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Factorials of Non-Integral Values

Asked by Richard Gillion on January 19, 1998:
Someone said that the "factorial" of 0.5 is  (IMAGE) . It is indeed the only value which gives a smooth curve for "factorials" 1.5, 2.5, 3.5, . . . continuous with the more familiar integer values. What possible connection is there between factorials and pi?
The "factorial" concept applies only to non-negative integers, so strictly speaking there is no such thing as the factorial of 0.5. However, there is an important mathematical function called the Gamma function, defined by


and if you define  (IMAGE) (the integral of  (IMAGE) ), then f(x) has the same fundamental property that factorials have: f(x) = x f(x-1) for all x > 0. (The reason is given below). Using this, and the fact that when you calculate f(0) you get 1 which is the same as 0!, you can prove by induction that f(x) = x! when x is a non-negative integer.

Therefore, this function f(x) provides a natural extension of the factorial concept to all non-negative real numbers. What is meant by "the factorial of 0.5" is really


This function f(x) is by no means the only possible extension of the factorial concept. You can construct infinitely many different continuous, infinitely-differentiable functions f(x) that have the properties that f(x) = x f(x-1) for all x and f(x) = x! when x is a non-negative integer. However, these other functions involve much more complicated definitions rather than a simple, elegant formula. The definition given above is the most "natural" (but not the only) way to extend the meaning of "factorial" to non-integer values.

Two questions now arise: why does the above definition have the property that f(x) = x f(x-1), and secondly, the heart of your question, why does  (IMAGE) ?

The first question is answered through the technique of integration by parts. (I hope you've seen some calculus or this explanation won't mean much to you). If u and v are two functions, the product rule tells us that (uv)' = u'v + uv', so uv' = (uv)' - u'v. Therefore, the integral of uv' is the same as the integral of (uv)' minus the integral of u'v. The first integral is, by the fundamental theorem of calculus that says the integral of a derivative is the original function evaluated at the endpoints, uv evaluated at the endpoints of integration.

Applying this to f(x) which is the integral of  (IMAGE) , and letting  (IMAGE) ,  (IMAGE) , we have  (IMAGE) and  (IMAGE) . We therefore have


(the first term can be shown to be zero through l'Hôpital's rule; intutitively,  (IMAGE) goes to zero much faster than  (IMAGE) blows up).

Now, as to why a  (IMAGE) is involved in the value of f(0.5). We calculate the integral by making the substitution  (IMAGE) , so  (IMAGE) and dt = 2x dx:


(The last equality is by symmetry).

Next, we do an integration by parts with u = x and  (IMAGE) to transform this integral into


The trick here is to square the integral and use y instead of x as the variable of integration in the second factor:


This is the same as one quarter times the "double integral" of the function  (IMAGE) over the entire xy-plane. Now, because of the properties of exponentials, this equals  (IMAGE) where r is distance to the origin. This function is radially symmetric; it depends only on distance to the origin and not on the angle. Therefore, on a small ring of radius r and small thickness  (IMAGE) this function is roughly constant, with the value  (IMAGE) . The double integral over the ring is therefore approximately  (IMAGE) times the area of the ring, which is approximately  (IMAGE) : circumference times thickness.

If we add this up over a series of concentric rings, and take the limit as the thickness of each ring goes to zero, the errors in the approximations go to zero and we end up with


from which one sees that  (IMAGE) .

One reason for the appearance of the  (IMAGE) is that the properties of exponentials give you a radially symmetric function when doing the above calculation, and the  (IMAGE) appears as part of the formula for the circumference of the rings on which that function is roughly constant.

Another reason the appearance of the  (IMAGE) should not be surprising is that  (IMAGE) is intimately connected to the properties of exponentials: when the exponential function is extended to complex numbers, its period is  (IMAGE) (in the sense that, if you add  (IMAGE) to the input, the output remains unchanged). This in turn is because  (IMAGE) , which is described in the answer to another question.

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