# Factorial of 3/2? [closed]

How do you compute the factorial of something like $3/2$ or $-2$? Wolfram Alpha gives an answer, but how does it arrive at that point?

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## closed as off-topic by abx, Andreas Blass, Emil Jeřábek, Gjergji Zaimi, Dima PasechnikApr 22 at 13:53

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With all due respect, I think this question is not of research level, so it would be more suitable for math.stackexchange.com – GH from MO Apr 22 at 13:15
Indeed, but the question is over five years old. Imagine the trauma induced in migrating a five year old. – The Masked Avenger Apr 22 at 13:43
This is what happens when quid edits old questions... – Gerald Edgar Apr 22 at 14:12
@GeraldEdgar what is that supposed to mean exactly? In addition you could have at least the courtesy to notify me when talking about me. (Please reply on meta or in chat.) – quid Apr 22 at 16:35
@GeraldEdgar it appears you are unable or unwilling to clarify your remark, but still maintain it. FYI, almost all my recent edits before this one were done following a request of a moderator. This one and the subsequent one are also in line with existing retagging policy. Besides, you can compare my actions to those of some other long-term users to see that they are not that unusual (I can provide details on request). In any case, I hope you making this type of remark will stay an isolate event. (In this hope I will not escalate this matter to the moderators right away.) – quid Apr 23 at 11:15

This is the Gamma function. The gamma function is defined by an integral, and we define n! = Γ(n+1).

For factorials of half-integers, you can start with (1/2)! = Γ(3/2) = ∫0 t1/2 e-t dt. Substituting u = t1/2 turns this into the "Gaussian integral" (integral of e^(-x^2), possibly with some constants), and you get (1/2)! = sqrt(π)/2.

The recurrence n! = n * (n-1)! holds for the Γ function, so you get (3/2)! = (3/2) * (1/2)! = 3 sqrt(π)/4.

(-2)! actually doesn't exist. If it did, then we'd have

1! = 1 * 0! = 1 * 0 * (-1)! = 1 * 0 * (-1) * (-2)!

so 1! = 0 * (-2)!, but 1! = 1.

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Just to add a footnote, one might ask why the gamma function and not some other function that agrees with factorial (when shifted by one). There are other ways one might interpolate between the values of factorial. One justification is the Bohr-Mollerup theorem.

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What if we relax log convex requirement? – Turbo Apr 22 at 20:14