Anyone who might be sophisticated enough to do this Feynman trick would already know how to do the contour integral, which gives the result in just a couple of lines.

Amazing result, by the way, you can actually get a better one by applying the fundamental theorem of engineering saying that π=e therefore getting I=1.

9:03 Our solution for I’(a) relies on the dirichlet integral evaluating to pi/2 however that doesn’t work for a = 0 which would give \int_{0}^{\infty} sin(0x)/x dx = 0 making I’(a) discontinuous at a = 0.

Très belle démonstration de la puissance, de l'efficacité et des immenses perspectives offertes par cette méthode ingénieuse pour se sortir de situations apparemment inextricables. Merci pour votre travail

Great video Kamal! Isn't that simply the real part of the Fourier Transform of 1/(1+x^2) evaluated at 1? This Fourier Transform is well known (exp of abs) 😊

Amazing integration, just want to point out when solving a differential equation you must first find the characteristic equation for the homogeneous part 1st and then find the particular solution. In this specific case this is a strictly homogeneous equation so there is only the characteristic equation to be solved which is r^2-1=0 and hence r=1 or -1 and the form is c1e^(r1a)+c2e^(r2a). This only works when r1 and r2 are different from each other and also real. For repeated roots and complex roots the form of the solution will change. By the way r1=1 and r2=-1. Also it’s important that the solution the second order differential is not 0 because that would imply are original integral is equal to 0 even though 0 is a possible solution to the differential equation. But when graphing cos(x)/(x^2+1) clearly the area is not 0.

Another cool integral, presented by the best teacher I‘ve ever had - excellent 👌

Great, thank you. I will check out the sinx/x integral.

I really love this technique

What a crazily beautiful result!

really nice result and nice video.

Excellent work sir

Honestly seems to me like an overkill, I think it is more straightforward to just solve this using complex analysis + residue theorem (with a route around the i pole)

Great video again !!

How do we know in instances like these to make it match the bottom by multiplying by 1 and adding 0 like it sometimes seems to be necessary? Is there an alternative way that doesn't require this?

Here's what's bugging me about the derivation. Let's look again at I'(a). Before we manipulate it, it's the integral of -x sin(ax)/(x^2 + 1) dx. Now, if we take the limit as a --> 0, it looks like we get 0, since sin(0) = 0. But in your derivation, you get I'(0) = -pi/2.

I was wondering why at 4:58, you can take the constant "a" inside the differential? I've never seen this done before so any explanation for why this works would be much appreciated. Thanks!

At 2:35 I’(0)=0 because the integrand is 0. But at 5:56, I’(0) is -pi/2. Please clarify.

It can be done easily using cauchy residue theorem

Why can't I deal with I(a)=I(-a) that means c1=c2???? I(a)=pi/2(e^a+e(-a))->I(1)=pi/2(e+1/e)

Good work

I will try this question by solving INT exp(iax)/(1+x²).

8:30 u can't take a=0 because previously u divided by a. Notice if u take the expression 2:48 I'(0)=0

I don’t think you can switch the differential and the integral because the integral of the derivative doesn’t absolutely converge. In order to do that you would need to do an integration by parts to increase the degree of the denominator thus when you differentiate you get something absolutely integrable

By the definition of I we should have I(-a)=I(a) which is not satisfied by I(a)=pi/2.exp(-a).

If you apply the fundamental theorem of engineering, the final answer is 1, as pi = e

why is a partial derivative when it is after integral sign but normal derivative when it is before integral sign? what is the difference in which order we put these operators?

After watching this I wonder how many people could actually replicate the solution on their own. I think not many.

At 1:10 you show I(a) equal to an integral (from 0 to infinity) with "a" plugged in, but you dropped the factor of 2, which you showed immediately above. As this percolates down to the end, this would make the final solution 2*pi/e rather than pi/e. Right?

is there not a contradiction between your statement about I'(a)=-pi/2 at 8:55 and your earlier initial statement about I'(a) earlier at 2:30, which would surely collapse to 0 when a=0?

Cool! Now, which definite integral gives pi e?

Can you please explain how I(a) = C1•e^(-a) + C2•e^a ?

not veryfing that we can change integral and derivative - wrong

8:57 okay nice nice very nice😂 Btw that was very cool integral

7:08 That is indeed a solution to the differential equation, but how do you know that it's the correct one? Put another way: the desired function I satisfies I''(a) = I(a), but that does not mean that every function satisfying this condition is necessarily I. Here, you would also have to argue that NO function other than the one you give solves I''(a) = I(a). Can this be done? If so, how?

Nicely done! But what will learning these integrals of such specific mathematical operators lead us to in real world? Request you to corelate integrals with real world applications. That would make this video even more explosive!

That's a fantastic computation and an amazing result. I wonder if this result could be used to prove something about the number π/e, for example, if it's transcendental or not.

Please which app are u using to solve this problem

Maths 505 could you please prove that integral from 0 to inf cos(mx)/(1+x^2)dx = integral from 0 to inf xsin(mx)/(1+x^2)dx

Could someone explain how did he solve the differential equation part ? How to solve f"(x)=f(x)

Can anyone suggest me a book to start with feynmanns integrals???

Любопытно! 👍

this was a journey

Teacher: solve this one integral. Student: casually solves infinitely many integrals.

nice

I am in my first year of a bachelor's degree in mathematics and I love these integrals but I don't like the way that Feynman do these. Not for me, looks like a physicist thing.

Use contour integration and solve it 1/2 real part of e^iz/z2+1

If there is another high-intelligence species in this universe, do they know Feynman's technique too?

If I(a) = I' '(a), do you need all that extra working to arrive at e^a as the solution? It's basically the definition of e^a, right?

Integration of functiins with compkex variable is much more straigtforward

The technique is elegant, no shadows of doubt about it. But would it be possible to identify cases where that technique would be the best approach or something like we do when we learn other integration techniques such as integration by parts, variable substitution, trigonometric integration?

Incredible hard question. Even you know Feynman method, you need to have knowledge of PDE to solve this question.

I will solve it by Residue Theorem

Except the switch up of the integration and differentiation was not justified, and actually the mentioned "trickery" was done to avoid getting a nonconvergent integral without mentioning why one does that

good day. You are multipying by X both of numerator and denominator and in the same time calculating integral from "0" to infinity which means that "0" is included....can you?

. |=|cos x{dx/(1+x^2)}= =|cos x d(arctan x)= =arctan x.cos x-|arctan x (-sin x dx)= =arctan x.cos x+|arctan x.d(cos x)= =2 arctan x.cos x-|cos x d(arctan x). Maka 2|cos x dx/(1+x^2)= 2 arctan x.cos x Jadi |=cos x.arctan x+C

He forgot the dx at the end

That is actually mad

1. This integral can be found with residues in one line. 2. If you do not know complex integration, you may know Fourier transform. The parametric integral is pi/2 exp(-|a|) by taking the Fourier transform of a symmetric decaying exponential. Your solution ignores the case a

WOW! pi / e !!!!!

If you had something like "cosaxsinax", then would you treat it as "cos(ax sin (a)x)", "cos(a) x sin(ax)", "cos(a) sin(a) x^2" etc. or something else? Even if here one can 'guess' that you secretly mean that "sin ax = sin(ax)", it's often problematic, if people just skip brackets like this and we are working with many different variables or expressions that are multiplied together... I never understood why some physicists purposefully keep skipping brackets to create ambiguous expressions and sometimes even technically invalidate correctness of what they do, as if they loved to see the world BURN! :D :D Edit: 5:15 why not just "dax" instead of d(ax)? Let's be consistent in skipping the brackets 🙂

Very cumbersome to say the least.

Great trickery, but you really need to work on the readability your π

how is x^2+1 an even function? if x=2, result is 5.