- #1
mjordan2nd
- 177
- 1
If I want to find [itex] \int f(x) dx [/itex] where [itex] x=a+b [/itex] and a and b are both variables how would I do this in terms of a and b. Let me give you some context about this question.
I'm trying to understand the following in my statistical mechanics class. My book states that if I have a random variable x that follows a gaussian distribution then
[tex]
<e^{iqx}> = \int_\infty^\infty dx \frac{ e^{ - \frac{ \beta (x-<x>)^2}{2}}} { \sqrt{ 2 \pi / \beta}} e^{iqx} = e^{iq<x> - \frac{q^2}{2 \beta}} .
[/tex]
where q is real. It then seems to pull the following out from the deepest regions of the netherworld using sorcery and other very, very dark arts:
[tex]
\langle e^{i \sum_{j=1}^N q_j x_j} \rangle = exp \left( i \sum_{j=1}^N q_j <x_j> - \sum_{j,k=1}^N q_j G_{jk} \frac{q_k}{2} \right).
[/tex]
In order to combat the dark arts, I'm trying to prove this to myself. That is the context that the integral I originally asked about came up.
Any help is appreciated. Thanks.
I'm trying to understand the following in my statistical mechanics class. My book states that if I have a random variable x that follows a gaussian distribution then
[tex]
<e^{iqx}> = \int_\infty^\infty dx \frac{ e^{ - \frac{ \beta (x-<x>)^2}{2}}} { \sqrt{ 2 \pi / \beta}} e^{iqx} = e^{iq<x> - \frac{q^2}{2 \beta}} .
[/tex]
where q is real. It then seems to pull the following out from the deepest regions of the netherworld using sorcery and other very, very dark arts:
[tex]
\langle e^{i \sum_{j=1}^N q_j x_j} \rangle = exp \left( i \sum_{j=1}^N q_j <x_j> - \sum_{j,k=1}^N q_j G_{jk} \frac{q_k}{2} \right).
[/tex]
In order to combat the dark arts, I'm trying to prove this to myself. That is the context that the integral I originally asked about came up.
Any help is appreciated. Thanks.
Last edited: