very good as usual, but I would appreciate if you could mention from time to time "we can do XXX because the integral converges because of YYY". For example right at the beginning (not at all obvious it converges at 0 !) or at ~8:30 when you cut the integral in 3 small pieces, it was a scary move and that would also have explained why you kept piece 2 and 4 together
The second comment just explains itself, right? We do it that way because it works. Because then we get 2 "known" integrals and the rest we have to work out still.
As a physicist i should call you out; you just used a technique that is used often to calculate the normalization of certain functions (hypergeometric and spherical Bessel functions). That works because the involved integrals can be awful and the easiest option is going to an asymptotic case (or go complex and work with an adequate integral representation; in this case you should now what is the most adequate representation but at the end the result is the same). Also, we use and abuse of the Stirling approximation in statistical mechanics. Just check how the partition function of a canonical ensemble is calculated, or how the Fermi-Dirac and Bose-Einstein distributions are obtain from statistical arguments.
as an engineering student I'm very glad to see some of these devilish approximations not being used in Matlab 😈 (please never use Cavalieri-Simpson approximation)
Indeed, it was 2, which means I = 4 - log(16π). Now I'm not sure if Wolfram Alpha is wrong or not, but that value is bigger than what the integral is, but it's close enough
Forse lo sviluppo in serie di cothx toglie 1/x,poi però ci sono i numeri di Bernoulli...mi sembra in po complicata..I=(α^3)ΣB(2n)((2/α)^2n)Γ(2n-1)/(2n)!
love it when the solution is via general case
This is awesome. thank you.
very good as usual, but I would appreciate if you could mention from time to time "we can do XXX because the integral converges because of YYY". For example right at the beginning (not at all obvious it converges at 0 !) or at ~8:30 when you cut the integral in 3 small pieces, it was a scary move and that would also have explained why you kept piece 2 and 4 together
The second comment just explains itself, right? We do it that way because it works. Because then we get 2 "known" integrals and the rest we have to work out still.
As a physicist i should call you out; you just used a technique that is used often to calculate the normalization of certain functions (hypergeometric and spherical Bessel functions). That works because the involved integrals can be awful and the easiest option is going to an asymptotic case (or go complex and work with an adequate integral representation; in this case you should now what is the most adequate representation but at the end the result is the same). Also, we use and abuse of the Stirling approximation in statistical mechanics. Just check how the partition function of a canonical ensemble is calculated, or how the Fermi-Dirac and Bose-Einstein distributions are obtain from statistical arguments.
as an engineering student I'm very glad to see some of these devilish approximations not being used in Matlab 😈 (please never use Cavalieri-Simpson approximation)
Stirling approach? OK cool , the result is nice ! !
Engineers saved the day, once again! Great video.
This was monstrous I really enjoyed the solution development and I enjoyed the calculation of c which was quite painful 😁🤞
i saw hyperbolic cotangent and actually loled
If you use taylor for the coth you cancel the 1/x term and then you get an infinite series of x^ne^-αx
December 17, 2024
December 17, 2024
December 17
2024
when you use stirlings approximation does it make the answer exact or just an approximation?
Well we are talking about the limit as alpha goes to infinity and c is a constant so yes the solution is exact
Missing a 2 in the thumbnail. Wonder what’s the answer to that integral
another monster!
I think you missed a 2, so wouldn't 2 * log (sqrt (2*pi) = log (2 *pi)? C = -log (2 * pi)?
Wow! If that's not a monster integral so what is? 💯💯💥💥
What about using feynmans trick twice on the second term times the exp?
The most fun part was Sterling.
Indeed
wasn't alpha=2?
Indeed, it was 2, which means I = 4 - log(16π). Now I'm not sure if Wolfram Alpha is wrong or not, but that value is bigger than what the integral is, but it's close enough
Namârië
Forse lo sviluppo in serie di cothx toglie 1/x,poi però ci sono i numeri di Bernoulli...mi sembra in po complicata..I=(α^3)ΣB(2n)((2/α)^2n)Γ(2n-1)/(2n)!
Solve questions based on melin & Z transformation