Hey friend 😊. I solved it . Just the difference is I took I(a) = - ln(1-a(y-y^)) /(y-y^2) dy from 0 to 1 I'(a) = 1 /(1-ay+ay^2) dy from 0 to 1 After basic integration I'(a) = 4/(sqrt(a)*sqrt(4-a)) arctan (sqrt(a) /sqrt(4-a)) We need I(1) that is, intgrl 4/(sqrt(a)*sqrt(4-a)) arctan (sqrt(a) /sqrt(4-a)) da from 0 to 1 So here take sqrt(a) = 2sint It reduces to intgrl 8 t dt from 0 to pi/6 8 pi^2 /72 =pi^2 /9 私はあなたのヴェディオスの友人カマルを愛しています 😊
More straight forward to integrate first with respect to y., by factoring out a 1/x. Then change of variable for y to integrate and get arc tan function of x.
@@maths_505 I suspected it, unfortunately I am very late , "terribly sorry about that" 😅. Could you make a video about calculating the residu of the pole -n of the gamma function (or fatorial) ? I think it is equal to some thing like (-1)^n / n! but... the proof.
Σ(1/(k+1))β(k+1,k+1)=(√π/2)Σ(1/(k+1)4^k)*Γ(k+1)/Γ(k+3/2)=???..il mio metodo è ovviamente molto semplice..Σβ(k,k)/k..(k=1...inf)si vede anche a occhio..non riesco però a sintetizzare il risultato come te..ma è ovvio..I=1+(1/6)/2+(1/30)/3+(1/140)/4+(1/630)/5+....π^2/9
since pi= 3, the final result is simply 1
Engineers be like :
Hey friend 😊. I solved it . Just the difference is I took
I(a) = - ln(1-a(y-y^)) /(y-y^2) dy from 0 to 1
I'(a) = 1 /(1-ay+ay^2) dy from 0 to 1
After basic integration
I'(a) = 4/(sqrt(a)*sqrt(4-a)) arctan (sqrt(a) /sqrt(4-a))
We need I(1) that is,
intgrl 4/(sqrt(a)*sqrt(4-a)) arctan (sqrt(a) /sqrt(4-a)) da from 0 to 1
So here take sqrt(a) = 2sint
It reduces to intgrl 8 t dt from 0 to pi/6
8 pi^2 /72 =pi^2 /9
私はあなたのヴェディオスの友人カマルを愛しています 😊
It is very interesting. Thank you for your featured effort.
More straight forward to integrate first with respect to y., by factoring out a 1/x. Then change of variable for y to integrate and get arc tan function of x.
I'm doin calc2 and double integrals are fun and cool as long as it's Fubinis theorem applicable ,best example is ur integral❤
Slight pen squeak makes it even more realistic!
I would love to see more mechanics and and maybe some Calc 3? Jaboians?
Also vid was great today! What a nice solution
Marvelous!
awesome Kamaaal , you inspired me to start youtube, may Allah bless you.
Thank you brother
Hi,
"ok, cool" : 4:03 , 4:44 , 5:37 , 9:56 ,
"terribly sorry about that" : 7:33 .
I literally wait for this comment everytime I post a video😂
@@maths_505 I suspected it, unfortunately I am very late , "terribly sorry about that" 😅.
Could you make a video about calculating the residu of the pole -n of the gamma function (or fatorial) ? I think it is equal to some thing like (-1)^n / n! but... the proof.
@@CM63_France you'll find it in the complex analysis playlist
Good video
Do u need a hug brotha
Σ(1/(k+1))β(k+1,k+1)=(√π/2)Σ(1/(k+1)4^k)*Γ(k+1)/Γ(k+3/2)=???..il mio metodo è ovviamente molto semplice..Σβ(k,k)/k..(k=1...inf)si vede anche a occhio..non riesco però a sintetizzare il risultato come te..ma è ovvio..I=1+(1/6)/2+(1/30)/3+(1/140)/4+(1/630)/5+....π^2/9