Volume of a simplex using delta function

[RedX]

I tried to calculate the volume of a simplex, but got an integral I couldn't do.

For simplicity take a 2-simplex (the volume of a 2-simplex is 1/6)

$$V=\int da \int dx \int dy \int dz \mbox{ } \delta(1-a-x-y-z)$$

where the integration limits are over the 4-cube.

My reasoning for this formula is that as (1-a) varies from 0 to 1, the integral counts the volume when x+y+z=(1-a), and gives zero otherwise, so you end up measuring the volume 0<x+y+z<1.

I then wrote the delta function as a Fourier integral:

$$V=\frac{1}{2\pi}\int dk \int da \int dx \int dy \int dz \mbox{ }e^{ik(1-a-x-y-z)}=<br /> \frac{1}{2\pi}\int dk \mbox{ }e^{ik} \int da \mbox{ }e^{-ika} \int dx \mbox{ }e^{-ikx}\int dy \mbox{ }e^{-iky}\int dz \mbox{ }e^{-ikz}$$

The integral of a,x,y,z all give the same value of $$\frac{1}{-ik}\left(e^{-ik}-1 \right)$$ since the limits are from 0 to 1. There are (d+2) of these terms for a d-simplex. So you end up with the integral:

$$\frac{i^{d+2}}{2\pi} \int dk \mbox{ } e^{ik} \left(\frac{1}{k}\right)^{d+2}\left(e^{-ik}-1 \right)^{d+2}$$

I'm aware there are easier ways to calculate the volume of a d-simplex.

But I want to know if in principle this method works, and why is the integral so hard when the result should be easy, since the volume of a d-simplex is a simple expression.

 

[RedX]

I figured it out. The method I outlined is correct.

So we had this:

$$\frac{i^{d+2}}{2\pi} \int dk \mbox{ } e^{ik} \left(\frac{1}{k}\right)^{d+2}\left(e^{-ik}-1 \right)^{d+2}$$

Manipulating:

$$\frac{2^{d+1}}{\pi} \int dk \left(\frac{1}{k}\right)^{d+2}sin^{d+2}\left(\frac{k}{2}\right)\left[cos\left(\frac{dk}{2}\right)-isin\left(\frac{dk}{2}\right) \right]$$

For some reason Mathematica can't handle this expression (or maybe just the online free version can't handle this). So what I did was that I inserted by hand several different values of d, and thankfully, the imaginary part vanished (as it should), and you get:

$$\frac{1}{(d+1)!}$$

which is the correct answer.

I'm quite interested in seeing how the T.A. solves this problem next Monday in class...because a T.A. ought to spend more time on the homework than even the students, because the T.A. has a responsibility to be even more prepared, so his answer ought to be really good too.