This is the one-particle partition sum for a free non-relativistic particle. Of course, ##\alpha=1/(2m)##. For a classical ideal gas of ##N## partices you simply have ##Z_N=Z^N/N!##, where I borrowed the ##1/N!## from the quantum theoretical indistinguishability of particles; this is the minimal input from quantum mechanics you need to avoid inconsistencies like Gibbs's paradoxon.
For the full quantum-mechanical treatment of the ideal gas you need to take into account the bosonic or fermionic nature of the particles. That's best done in the occupation-number representation and by first using a finite volume. You can take any boundary conditions you like, because those get unimportant in the infinite-volume/thermodynamical limit). Choosing periodic boundary conditions with the volume being a cube of length, ##L## your single-particle momentum spectrum is ##\vec{k} \in 2 \pi \hbar/L \mathbb{Z}##, and in the grand-canonical ensemble you get:
Fermions:
There for each ##\vec{k}## and each spin you have ##N(\vec{k},\sigma) \in \{0,1 \}##. So your partition sum is
$$Z(\beta,\mu)=\prod_{\vec{k},\sigma} \sum_{N=0}^1 \exp[N(-\beta \vec{k}^2/(2m) + \beta \mu)] = \prod_{\vec{k}} [1+\exp(-\beta \vec{k}^2/(2m)+\beta \mu)]^{2s+1}.$$
Where ##s## is the spin of the particle (from relativistic QFT we know it must be half-integer for fermions).
Bosons:
For bosons the calculation is the same but for each ##(\vec{k},\sigma)## the occupation numbers ##N(\vec{k},\sigma) \in \mathbb{N}_0##. This gives
$$Z(\beta,\mu)=\prod_{\vec{k},\sigma} \sum_{N=0}^{\infty} \exp[N(-\beta \vec{k}^2/(2m) + \beta \mu)] = \prod_{\vec{k}} [1-\exp(-\beta \vec{k}^2/(2m)+\beta \mu)]^{-(2s+1)}.$$
Note that here you necessarily must have ##\mu<0##, because otherwise the geometric series wouldn't converge (at least for ##\vec{k}=0##).
The thermodynamic/infinite-volume limit is simple for fermions. You consider the logarithm of ##Z##, and the resulting sum over the momenta can be approximated by an integral:
$$\sum_{\vec{k}} \simeq L^3 \int \mathrm{d}^3 k/(2 \pi \hbar)^3,$$
because in a momentum-volume element ##\Delta^3 k## there are ##\Delta^3 k L^3/(2 \pi \hbar)^3## single-particle momentum eigenstates.
This leads to
$$\ln Z=\frac{(2s+1) V}{(2 \pi \hbar)^3} \int_{\mathbb{R}^3} \mathrm{d}^3 k \ln[1+\exp(-\beta \vec{k}^2/(2m)+\beta \mu)].$$
For bosons the thermodynamic limit is subtle, because the ground state must be considered separately. Before you go the infinite-volume limit, keeping the discrete momenta there is no problem. You just get a sum, and if you try to fix the particle number at a given temperature by choosing an appropriate ##\mu<0##, you can get any particle number you want, because for ##\mu \rightarrow -0^{+}## the particle number diverges, i.e., you can accommodate as many particle you like at any given ##T##. In the limit ##T \rightarrow 0## you get that all particles are in the ground state (Bose-Einstein condensate).
If you go over to the infinite-volume limit, the total particle number does not diverge anymore for ##\mu \rightarrow -0^+##. So if you naively approximate the sum with the integral you miss the zero-mode contribution. So the correct approximation is to take the zero-mode out and only for the rest you approximate with the integral. Then you get
$$\ln Z=-(2s+1) \ln[1-\exp(\beta \mu)] -\frac{(2s+1)V}{(2 \pi \hbar)^3} \int_{\mathbb{R}^3} \ln [1-\exp(-\vec{k}^2/(2m)+\mu \beta].$$
To get the (average) particle number you have to set ##\mu \beta=\alpha##, differentiate wrt. ##\alpha## and then set ##\alpha=\mu \beta## again. This gives
$$N=\frac{2s+1}{\exp(-\mu \beta)-1} + \frac{(2s+1) V}{(2 \pi \hbar)^3} \int_{\mathbb{R}^3} \mathrm{d}^3 k \frac{1}{\exp(\beta \vec{k}^2/(2m)-\mu \beta)-1}.$$
The correct infinite-volume limit now is to divide this by ##V## to get the particle-number density ##n=N/V##. Then for a given ##T## you keep ##n## fixed and take ##V \rightarrow \infty##. Then there are two cases
(a) "Normal state"
$$n < \frac{(2s+1) V}{(2 \pi \hbar)^3} \int_{\mathbb{R}^3} \mathrm{d}^3 k \frac{1}{\exp(\beta \vec{k}^2/(2m))-1} = n_{>}. \qquad (*)$$
Then you choose ##\mu<0## such that
$$n=\frac{(2s+1) V}{(2 \pi \hbar)^3} \int_{\mathbb{R}^3} \mathrm{d}^3 k \frac{1}{\exp(\beta \vec{k}^2/(2m)-\mu \beta)-1}.$$
Making then ##V \rightarrow \infty## at fixed ##\beta## and ##\mu##, the contribution from the zero mode obviously vanishes.
(b) "Condensate state"
If (*) is not fulfilled, then you set ##\mu=0## in the integral. In the zero-mode contribution you make ##V \rightarrow \infty## and ##\mu \rightarrow -0^+## such that this expression stays constant at
$$n_0=n-n_{>}.$$
This is a state, where a macroscopically relevant portion of the particles occupies the ground state at ##\vec{k}=0##, i.e., you have a Bose-Einstein condensate in addition to a "normal" (degenerate) Bose gas. For ##T \rightarrow 0## all particles occupy the ground state.
This problem with the particle-number density never occurs of course for fermions, because there you can choose ##\mu \in \mathbb{R}## arbitrarily, and thus you can choose ##\mu## always such that the integral at a given ##T## gives any particle-number density you like by adjustint ##\mu##.
