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
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?
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.
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 ?
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
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
These just keep getting better and better, thankes for accepting my request!
at 7:40 multiplie denominator and numerator inside logarithm by alpha and make substitution alpha=sinh(y) leads Quickly to the result
A work of art.
I'm proud for attending this video. Best regards.
Absolutle great solution, well spended first 20 min after woken up , thanks for your hard work and waitng for next video
Have you ever dealt with integrals involving elliptic functions?
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 .
“Terribly sorry about that” it’s fine bruh nobody cares if u accidentally write something weird
A good day when you upload Kamal.
watching this feels like watching teacher solving a "simple question" and ends up with me not knowing where am i
can you do the integral from 0 to infinity of 1/x!
I haven’t finished the entire thing yet, but why don’t you jut use inverse hyperbolic tan for the form dx/(a-x2)?
Yo bro do you like algebraic or complex geometry problems/topics? Also algebraic number theory would be awesome, think about it
As the saying goes: It's worth it's weight in Golden Ratios
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?
its very early for you to start with integration first learn functions, inverse trigno, other topics
could've used a particular formula at the end to remove the dilogarithms but cool nonetheless
He picked it up by 10:48.
ah yes, the leibniz trick
Can you do it using double intrgral?
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.
Yes indeed
I was busy for a few months and after coming back i sense a change in style of thumbnails, or is it only me?
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 ?
I actually plan on presenting my solution soon
@@maths_505what method did you use ? To see if it differ from mine
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
You didn't pick up the error by 8:35!
At timestep 6:39 you left out an alpha in the denominator.
I learned that my calculator can be deceptive.
Ohhhkay cruel!
ok now in terms of alpha 😳
No views in 12 seconds? Bro fell off
Indeed
I’m bored of this kind of comment. Who even started it in the first place
not surprising. but i've been out of the game for 20 years.
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
@@alexkaralekas4060 we still can since the only change would be the upper limit of the integral (replacing 2 by, say, x)
Hello