To know what one-half factorial is you need to have a good understanding of how you work out a given factorial, i.e. you need a definition of factorial. The first definition you usually come across is something like:

n!=n(n-1)! with 0!=1.

This is the recursive equivalent of n!=n(n-1)(n-2)... 2. 1.

Using these definitions it is clear that there is no such thing as (1/2)!, or in formal language, (1/2)! is not defined.

There is however another common definition of factorial; the gamma function:

G(n) = ò₀^¥t^n-1 e^-t dt

(That's integral from 0 to infinity over t in case it's not clear.)

n! is now defined simply by n!=G(n+1). If you've done integration by parts you will be able to evaluate this for positive integers, although it gets very tedious for n much above 2. A better way is to find a recurrence relation for G(n) in terms of G(n-1). Integrate G(n) by parts and you should find that it equals nG(n-1). Hence n!=n(n-1)! and since G (1)=1 we have the usual definition of factorial for positive integers.

However, note that the n in G(n) need not be a natural number or even a real number. The gamma function defines factorials for all complex numbers.

Using this definition then, the question "What is (1/2)!?" becomes meaningful. The integral G(1/2) is not trivial however. For those interested it requires squaring the integral to make it an area integral, and using the substitution x=u², y=v², with appropriate Jacobian. This is degree level so don't worry if you haven't heard of this sort of thing (unless you're doing a maths degree that is). It turns out that G(1/2) does indeed equal Ö(p)/2.

Hope this answers your question. As always, if there is anything above that you would like me to go over in more detail, or anything you didn't understand, please ask.

Sorry about the (very) tardy reply to this question, but I've been a little busy lately. Anyway here goes...

When you do your usual integration trick, you're integrating a function along a line, conventionally the x-axis, and in so doing finding the area under the graph of the function. Area integration is exactly the same only you are integrating over an area (in the x-y plane say) and finding the volume under the function you're integrating. This function might be something like z=x y+x+y+1, it's a function of two variables.

Integrating these things is dead easy. If you want to integrate the above function over the square from x=0 to 1 and y=0 to 1 (note that you must choose an area for the region of integration) you write something like this:

ò₀¹ò₀¹ x y + x + y + 1 dx dy

The first integral sign corresponds to the dy and the second to the dx.

To evaluate it you simply integrate by one variable at a time treating the other variable as a constant. It doesn't matter which one you do first either, choose the one that is easiest first or whatever. I'll do x first to get:

ò₀¹ [(1/2)x² y+(1/2) x²+x y+x]₀¹ dy=ò₀¹ (1/2) y+(1/2)+y+1 dy = 1/4+1/2+1/2+1=2.25

and that's your volume.

This process is fine if your area is a rectangle, but it gets rather more complicated when you try to integrate over odd shaped regions. If you wanted to integrate over a triangle for example you could do it by putting variables in the limits like so:

ò₀¹ò₀^1-y x y + x + y + 1 dx dy

Can you see what the region of integration will be?

Well it will be the right angled triangle with two equal sides of length 1 on the positive x and y axes. The limits for x are 0 (so it's bounded at the y-axis) and 1-y (so there is a bound at x=1-y, the hypotenuse of the triangle. y only goes from 0 to 1 so you just get the triangle. You might like to try constructing some other shaped regions.

It's now not irrelevant in what order you do the integrations. If you integrate y first in the above example, after you have integrated x you will have y's left in the result, as the limits for x have a y in them, but the volume enclosed will be a real number, not dependant on x or y. In general if a particular variable has limits that depend upon y (say) you must do this integral before you integrate y.

This is still not sufficient to do all the integrals we want to though. For example it may be convenient to use radial coordinates (where you define position in terms of distance from the origin and angle subtended at the x-axis). We need some idea of substitution for these new double integrals, and it isn't as simple as in the single integral case. You might like to think why. Consider the elemental segments we are integrating over, in the one dimensional case our element was a line dx, but now we have an elemental area dx×dy. The Jacobian is to do with this substitution.

To make it quick, if you have two variables, x and y, and do the substitution u = f(x,y) and v = g(x,y) (u and v are the new variables, f and g are functions of x and y) then "to change the dx×dy into du×dv" you must also multiply by the Jacobian which is defined as the determinant of the matrix of partial derivatives.

You may have gathered by now that when I squared the integral in the (1/2)! question, what I was doing was integrating over a square rather than a line segment and squaring the function at the same time so for example:

I=ò₀¹ x dx

would square to

I²=ò₀¹ ò₀¹ x y dx dy

Hope all this has made sense, don't worry if you can't do area integrals yet. This isn't the hard bit of the subject however. What I have told you so far has been mostly methods for evaluating integrals and you should note that I have given only vague justification for most of it. It's important that all that I have said is set on firmer ground (i.e. can be proven) but this is difficult and time-consuming, so I suggest you ether leave this until later in your mathematical career or get a very good book. Ether way you should bear in mind that the above only constitutes an overview of the easier bits of the subject.

Richard.