evaluation. Really enjoyed this video.

[Jagoalexander]

It has also just occurred to me that I didn’t explicitly state the integrals we will be solving, we are trying to solve Sin(x^2) and Cos(x^2)

[ivan-nmxn]

nice! By the way, what’s the song in the beginning?

[heyathereboi]

Hey there, loved the part where you substitute the (2t*exponential) by the derivative, really reminded me of solving for the expectation value of x^2 in the introductory QM, but in this case specifically is taking the derivative with respect to t justified, as u has a t dependence itself (u=x/t) so the derivative also acts on it, doesn't it? Otherwise - great video!

[abhinavjuly]

Nice video i knew it this prob is going to solve by fey.. method but i was stuck into some steps.you explained well now i understood where i was making mistake.😊

[NickNicholas]

When you plugged in I(0) wouldnt it also make U zero since it is a function of I?

[YaleQuatts]

I’m missing that last bit…. How did the imaginary bit (attributed to sin) become real? What happened to the “ i “ in the last step?

[philipframpton]

Can you explain 5:47, why did the derivative of the integral simplify so easily.

[christoffel]

I don’t see how the integral goes to zero when you take the infinite limit. If the i wasn’t in the exponential it definitely would, but with the i you would just get oscillations. The integral of the magnitude as far as I can tell doesn’t go to zero either.

[BoringExtrovert]

Hey man, you should give a good argument on why you only took the principle square root. Other than that good video

[marceskandar]

The root of « i »?? Is it a problem no?

[lucapolidori]

When a SIN is extended to infinity its area is equal to the COS, just looking at the waveform. Having the first peak of the curve at 0 or at 1, at the infinity doesn't change anything. Am I wrong?

[RashmiRay-cy]

Couldn't we just say I = Im \int^{\infty}_{0} dx e^-x^2/i = Im 1/2 √(pi i), (using the result for integrating a Gaussian) = √(pi)/2 Im e^(i pi/4) = √(pi)/2 sin(pi/4) = √(pi)/2 (1/√2) = √(pi/8)?

[topquark]

Nice work. By the way, "Fresnel" is pronounced "fre-NELL" not "FRES-nel"

[maths_]

Bro you totally butchered fresnels name😂