- The problem statement, all variables and given/known data
With the Hamiltonian here:
Compute the cananonical ensemble partition function given by $\frac{1}{h} \int dq dp \exp^{-\beta(H(p,q)}$
for 1-d , where $h$ is planks constant
Relevant equations
The attempt at a solution I am okay for the $p^2/2m$ term and the $aq^2$ term via a simple change of variables and using the gaussian integral result $\int e^{-x^2} dx = \sqrt{\pi}$
I am stuck on the $ \int dq e^{\beta b q^{3}}$ and $ \int dq e^{\beta c q^{3}}$ terms.
If these were of the form $ \int dq e^{-\beta b q^{3}}$ I could evaluate via $\int dx e^{-x^n} = \frac{1}{n} \Gamma (1/n) $ where $ \Gamma(1/n) $ is the gamma function;
however because it is a plus sign I have no idea how to integrate forms of $ \int dq e^{x^n}$
Or should I be considering the integral over $q$ all together and there is another way to simply:
$\int dq e^{-\beta(aq^2-bq^3-cq^4)}$
PART B)
To compute the grand canonical ensemble of $N$ oscillators where
$ E_n=((n+\frac{1}{2})−x(n+\frac{1}{2})^2)\bar{h}ω $
to leading order in $x$
I have that
$\zeta(β,μ)=∑\lim^{N=∞}_{N=0}z^NZ(β,N)$
where $\zeta(β,μ)$ is the grand canonical ensemble, $Z(β,N)$ is the canonical ensemble and $z=e^{βμ} $ is the fugacity.
I have attempted to answer this question via summing first over $N$ in hope of getting a simplified expression for $Z_1$ to leading order in $x$ and then I will raise this to the power of $N$ , dividing by $N!$ (Gibbs factor) and then use (1) above to get the grand canonical partition function.
My attempt is as follows:
$Z_1 = \sum_n e^{-\beta((n+\frac{1}{2})\bar{h}\omega-x(n+\frac{1}{2})^2\bar{h}\omega)}=e^{-\frac{\beta \bar{h} \omega}{2}}\sum_n e^{-\beta \bar{h} \omega n} e^{\beta \bar{h} \omega x (n+\frac{1}{2})^2}$
I will expand out the exponential $x$ term to get
$ e^{-\frac{\beta \bar{h} \omega}{2}} \sum_n e^{-\beta \bar{h} \omega n} (1+\beta\bar{h}\omega x(n+\frac{1}{2})^2) + O(x^2)$
$e^{-\frac{\beta \bar{h} \omega}{2}}\sum_n e^{-\beta \bar{h} \omega n} + e^{-\frac{\beta \bar{h} \omega}{2}}e^{-\beta \bar{h} \omega n} (\beta \bar{h} \omega x) \sum_{n=0}^{n=\infty} (n^2+ \frac{1}{4} + n)$
which diverges...
Many thanks in advance
This is my first answer, so I hope I'm doing it right.
As pointed out in an earlier comment, I think you need to start of by getting the limits straight, which will answer a couple of your questions. The integral over $p$ is independent and easily done as you've stated yourself. The integral over $q$ goes from $-\infty$ to $+\infty$, as it is the position in one dimension.
Note in passing that it is $$\int_0^{\infty} e^{-x^n} = \frac{1}{n}\Gamma\left(\frac{1}{n}\right)$$ but your lower limit is $-\infty$, and so this cannot be used.
(Incidentally, $\int_{-\infty}^{+\infty}e^{\pm x^3}dx$ does not converge to the best of my knowledge).
But all of this is beside the point: unless I've misunderstood you (please correct me if I'm wrong!), you're claiming that
$$\int_{-\infty}^{+\infty}dq \,\,e^{-\beta a q^2 + \beta b q^3 + \beta c q^4} = \int_{-\infty}^{+\infty}dq \,\,e^{-\beta a q^2} \int_{-\infty}^{+\infty}dq \,\,e^{\beta b q^3} \int_{-\infty}^{+\infty}dq \,\,e^{ \beta c q^4}$$
which is clearly not true. So performing the integrals separately is not the way to go and you must consider the integral over all the functions of $q$ together. If the extra terms had been linear in $q$ you could have used the "completing the square" trick, but I don't there is anything similar for higher powers. For an exact (and possibly useless) answer, you might be able to use this formula that is completely beyond me, but I don't think it's helpful.
With a little research I found the original question online. If it's the same question, it says
In other words, when
Which leads me to believe that they require you to perform some sort of series in the two "problematic" terms.
If I were to solve this question, I would do the following. I'd begin by making a simple variable substitution: $u = \sqrt{a} q$, which makes my integral
$$\frac{1}{\sqrt{a}} \int_{-\infty}^\infty du\,\,\exp{\left(-\beta u^2 + \frac{\beta b}{a\sqrt{a}}u^3 + \frac{\beta c}{a^2} u^4\right)} = \frac{1}{\sqrt{a}} \int_{-\infty}^\infty du\,\, e^{-\beta u^2}e^{\frac{\beta b}{a\sqrt{a}}u^3} e^{\frac{\beta c}{a^2} u^4}$$
I would then approximate the anharmonic terms using $e^{ax}\approx 1 + ax + \frac{a^2 x^2}{2}...$
$$I = \frac{1}{\sqrt{a}} \int_{-\infty}^\infty du\,\, e^{-\beta u^2} \left(1 + \frac{\beta b}{a\sqrt{a}}u^3 + \frac{\beta^2 b^2}{2 a^3}u^6\right) \left(1+ \frac{\beta c}{a^2} u^4\right)$$
You'll notice I took three terms for the first approximation and only two for the second. Why I did this will become clear in a moment. This is now just a sum of integrals of the form
$$\int_{-\infty}^\infty du\,\, e^{-\beta u^2}u^n$$
Clearly, when $n$ is an odd integer, these integrals are zero, since the function is odd and its positive and negative contributions cancel each other. This is why when expanding the series earlier I stopped at the first even power, to get the leading dependencies in $b$ and $c$.
We are thus left with three integrals (the fourth term involving $\frac{b^2 c}{a^5}$ can be neglected since it is of a much higher order.)
$$I = \frac{1}{\sqrt{a}}\left( \int_{-\infty}^\infty du\,\, e^{-\beta u^2} + \frac{\beta c}{a^2} \int_{-\infty}^\infty du\,\, e^{-\beta u^2}u^4 + \frac{\beta^2 b^2}{2 a^3}\int_{-\infty}^\infty du\,\, e^{-\beta u^2}u^6\right)$$
The remaining (even) integrals can be easily computed realising that whenever $2z-1$ is even,
$$\int_{-\infty}^{\infty} u^{2z-1} e^{-\beta u^2}du = \beta^{-z} \Gamma[z]$$
Thus,
$$I \approx \sqrt{\frac{\pi}{\beta a}}\left( 1 + \frac{3 c}{4 a^2 \beta} + \frac{15 b^2}{16 a^3 \beta}\right)$$
EDIT:
Upon some introspection I've decided that it does indeed make sense to have a term in $\frac{b^2}{a^3}$. I've made the necessary changes.
Furthermore, the complete canonical partition function (including the integral over $p$) is thus
$$Z = \frac{\pi}{\beta}\sqrt{\frac{2 m}{a}}\left( 1 + \frac{3 c}{4 a^2 \beta} + \frac{15 b^2}{16 a^3 \beta}\right)$$