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
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 ❤
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
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.
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
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
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)😅 😊😂😂
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
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_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".
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
Did you find it?
Bro solved a monster integral and when he had to evaluate 1/2-1/3 he said this is the hard part 😭💀
the "constant of integration C" line made me piss my trousers
😂😂😂
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
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
12：42 Fourrier series:
1/1²+1/2²+1/3²+1/4²+....=pi²/6
I went through all your integration problems but first time I understood it
I think I missed the part of the solution where you got the cosecant and cotangent functions in as well.
can you do a follow up video on this very modern technique at 9:20?
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)
@@blibilbbro he is joking
don't u get it
that he understands all the other awesome tricks but not this single step😂😂😂
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
Yes, ζ(s) = π^2 /2 - π arcsec(s)
Additional observation: when we take the integration limits from 1 to 2 we get -5/3*i*Gieseking's constant.
so does I(z) = -pi*sec-1(z) + pi^2/2 = Z(z) for any complex z?
Real beauty!
isnt there a mistake at 5:30? why do we just assume we can distribute a sec^2 over (cos^2 +a^2-1)?
Bro there is a u beside the left parentheses
@@venkatamarutiramtarigoppul2078 oh my bad i just misread it
Bro can you try proving this one? integral 0 to pi arctan(ln(sinx)/x) dx = - pi*arctan[2ln(2)/pi]
so coooool！
Ohh My god!😮
Thumbnail goes CRAZY 🥵🥵
Thanks for helping me design these bro.....they really make the video stand out.
Goes hard
@@U77866I'm hard.
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 ❤
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
ua-cam.com/video/4bxL0Tow7wo/v-deo.htmlsi=QuVpUIWgmm-lhL2H
Oops sorry wrong link
ua-cam.com/video/fcBdE1bkQhg/v-deo.htmlsi=RYLv8zsuRNBYtkgh
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.
Wow now that's some roughhousing😂
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.
*uses the overpowered technique of looking up a table of antiderivatives*
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
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
isnt x = e^y-1/e^y+1 ?
@@anupamamehra6068 oh true sign error mb
@@anupamamehra6068 fixed
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)😅
😊😂😂
Bro why do some of your video titles read like they're from the hub.
Beautifull
❤❤❤
1:10 when alpha=inf it approch zero
Hence you can straight away intergrate the whole thing from [2,inf)11:43
asnwer=1+2 /1-/2
Very nice solution. Thank you
8:58 He got me here
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
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.
@@zunaidparkerbut here is the problem , which number i substitute in ln(a(x^2+1)) so i get the 0 for the limit
@@zunaidparker maybe a=1 for ln(a) + ln(x^2+1) but what do i do for ln(x^2+1)
@@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".
@@zunaidparker integral from 0 to infinity of ln(x^2+1)/x^2+1