Pi^2/6 makes me think theres gotta be a way using series to turn it into basel somehow and now i need to find it
@harshitsingh630323 күн бұрын
Did you find it?
@DDroog-eq7tw8 ай бұрын
I tried it before watching. If you want to solve it on hard mode, introduce α in the ln with 2+αx only in the numerator (so I(-1) = 0) then try to solve it. I went through two separate Feynman techniques then did an integration by parts to get dialogarithms before I found an error in the beginning of my work, where i forgot to divide by two. I don't want to re-do most of the work, but it seems possible that way.
@maths_5058 ай бұрын
Wow now that's some roughhousing😂
@smsofisami7248 ай бұрын
Bro solved a monster integral and when he had to evaluate 1/2-1/3 he said this is the hard part 😭💀
@GearsScrewlose8 ай бұрын
If you rewrite ln[(2+x)/(2-x)] as a series combine and switch them sums, we can rewrite the integral as the beta function.
@srinjansingharoy2028 ай бұрын
the "constant of integration C" line made me piss my trousers
@maths_5058 ай бұрын
😂😂😂
@thierrytitou37098 ай бұрын
Yes, indeed, write ln[(2+x)/(2-x)] = ln(1+ x/2) - ln(1-x/2) = 2*argth(x/2) and expand those ln's in taylor series as for 0
@jieyuenlee17586 ай бұрын
1:10 when alpha=inf it approch zero Hence you can straight away intergrate the whole thing from [2,inf)11:43
@pandavroomvroom8 ай бұрын
*uses the overpowered technique of looking up a table of antiderivatives*
@MrWael19708 ай бұрын
Very nice solution. Thank you
@anupamamehra60688 ай бұрын
use the fact that integral from 0 to 1 of f(x) = integral from 0 to 1 of f(1-x) to get 1-x in denominator - then use the series expansion of 1/1-x and maybe we get something
Euler substitution sqrt(1-x^2) = 1 - xt We will have Int(ln((t^2+t+1)/(t^2-t+1))/t,t=0..1) Integration by parts with u = ln((t^2+t+1)/(t^2-t+1)) dv = 1/t dt We have integral 2Int(ln(t)*(t^2-1)/(t^4+t^2+1),t=0..1) Expand rational factor 2Int(ln(t)*(t^2-1)^2/((t^4+t^2+1)(t^2-1)),t=0..1) 2Int((t^4-2t^2+1)*ln(t)/(t^6-1),t=0..1) =-2Int((t^4-2t^2+1)*ln(t)/(1-t^6),t=0..1) Now we expand 1/(1-t^6) as geometric series with common ratio t^6 -2(Int(sum(t^(6n+4)*ln(t),n=0..infinity) -sum(t^(6n+2)*ln(t),n=0..infinity) + sum(t^(6n)*ln(t),n=0..infinity) ,t=0..1) Here we change order of integration and summation then we calculate integrals by parts Finally we will have three sums to evaluate -2(sum(-1/(6n+1)^2,n=0..infinity)-2*sum(-1/(6n+3)^2,n=0..infinity)+sum(-1/(6n+5)^2,n=0..infinity)) -2(sum(-1/(6n+1)^2,n=0..infinity)+sum(-1/(6n+3)^2,n=0..infinity)+sum(-1/(6n+5)^2,n=0..infinity) - 3sum(-1/(6n+3)^2,n=0..infinity)) -2(sum(-1/(2n+1)^2,n=0..infinity)-1/3sum(-1/(2n+1)^2,n=0..infinity)) 2(sum(1/(2n+1)^2,n=0..infinity)-1/3sum(1/(2n+1)^2,n=0..infinity)) 4/3*sum(1/(2n+1)^2,n=0..infinity)=4/3*(sum(1/(n+1)^2,n=0..infinity) - sum(1/(4(n+1)^2),n=0..infinity)) 4/3*sum(1/(2n+1)^2,n=0..infinity)=4/3*(sum(1/(n+1)^2,n=0..infinity) - 1/4*sum(1/(n+1)^2,n=0..infinity)) 4/3*sum(1/(2n+1)^2,n=0..infinity)=4/3*3/4*sum(1/(n+1)^2,n=0..infinity) 4/3*sum(1/(2n+1)^2,n=0..infinity)=sum(1/n^2,n=1..infinity) =π^2/6
@giuseppemalaguti4358 ай бұрын
1operazione..x=sinθ...{...ln (2+sinθ/2-sinθ)*1/sinθ..}….uso feyman con I(a)={...2-asinθ/2+asinθ...}..I(0)=0,I=I(1)...risulta I'(a)=π/√(4-a^2),I=πarcsin(a/2)...I=I(1)=πarcsin(1/2)=π^2/6
@maxvangulik19888 ай бұрын
y=ln((1+x)/(1-x)) e^y=(1+x)/(1-x) e^y-xe^y=1+x (1+e^y)x+(1-e^y)=0 x=(e^y-1)/(e^y+1) x=tanh(y/2) y=2artanh(x) ln((1+x)/(1-x))=2artanh(x) not sure what you can do with this but it's something i noticed
@anupamamehra60688 ай бұрын
isnt x = e^y-1/e^y+1 ?
@maxvangulik19888 ай бұрын
@@anupamamehra6068 oh true sign error mb
@maxvangulik19888 ай бұрын
@@anupamamehra6068 fixed
@aravindakannank.s.8 ай бұрын
9:20 my mind blown bro after seeing the trick of looking up anti derivatives which is not happened quite a while pardon me not quite almost never (except some differential equation videos)😅 😊😂😂
@daddy_myers8 ай бұрын
Thumbnail goes CRAZY 🥵🥵
@maths_5058 ай бұрын
Thanks for helping me design these bro.....they really make the video stand out.
@U778668 ай бұрын
Goes hard
@daddy_myers8 ай бұрын
@@U77866I'm hard.
@threepointone4158 ай бұрын
Yes, ζ(s) = π^2 /2 - π arcsec(s)
@HaliPuppeh8 ай бұрын
I think I missed the part of the solution where you got the cosecant and cotangent functions in as well.
@vcvartak71117 ай бұрын
I went through all your integration problems but first time I understood it
@DD-ce4nd8 ай бұрын
Additional observation: when we take the integration limits from 1 to 2 we get -5/3*i*Gieseking's constant.
@MatthisDayer8 ай бұрын
can you do a follow up video on this very modern technique at 9:20?
@blibilb8 ай бұрын
looking up the table on anti derivatives? its basically a result you have to keep in mind like how integral of arcsin x is 1/sqrt(1-x^2)
@aravindakannank.s.8 ай бұрын
@@blibilbbro he is joking don't u get it that he understands all the other awesome tricks but not this single step😂😂😂
@Kurama.007 ай бұрын
8:58 He got me here
@ericthegreat78058 ай бұрын
so does I(z) = -pi*sec-1(z) + pi^2/2 = Z(z) for any complex z?
@trelosyiaellinika5 ай бұрын
Real beauty!
@sadi_supercell21328 ай бұрын
Beautifull
@superfilmologer8 ай бұрын
isnt there a mistake at 5:30? why do we just assume we can distribute a sec^2 over (cos^2 +a^2-1)?
@venkatamarutiramtarigoppul20788 ай бұрын
Bro there is a u beside the left parentheses
@superfilmologer8 ай бұрын
@@venkatamarutiramtarigoppul2078 oh my bad i just misread it
@sidhantmohanty52568 ай бұрын
Bro can you try proving this one? integral 0 to pi arctan(ln(sinx)/x) dx = - pi*arctan[2ln(2)/pi]
@comdo7778 ай бұрын
asnwer=1+2 /1-/2
@oatq1758 ай бұрын
so coooool!
@BladimirRemon8 ай бұрын
Ohh My god!😮
@sadi_supercell21328 ай бұрын
Btw do i always use lim as parameter goes to infinity for the part of the function where i inserted parameter or i can use whatever number for example limit as parameter goes to 0,1,-2... edit: when i want to find the missing constant c
@zunaidparker8 ай бұрын
You should use any "convenient" number for that purpose. In this case it happened to be infinity, which isn't a number you can just substitute in so you have to consider it as a limit.
@sadi_supercell21328 ай бұрын
@@zunaidparkerbut here is the problem , which number i substitute in ln(a(x^2+1)) so i get the 0 for the limit
@sadi_supercell21328 ай бұрын
@@zunaidparker maybe a=1 for ln(a) + ln(x^2+1) but what do i do for ln(x^2+1)
@zunaidparker8 ай бұрын
@@sadi_supercell2132 if you're left with an expression like that it means that you didn't choose a good parameterization to begin with. You should always choose where to insert the "a" to end up with an expression that is "nice".
@sadi_supercell21328 ай бұрын
@@zunaidparker integral from 0 to infinity of ln(x^2+1)/x^2+1
@ThanhNhan_GiaSu2348 ай бұрын
❤❤❤
@xleph25257 ай бұрын
Bro why do some of your video titles read like they're from the hub.
@retarded.pigeon8 ай бұрын
Hi, it's not obvious to me why you can switch the order of derivation and integration operations in the Feynman technique, can you give an explanation? Ty ❤
@dacomputernerd40968 ай бұрын
I gather that is just what the Feynman technique does. It states that the derivative with respect to one variable of a definite integral with constant limits with respect to another is the definite integral of the partial derivative, probably assuming some stuff about continuity and convergence. The rule this stems from also allows the limits to be functions of the variable you differentiate with respect to, but it just adds additional terms to the result which go to zero when the limits are constants due to including partial derivatives of the limits