I have the thermal partition function and the density of states for the 3D simple harmonic oscillator, which are given below $$ Z(\beta) = \frac { 1 } { \left( 2 \sinh \left( \frac { \beta \omega } { 2 } \right) \right) ^ { 3 } } $$ and $$ \rho ( E ) = \frac { \left( \frac { E } { \omega } - \frac { 1 } { 2 } \right) \left( \frac { E } { \omega } + \frac { 1 } { 2 } \right) } { 2 \omega } $$ where $\beta$ is the inverse temperature and $\omega$ is the natural frequency of the SHO.
Now, the partition function can also be written as the following integral $$ Z(\beta) = \int_{0}^{\infty} dE \: \rho(E) e^{-\beta E} $$ which is a Laplace transform, hence we can calculate the asymptotic of the density of states by inverting this Laplace transform. The inverse Laplace transform can be done using the Bromwich contour integral which is $$ \rho(E) = \frac{1}{2\pi i} \int_{\gamma - i \infty}^{\gamma + i \infty} d\beta \: Z(\beta) e^{\beta E} $$ where $\gamma$ is greater than the real part of all the singularities of $Z(\beta)$. Now, I tried doing this integral using the residue theorem by closing the contour on the left half of the complex plane. But somehow I am getting the correct density of states using the residue of just one pole, whereas there is an infinite number of poles on the imaginary axis due to the periodicity of the hyperbolic sine function (on the imaginary axis).
Can someone explain where I am wrong in this process?
P.S. We also know that $n = \frac{E}{\omega} - \frac{3}{2}$ is a positive integer.