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
@Nottherealbegula43 ай бұрын
These just keep getting better and better, thankes for accepting my request!
@ranaranino47313 ай бұрын
at 7:40 multiplie denominator and numerator inside logarithm by alpha and make substitution alpha=sinh(y) leads Quickly to the result
@cosimo77702 ай бұрын
A work of art.
@MrWael19703 ай бұрын
I'm proud for attending this video. Best regards.
@xanterrx97412 ай бұрын
Absolutle great solution, well spended first 20 min after woken up , thanks for your hard work and waitng for next video
@Mosux20073 ай бұрын
Have you ever dealt with integrals involving elliptic functions?
@CM63_France2 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-ns4yu3 ай бұрын
“Terribly sorry about that” it’s fine bruh nobody cares if u accidentally write something weird
@mcalkis57712 ай бұрын
A good day when you upload Kamal.
@symmetricfivefold3 ай бұрын
watching this feels like watching teacher solving a "simple question" and ends up with me not knowing where am i
@barryfortnite38492 ай бұрын
can you do the integral from 0 to infinity of 1/x!
@BoringExtrovert3 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)?
@Circuito282 ай бұрын
Yo bro do you like algebraic or complex geometry problems/topics? Also algebraic number theory would be awesome, think about it
@threepointone4153 ай бұрын
As the saying goes: It's worth it's weight in Golden Ratios
@daveydd2 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?
@utkarshsaini5650Ай бұрын
its very early for you to start with integration first learn functions, inverse trigno, other topics
@Tosi314153 ай бұрын
could've used a particular formula at the end to remove the dilogarithms but cool nonetheless
@willemesterhuyse25472 ай бұрын
He picked it up by 10:48.
@kingzenoiii3 ай бұрын
ah yes, the leibniz trick
@jejnsndn3 ай бұрын
Can you do it using double intrgral?
@Rundas694203 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_5053 ай бұрын
Yes indeed
@TMH20072 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-grandcross3 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_5053 ай бұрын
I actually plan on presenting my solution soon
@gowipe-grandcross3 ай бұрын
@@maths_505what method did you use ? To see if it differ from mine
@Mario_Altare3 ай бұрын
I'm afraid I've got the sub wrong: after considering that sin x = tan x/√(1+tan^2x), 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
@willemesterhuyse25472 ай бұрын
You didn't pick up the error by 8:35!
@willemesterhuyse25472 ай бұрын
At timestep 6:39 you left out an alpha in the denominator.
@jackkalver46443 ай бұрын
I learned that my calculator can be deceptive.
@orionspur3 ай бұрын
Ohhhkay cruel!
@MathematicFanatic2 ай бұрын
ok now in terms of alpha 😳
@Babyshark-co8ks3 ай бұрын
No views in 12 seconds? Bro fell off
@maths_5053 ай бұрын
Indeed
@Bruh-lb5ck3 ай бұрын
I’m bored of this kind of comment. Who even started it in the first place
@stuffthings14173 ай бұрын
not surprising. but i've been out of the game for 20 years.
@alexkaralekas40603 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_5053 ай бұрын
@@alexkaralekas4060 we still can since the only change would be the upper limit of the integral (replacing 2 by, say, x)