Partition function of an Ideal monoatomic relativistic gas | Solving an integral I'm interested on finding the partition function $Z(\beta)$ of an ideal monoatomic relativistic gas.  The partition function $Z(\beta)$ is given for this case as
$$Z(\beta) \equiv \prod\limits_i^{N_A} Z_i(\beta) \qquad \wedge \qquad  Z_i(\beta) = k \int dq_1 dq_2 ... dq_N \int dp_1 dp_2 ... dp_N e^{-\beta H(\{\ q_i, p_i\})}$$
For one particle moving in coordinates $q_i$ with momentum $p_i$. The parameter $\beta$ is simply $\beta \equiv 1/k_B T$, and $H$, the Hamiltonian, follows the form $H = \sqrt{m^2c^4 + m^2p^2}$  , where $\vec{p} = \sum\limits_i^N p_i \hat{x_i}$  , e.g., in 2-dim $\vec{p}\cdot \vec{p} = (p_x \hat{i} + p_y \hat{j})\cdot (p_x \hat{i} + p_y \hat{j}) = p_x^2 + p_y^2$.
In the simplest case the integral is
$$\int\limits_{-\infty}^{\infty} \exp{\left(-\beta \sqrt{m^2c^4 + p_x^2c^2}\right)} \,dp_x$$
As $\vec{p} = \gamma m \vec{v} \; \| \; \gamma \equiv (1-v^2/c^2)^{-1/2}$ with $v<c$, in the limit $v \to c$, $p$ goes to $\infty$
Do you know any "trick" to tackle the problem?
I'm thinking on expressing $H$ as $H = mc^2 \sqrt{1+(p/mc)^2}$ and replacing $x\equiv p/mc$. Then, assuming that $p << mc$ I could expand to 2nd order in a Taylor series around $x_0 =0$ but I find this not as rigurous as I'd like it to be.
Any hint?
 A: In the 2-dim case, as @ChristopherA.Wong mentioned, it's simple to evaluate directly the integral by substitution.
Let
$$Z_i(\beta) = \frac{1}{h^2} \int dq_x dq_y \, \int dp_x dp_y \, e^{-\beta\,\sqrt{m^2 c^4 + m^2 p^2}}$$
where $dq_x dq_y$ is simply $dA$. Changing to polar coordinates:
$$p_x = r\,\cos{\theta} \qquad p_y = r\, \sin{\theta} \qquad \rightarrow \vec{p} \cdot \vec{p} = r^2 \, \big| \, r>0$$
it follows:
$$Z_i(\beta) = \frac{1}{h^2}\int dA \,\int_0^{2\pi} \int_0^{\infty} r dr d\theta \, e^{-\beta mc^2 \sqrt{1 + (\vec{p}/mc)^2}} = \frac{A}{h^2} \int_0^{2\pi} d\theta \, \int_0^{\infty} e^{-\beta mc^2 \sqrt{1+(r/mc)^2}} r dr$$
and from u-sub: $1 + (r/mc)^2 \to x$:
$$Z_i(\beta) = \frac{2\pi A}{h^2} \int_1^\infty \frac{m^2 c^2}{2} e^{-\beta mc^2 \sqrt{x}}  \, dx = \frac{\pi A m^2 c^2}{h^2} \int_1^{\infty} e^{\zeta \sqrt{x}} dx \; \Big| \; \zeta \equiv -\beta mc^2$$
Finally, integrating by parts:
\begin{align*}
    \int_1^{\infty} e^{\zeta \sqrt{x}} dx &= \int_\zeta^{\infty} e^u \, \frac{2 u}{\zeta^2 } \,du = \frac{2}{\zeta^2} \int_\zeta^\infty u e^u du\\
    &= \frac{2}{\zeta^2} \left(u\, e^u \Big|_\zeta^\infty - \int_\zeta^\infty e^u du\right) = \frac{2}{\zeta^2}\left(e^u (u -1)\right)\big|_\zeta^\infty \\
    &= - \frac{2\, e^\zeta\, (\zeta-1)}{\zeta^2} = \frac{2\, e^{-\beta mc^2 }(1+ mc^2 \beta)}{(\beta m c^2)^2}
\end{align*}
from where $Z(\beta)$ can be easily computed.
then
$$\boxed{Z_i(\beta) = \frac{2\pi A}{h^2 \beta^2 c^2}e^{-\beta mc^2 }(1+ mc^2 \beta)}$$
