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Hello everyone
Can someone help me out solving this integral:
\begin{equation}
S_T(\omega)=\frac{2k_BT^2g}{4\pi^2c^2}\int_0^{\infty}\frac{sin^2(kl)}{k^2l^2}\frac{k^2}{D^2k^4+\omega^2}dk
\end{equation}
Where $$D=g/c$$
According to this paper https://doi.org/10.1103/PhysRevB.13.556. The solution to the integral is:
\begin{equation}
S_T(\omega)=\frac{k_BT^2D^{1/2}}{4\sqrt{2}l^2c\pi\omega^{3/2}}(1-e^{-\theta}(sin(\theta)+cos(\theta)))
\end{equation}
Where $$\theta=(\omega/\omega_0)^{1/2}$$ and $$\omega_0=D/2l^2$$
I am not able to reach the result they present in the paper. Hence any help will be very much appreciated.
Can someone help me out solving this integral:
\begin{equation}
S_T(\omega)=\frac{2k_BT^2g}{4\pi^2c^2}\int_0^{\infty}\frac{sin^2(kl)}{k^2l^2}\frac{k^2}{D^2k^4+\omega^2}dk
\end{equation}
Where $$D=g/c$$
According to this paper https://doi.org/10.1103/PhysRevB.13.556. The solution to the integral is:
\begin{equation}
S_T(\omega)=\frac{k_BT^2D^{1/2}}{4\sqrt{2}l^2c\pi\omega^{3/2}}(1-e^{-\theta}(sin(\theta)+cos(\theta)))
\end{equation}
Where $$\theta=(\omega/\omega_0)^{1/2}$$ and $$\omega_0=D/2l^2$$
I am not able to reach the result they present in the paper. Hence any help will be very much appreciated.