Talk:Gamma function

From Rosetta Code

Integrals

I prefer the form (operator-like) instead of the form ; both are correct; no need to change it the next time:D --ShinTakezou 20:01, 5 March 2009 (UTC)

Hmm, I never saw this form in mathematical literature. Technically, there is no any multiplication of f(x) by dx. You cannot commute them, if you meant that. And integral operator is not equal to definite integral. The definite integral using the integral operator would be sort of: . --Dmitry-kazakov 20:55, 5 March 2009 (UTC)
I did my math in my life, and I've seen it. Seen or not, there's a simple analogy between and . Going into definite thing, there's no a rule stating that can't mean the integral of f(x) computed between x0 and x1; the analogy with derivative could be (or any of the form you prefer to say the derivative of f(x) computed in x0); no need for the integral to put limits "outside" the "operator boundary" like your attempt . More close analogy would be with the sum (the integral sign is nothing but an S). But this discussion is OT for RC. Mine was not an error, I will write it the same way for other "math" tasks; no need to fix it. --ShinTakezou 11:34, 6 March 2009 (UTC)

Complex field

Actually Gamma is defined on complex numbers. Is the task about its real part only? --Dmitry-kazakov 22:25, 5 March 2009 (UTC)