- #1
RedX
- 970
- 3
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)
[tex]V=\int da \int dx \int dy \int dz \mbox{ } \delta(1-a-x-y-z) [/tex]
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:
[tex]V=\frac{1}{2\pi}\int dk \int da \int dx \int dy \int dz \mbox{ }e^{ik(1-a-x-y-z)}=
\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}[/tex]
The integral of a,x,y,z all give the same value of [tex] \frac{1}{-ik}\left(e^{-ik}-1 \right) [/tex] 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:
[tex]\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} [/tex]
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.
For simplicity take a 2-simplex (the volume of a 2-simplex is 1/6)
[tex]V=\int da \int dx \int dy \int dz \mbox{ } \delta(1-a-x-y-z) [/tex]
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:
[tex]V=\frac{1}{2\pi}\int dk \int da \int dx \int dy \int dz \mbox{ }e^{ik(1-a-x-y-z)}=
\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}[/tex]
The integral of a,x,y,z all give the same value of [tex] \frac{1}{-ik}\left(e^{-ik}-1 \right) [/tex] 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:
[tex]\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} [/tex]
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.