a neat parameterized integral identity

[@bernieg5874]

With problems so un-natural and surprising as this it would be nice to have a moment to explore how it could arise from some situation or otherwise be motivated (e.g. in puzzle construction)

[@get2113]

I understand what you say. I made a similar comment a couple days ago. But in professor's defense, he is a teacher of methods and has a lot of video hours to fill with content. I think the pdf in the problem is a t on one degree of freedom, for whatever that's work. Nice channel prof.

[@MacHooolahan]

Agree - I was waiting for something "revelatory"... Oh well - just problem practice, I guess?

[@colbyforfun8028]

The real goal was the techniques we made along the way

[@two697]

I did a similar method but was able to arrive at 14:00 with a few less steps. If you let I=rcos(theta) and J=rcos(theta). dI/dx=-J and dJ/dx=I-1/x. r^2=I^2+J^2. d(r^2)/dx=2(I*dI/dx+J*dJ/dx)=-2J/x. I then spent a while trying to evaluate J exactly before checking the video

[@digxx]

You can also use some type of fubinis theorem. r^2=Int_0^\infty dt1 Int_0^\infty dt2 exp(-x*(t1+t2))/( (t1^2+1)(t2^2+1) )*(1+t1*t2). Then substituting t2=t-t1 gives r^2=Int_0^\infty dt1 Int_t1^\infty dt exp(-x*t)/( (t1^2+1)((t-t1)^2+1) )*(1+t1*(t-t1)) = Int_0^\infty dt exp(-x*t) Int_0^t dt1 (1+t1*(t-t1))/( (t1^2+1)((t-t1)^2+1) ). The inner integral looks complicated, but it is actually trivial after partial fraction decomposition giving ln(1+t^2)/t.

[@videolome]

Very nice video. Thank you.

[@sh335-s7x]

Love it!

[@tahirimathscienceonlinetea4273]

It's nice solution. In fact, this one is more beneficial thanks

[@demenion3521]

i think this problem can be much more easily solved by using pythagoras with r²=(rcos(θ))²+(rsin(θ))² and then smashing the integrals together. instead of generating some fairly complex differential equations

[@tronix2125]

How does one square an integral? Is integral multiplication even a defined thing?

[@demenion3521]

@@tronix2125 first of all, a definite integral is in this case just s real number (like π/4 or stuff like that), so of course you can multiply them. Then you usually want to rename the integration variable in one of the integrals to avoid confusion. Since both integrals are just numbers, you can then use the linearity of integrals to put one integral inside the other, just like you can pull constant factors out. Since the integration variables are different, you can then also put the integrand of the outer integral inside the inner integral to obtain a double integral (which is still factorable). Then you often use the transformation theorem to go to polar coordinates or intertwine the coordinates in some other way to get something that you can integrate better.

[@demenion3521]

@@tronix2125 as an example I can give you one of the standard proves of the gaussian integral: Let's say we want to know int(exp(-x²)dx, -inf, inf)=a. I'll use the second and third argument as bounds of integration and I'll call the desired result a. It turns out that calculating a² is easier, so let's do that. a² = int(exp(-x²)dx, -inf, inf)² = int(exp(-x²)dx, -inf, inf)*int(exp(-y²)dy, -inf, inf) = int[exp(-x²)*int(exp(-y²)dy, -inf, inf)dx, -inf, inf] = int[int(exp(-x²-y²)dydx, -inf, inf), -inf, inf] So far I'm exactly following the steps described above and have not done any calculation yet. In the last step I used exponent rules. Now I want to use polar coordinates since I recognize the x²+y²=r² in the exponent. The transformation theorem then tells me that the differential area changes as dxdy=rdrdθ. The area of integration is the whole real plane, so 0

[@mtaur4113]

Probably worth saying up front that x is not rcos(theta) as in polar coordinates here. It's playing the role of some parameter. I don't know the context or where these integral representations of a curve came from. Also not sure what happens if you just call them X and Y instead of polar form, no idea if the polar magic was necessary or easier or harder or what.

[@minwithoutintroduction]

ممتع.كانت الرحلة طويلة و النتيجة رائعة

[@Happy_Abe]

@14:35 can someone explain why the lower bound of the integral is 0? What argument going to infinity makes it 0?

[@davidcroft95]

The variable is x, and you can easily check that the two starting integral are going to zero as x goes to infinity

[@Happy_Abe]

@@davidcroft95 gotcha Got co fused between the x’s, u’s, and theta’s

[@davidcroft95]

@@Happy_Abe yeah, that can happen, sometimes MP goes too fast in the explanation

[@Numismathematics]

Very nice calculus.

[@psychSage]

20:29

[@ProfessorWumbo]

Nice

[@kmlhll2656]

Wow 😮

[@levonnigogoosian7547]

Can someone explain please. Why is r^2= 2rr'

[@sami-qv4st]

The derivative of r^2 with respect to x is 2rr' (chain rule).

[@szymonraczkowski9690]

cool

[@charleyhoward4594]

not sure what these complex proofs prove ?

[@levonnigogoosian7547]

Ignore my last comment. I misread the situation

[@Yangbowendw]

能听到m❤

[@curtmcd]

I don't get what is the application for this, or at least, what makes it interesting besides magic techniques and symbolic complexity? Why would I watch this rather than reading a proof on a web page?

[@tolberthobson2610]

I have to honestly say... the 'techniques' used here to 'build' the result WERE NOT interesting.

[@Devesteter252101]

frist

[@Decrupt]

trhird

[@buzzzysin]

frouth

[@yu6879ehng9005yu]

fitfth