In my opinion, I prefer to just use the actual factorial notation. Some people say "that's only defined for nonnegative integers" but I disagree. The Riemann zeta function, through analytic continuation, can be defined almost everywhere. Not all of it will satisfy the initial definition of ζ(x)=sum n≥1 of 1/n^x, but we still call it ζ. ζ(-1) is well defined and equal to -1/12, even if it's not the actual sum n≥1 of n. It's the same here. The factorial function has been analytically continued to all but countably many numbers. And whilst most don't satisfy the original definition of k!, the product n=1 to k of n, it's still the factorial function. (1/2)! is well defined and equal to √π/2, even if it's not the product from n=1 to 1/2 of n.
The factorial function n! is only defined on the non-negative integers n (0,1,2,...), and technically speaking the gamma/pi function is not an analytic continuation of the factorial, but rather an interpolation of the factorial function, as such an extension to the complex plane C is not unique (analytic continuation (AC) applies only to connected domains D, which is not the case here with the non-negative integers - as such, any AC must be unique). The gamma function is what we use by convention in place of this non-unique extension via the integral definition (satisfies gamma(z+1) = zgamma(z), gamma(1) = 1) which indeed only converges for Re(z) > 0. But it is not unique: for example, the so-called pseudogamma function also successfully interpolates the factorial. Hence, denoting something such as the equivalent of gamma(3/2) as (1/2)! can be ambiguous. The AC only applies to the gamma function itself, taking the domain to Re(z) < 0 (on which it is meromorphic, poles at non-positive integers); this is unique by the identity theorem, and this time our domains Di (from Re(z)>0, continued to slices -1 < Re(z) < 0, -2 < Re(z) < 0 etc... until ultimately onto Re(z)
Hi, would't the Pi-function have a z-1 in the exponent of t, because you substituted z -> z-1 with respect of the Gamma-funciotn? Greetings from Germany :)
He did something strange like evaluating π(Z) = √π/2 results instead. I was just as confused but just accepting that the "Indefinite integral" from 0 to infinity of √te^(-t)dt evaluates to √π/2 after substitution of those limits. Hopefully the ending part is correct! 😂
Your handwriting is seriously one of the prettiest I have ever seen. just one minor detail, in the end result, the 27 in the denominator should be +/- 27. Thanks a lot for these videos, amazing quality. As a mechanical engineer, I miss sometimes these math lessons.
