I quite enjoyed it when you didn't skip any steps or calculations:( I know we've all seen it a million times, but there was something mesmerizing in it
I went tan(theta) = x which after simplifying eventually gives an integral from 0 to pi/2 of ln^2(sin(theta)/cos(theta))/(1-sin.cos(theta)) which has a denominator you cn expand with a geometric series, after more simplifying you get a sum of k from 0 to infinity of integrals from 0 to pi/2 of ln^(sin(theta)/cos(theta))*sin^k(theta).cos^k(theta) which if you expand the ln(sin/cos) into ln(sin) - ln(cos) and then square that argument of logarithms and then use the linearity of the integral to split it up, you get a sum of k from 0 to infinity of 3 integrals that all look like sin^k(theta).cos^k(theta).[some logs of trig functions] and at that point you can consider each of the 3 integrals to be the second derivative of the beta function. I did not end up continuing the problem any further because I did not want to differentiate the beta function 3 times at least, but the simplification of that and then substituting in for k should give you the rest of what goes in the infinite summation. Yes this is my way of whining that I chose an overcomplicated solution
I did the beginning the same, but I derived that (x*sqrt(3)/2)/(x^2-x+1) = Sum k from 0 to infinity sin(k*pi/3)*x^k . Rewrote the sum as Sum (-1)^k/(k!) * gamma(k+1)*(-1)^k*sin(k*pi/3)*x^k to fit the Ramanujans master theorem. Kinda cheating I guess but the answer was close to the correct one, dropped a factor somewhere
I(a)=I(..x^a/...)..derivò 2 volte e risulta I=I"(0)..I(0)=4π/3√3..I'(0)=0..ma il problema è I(a)...boh... comunque le condizioni iniziali non servono,devo derivare non integrare...certo è che se proponi 2 integrali da 0->inf,e da 0->1..forse si arriva alla funzione Beta
For both cases multiply the num and deno with (1+x) and apply Feynman after then 😊.
Solvable using the Residue Theorem with a keyhole contour.
Beta function and Eulers reflection formula are the angular stones of crazy integrals, anyways great video as always :)
I quite enjoyed it when you didn't skip any steps or calculations:( I know we've all seen it a million times, but there was something mesmerizing in it
I went tan(theta) = x which after simplifying eventually gives an integral from 0 to pi/2 of ln^2(sin(theta)/cos(theta))/(1-sin.cos(theta)) which has a denominator you cn expand with a geometric series, after more simplifying you get a sum of k from 0 to infinity of integrals from 0 to pi/2 of ln^(sin(theta)/cos(theta))*sin^k(theta).cos^k(theta) which if you expand the ln(sin/cos) into ln(sin) - ln(cos) and then square that argument of logarithms and then use the linearity of the integral to split it up, you get a sum of k from 0 to infinity of 3 integrals that all look like sin^k(theta).cos^k(theta).[some logs of trig functions] and at that point you can consider each of the 3 integrals to be the second derivative of the beta function. I did not end up continuing the problem any further because I did not want to differentiate the beta function 3 times at least, but the simplification of that and then substituting in for k should give you the rest of what goes in the infinite summation.
Yes this is my way of whining that I chose an overcomplicated solution
Hi,
"terribly sorry about that" : 1:27 , 2:30 , 3:52 , 9:47 ,
"ok, cool" : 1.46 , 3:33 , 5:33 , 6:38 , 8:03 , 8:26 .
Math got talent 😊. Thank you indeed.
I did the beginning the same, but I derived that (x*sqrt(3)/2)/(x^2-x+1) = Sum k from 0 to infinity sin(k*pi/3)*x^k . Rewrote the sum as Sum (-1)^k/(k!) * gamma(k+1)*(-1)^k*sin(k*pi/3)*x^k to fit the Ramanujans master theorem. Kinda cheating I guess but the answer was close to the correct one, dropped a factor somewhere
The substitution x=1/t part I knew, relating both integrals but after that it became a struggle...:)
He is a 10 and he solves cool upper ESh symbols= He = ♾️ ❤
Now with ln(ln(x)) instead of ln^2(x). Malmsten would be proud.
Funny thing I was working on that last night after releasing the video 😂
Oookay cool
awesome :D
Genial
Cool
I(a)=I(..x^a/...)..derivò 2 volte e risulta I=I"(0)..I(0)=4π/3√3..I'(0)=0..ma il problema è I(a)...boh... comunque le condizioni iniziali non servono,devo derivare non integrare...certo è che se proponi 2 integrali da 0->inf,e da 0->1..forse si arriva alla funzione Beta
Irrationality in the denominator 😔
Dude, I told you to lay off using the infinity stones for math for a while 😒
Gotta put em to some use
@@maths_505 you should use The World so you can stop time and write and we cannot see him because we aren't stand users (I suppose)