This is great! I took a complex analysis class 41 years ago, but I was always weak on contour integration and applying the concept to real integrals (I was weak, or the course was weak). Inspired me to look at your complex analysis videos on MathMajor, and I'll probably go through those. Thank you for doing all of this! BTW, if you make or already have a video series on multi-variable calculus, I'll be reviewing that as well!
If the gamma function didn't have that pesky -1, the reflection formula would look like this: Γ(z)Γ(-z) = πz/sin(πz) You can easily see there is a reflection with the minus sign, and also the RHS looks like the reciprocal of sin(x)/x, an important function used in calculus to find trig derivatives.
@@fartoxedm5638 Γ(z) = (z-1)! with the usual definition (for positive integers, but let's extend it) The reflection formula is Γ(z)Γ(1-z) = π/sin(πz) Replace gammas with factorial: (z-1)!(-z)! = π/sin(πz) Multiply with z: (z)!(-z)! = πz/sin(πz) Change notation to make gamma match with factorial: Γ(z)Γ(-z) = πz/sin(πz)
@@f5673-t1h You have literally wrote the mapping from factorial world to the gamma world in your first line. You can't simply go with "Nah, let's take Г(x) = x!" In the end
It seems like we're doing something shady with the contour integral. In particular, the replacement u -> exp(i2pi) u is baffling, since there should be no change. I figure Michael is leaving out some t -> 0+ from some of these definitions and for C3 the replacement is actually u -> exp(i2pi - i2t) or something of that nature.
12:00 We have to restrict the value of z much earlier: The integral definitions of the Gamma function which Michael uses right from the start are only valid for Re(z) > 0 and Re(z) < 1, respectively. 15:00 Here it's not regardless of what z is, this only works for Re(z) > -1.
I would be very interested in a discussion of convergence on this integral. Normally, it’s not something I care about, but because the integral defining the gamma function is only defined for z>0, meaning this integral should diverge for z outside (0,1), meaning you actually sneakily did some analytic continuation here.
I used contour integration to derive this identity as well, but started with this representation of the beta function, the integral of t^(a-1)/(1+t)^(a+b) from 0 to infinity for t.
The restriction is necessary to evaluate the product Γ(z)Γ(1-z), because the integral representations of both Γ(z) and Γ(1-z) need to simultaneously converge and this only happens in the "critical strip" 0 < Re(z) < 1. Once this expression is evaluated, it turns out to simplify to π/sin(πz), giving us a valid /equation/ that holds in the critical strip. But once the equation is proved, it may be re-interpreted as a /formula/ for computing values of Γ in places where the integral representation does not converge (i.e. thereby "getting rid of the restriction"). In fact, treating the equation as a formula is the /unique/ way to extend Γ to the rest of the complex plane while maintaining its nature as an analytic function.
I make a Proof of this identity using contour integral on my notes. Later I made a Proof of the Riemann and Hurwitz Zeta Functional equation using complex contour integral (which have an infinite number of poles...)
As for moi - whenever the presentation goes off at an extreme tangent covering some gross new things they seem to be eminently forgettable. But I reserve the right to be wrong on this :-) Basis of my conjecture: math is not frightening, math is eminently doable. Nonetheless - great video, great swooping intro to some gigantic new things (I feel like calling them monsters and that is okay)
