A surprisingly difficult integral: int 0 to π/2 arctan(2sin(x)) solution using Feynman's trick

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Maths 505

Maths 505

Күн бұрын

Пікірлер: 41
@Maths-Integrals
@Maths-Integrals 3 ай бұрын
It's a very nice integral and the solution is interesting. There are also many very interesting properties of the Dilogarithm function. This simplifies the final result: I= ((pi)^2)/6 -(3/2)*(ln(phi))^2 where "phi" denote the golden ratio
@Nottherealbegula4
@Nottherealbegula4 3 ай бұрын
These just keep getting better and better, thankes for accepting my request!
@ranaranino4731
@ranaranino4731 3 ай бұрын
at 7:40 multiplie denominator and numerator inside logarithm by alpha and make substitution alpha=sinh(y) leads Quickly to the result
@cosimo7770
@cosimo7770 2 ай бұрын
A work of art.
@MrWael1970
@MrWael1970 3 ай бұрын
I'm proud for attending this video. Best regards.
@xanterrx9741
@xanterrx9741 2 ай бұрын
Absolutle great solution, well spended first 20 min after woken up , thanks for your hard work and waitng for next video
@Mosux2007
@Mosux2007 3 ай бұрын
Have you ever dealt with integrals involving elliptic functions?
@CM63_France
@CM63_France 2 ай бұрын
Hi, I know that tan^-1 (-2) = 2 tan^-1 (phi) (it's the angle between 2 adjacent faces of the dodecahedron) so tan^-1 (2) = pi - 2 tan^-1 (phi) , so tan^-1 (2) / 2 = pi/2 - tan^-1 (phi) , so tan^-1 (2) / 2 = tan^-1 (1/phi) . 13:43 : sin theta = 2 z / ( 1 - z^2) , minus and not plus, but fortunately this term cancelled out later . "ok, cool" : 1:33 , 10:01 , 14:24 , 19:21 , "terribly sorry about that" : 3:34 , 6:30 , 11:16 , 12:02 , 15:22 , 16:39 , 16:42 , 16:52 , 17:31 , 17:46 .
@JakePinedo-ns4yu
@JakePinedo-ns4yu 3 ай бұрын
“Terribly sorry about that” it’s fine bruh nobody cares if u accidentally write something weird
@mcalkis5771
@mcalkis5771 2 ай бұрын
A good day when you upload Kamal.
@symmetricfivefold
@symmetricfivefold 3 ай бұрын
watching this feels like watching teacher solving a "simple question" and ends up with me not knowing where am i
@barryfortnite3849
@barryfortnite3849 2 ай бұрын
can you do the integral from 0 to infinity of 1/x!
@BoringExtrovert
@BoringExtrovert 3 ай бұрын
I haven’t finished the entire thing yet, but why don’t you jut use inverse hyperbolic tan for the form dx/(a-x2)?
@Circuito28
@Circuito28 2 ай бұрын
Yo bro do you like algebraic or complex geometry problems/topics? Also algebraic number theory would be awesome, think about it
@threepointone415
@threepointone415 3 ай бұрын
As the saying goes: It's worth it's weight in Golden Ratios
@daveydd
@daveydd 2 ай бұрын
Sir I have a question, why or how is ln(2) so related to the trig functions? Especially when their integrals involve natural logarithms. What's its essence? What's its meaning behind it? Is it because of ln(sin(π/4)) and ln(cos(π/4)) being equal to -(1/2)(ln(2)) ? Perhaps that's only one of the bunch of reasons, I've tried googling but I haven't found much.. any ideas?
@utkarshsaini5650
@utkarshsaini5650 Ай бұрын
its very early for you to start with integration first learn functions, inverse trigno, other topics
@Tosi31415
@Tosi31415 3 ай бұрын
could've used a particular formula at the end to remove the dilogarithms but cool nonetheless
@willemesterhuyse2547
@willemesterhuyse2547 2 ай бұрын
He picked it up by 10:48.
@kingzenoiii
@kingzenoiii 3 ай бұрын
ah yes, the leibniz trick
@jejnsndn
@jejnsndn 3 ай бұрын
Can you do it using double intrgral?
@Rundas69420
@Rundas69420 3 ай бұрын
I think so, because the applicability of the Leibnitz rule is equivalent to transforming the single integral into a double integral by taking a derivative.
@maths_505
@maths_505 3 ай бұрын
Yes indeed
@TMH2007
@TMH2007 2 ай бұрын
I was busy for a few months and after coming back i sense a change in style of thumbnails, or is it only me?
@gowipe-grandcross
@gowipe-grandcross 3 ай бұрын
Hello, I would like to send you a result about a generalisation of the dirichlet integral (with the proof ofc) : the integral from 0 to infinity of sin(x^n)/x^n. I don't think you already present this on your channel. Is there a way to do it ?
@maths_505
@maths_505 3 ай бұрын
I actually plan on presenting my solution soon
@gowipe-grandcross
@gowipe-grandcross 3 ай бұрын
​@@maths_505what method did you use ? To see if it differ from mine
@Mario_Altare
@Mario_Altare 3 ай бұрын
I'm afraid I've got the sub wrong: after considering that sin x = tan x/√(1+tan^2⁡x), I let tan x = u, so I'(a) = ∫_0^∞ u du/{(√(1+u^2 )[1+(a^2+1) u^2])} Then, after letting 1+u^2 = v^2, I'(a) = ∫_0^1 dv/[1+(a^2+1)(1-v^2 )] I got stuck because I obtained the following integral: I(a) = 1/2 ∫_0^2 1/(√(1+a^2 ) √(2+a^2 )) * ln⁡((√(2+a^2 )+√(1+a^2 ))/(√(2+a^2 )-√(1+a^2 )))da which is a bit of a hell... Maybe there's a further sub to go on from here, but maybe there's not
@willemesterhuyse2547
@willemesterhuyse2547 2 ай бұрын
You didn't pick up the error by 8:35!
@willemesterhuyse2547
@willemesterhuyse2547 2 ай бұрын
At timestep 6:39 you left out an alpha in the denominator.
@jackkalver4644
@jackkalver4644 3 ай бұрын
I learned that my calculator can be deceptive.
@orionspur
@orionspur 3 ай бұрын
Ohhhkay cruel!
@MathematicFanatic
@MathematicFanatic 2 ай бұрын
ok now in terms of alpha 😳
@Babyshark-co8ks
@Babyshark-co8ks 3 ай бұрын
No views in 12 seconds? Bro fell off
@maths_505
@maths_505 3 ай бұрын
Indeed
@Bruh-lb5ck
@Bruh-lb5ck 3 ай бұрын
I’m bored of this kind of comment. Who even started it in the first place
@stuffthings1417
@stuffthings1417 3 ай бұрын
not surprising. but i've been out of the game for 20 years.
@alexkaralekas4060
@alexkaralekas4060 3 ай бұрын
I mean it feels a little cheated by not taking the definite integral at 7:40 because if you did you wouldn't be able to take the dilogarithm
@maths_505
@maths_505 3 ай бұрын
@@alexkaralekas4060 we still can since the only change would be the upper limit of the integral (replacing 2 by, say, x)
@dr.hanamantkarande9544
@dr.hanamantkarande9544 3 ай бұрын
Hello
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