I am currently working on this paper: http://web.calstatela.edu/faculty/rcooper2/article.pdf. I want to calculate $\phi_\nu$ on page 450. The author states that for $u_{\tau,\epsilon}$ ($\epsilon, \tau>0)$ and $\hat{\phi}_{\nu}= \frac{\hat{u}_{\tau,\epsilon}}{1+\nu(\xi_0^2+\tau^2)}$, $\phi_\nu$ has the following form
\begin{align*} \phi_\nu (x_0,x')= \frac{\pi}{1+\nu\tau^2} \int \limits_{-\infty}^{\infty} e^{-(\frac{1}{\nu}+\tau^2)|x_0-y_0|}u_{\tau,\epsilon}(y_0,x') dy_0. \end{align*}
I tried to calculate the inverse Fourier transform \begin{align*} \phi_\nu (x_0,x')= \frac{1}{2\pi} \int \limits_{-\infty}^{\infty}e^{ix_0\xi_0} \frac{1}{1+\nu(\xi_0^2+\tau^2)} \int \limits_{-\infty}^{\infty} e^{-\xi_0y_0} u_{\tau, \epsilon}(y_0,x') dy_0 d\xi_0\\= \frac{1}{2\pi} \int \limits_{-\infty}^{\infty} \int \limits_{-\infty}^{\infty} e^{i\xi_0(x_0-y_0)}\frac{u_{\tau, \epsilon}(y_0,x')}{1+\nu(\xi_0^2+\tau^2)} dy_0 d\xi_0 \end{align*} but I have to admit that I have no clue how to evaluate this integral.
Any help would be greatly appreciated.
Alright, I tried it on my own by using Jordan's Lemma. Could someone take a quick look if this is correct?
First I factorized $1+\nu(\xi_0^2+\tau^2) =\nu (\xi_0+i\sqrt{\frac{1}{\nu}+\tau^2})(\xi_0-i\sqrt{\frac{1}{\nu}+\tau^2})$. Hence $\frac{e^{i\xi_0(x_0-y_0})}{1+\nu(\xi_0^2+\tau^2)}$ has two simple poles. As we are only interested in residues satisfying $Im z>0$ we consider $i\sqrt{\frac{1}{\nu}+\tau^2}$.\begin{align*} \int \limits_{\mathbb{R}} \frac{e^{i\xi_0(x_0-y_0})}{1+\nu(\xi_0^2+\tau^2)} \mathrm{d}\xi_0= 2\pi i Res(\frac{e^{i\xi_0(x_0-y_0})}{1+\nu(\xi_0^2+\tau^2)},i\sqrt{\frac{1}{v}+\tau^2})= 2 \pi i \frac{e^{-\sqrt{\frac{1}{v}+\tau^2}(x_0-y_0}}{2i \nu \sqrt{\frac{1}{v}+\tau^2}} \end{align*} Thus \begin{align*} \phi_\nu (x_0,x') \frac{1}{2\pi \nu \sqrt{\frac{1}{v}+\tau^2}} \int \limits_R e^{-\sqrt{\frac{1}{v}+\tau^2}(x_0-y_0)} u_{\tau,\epsilon}(y_0,x') \mathrm{d}y_0 \end{align*}