When stuying a 1D differential equation, I tried to solve the problem by finding the Green's function and then solving with the inverse Fourier transform.
To make the problem well-behaved, I used a regularisation paremter $\epsilon$. My problem is that when $\epsilon\to 0$, the amplitude becomes constant whereas in the physical case, the amplitude is expected to decrease at some rate. I.e., I guess one could argue that the integral above doesn't converge to the desired solution.
As I am quite new to this topic, I am looking for help either for the specific case above or in general how one would go about to handle cases like these.
The integral I'm trying to calculate is then on the form
$$ I = \lim_{\epsilon\to 0^+}\int_{-\infty}^{\infty} \frac{\hat{f}(z)}{\Delta(z)} e^{izx} dz $$
where $\Delta(z) = z\cosh z-\sigma_0 \sinh z + i\epsilon$, $\hat{f}(z)$ is an entire function which goes to $0$ as $z\to \infty$, $\sigma_0>0$ and $x<0$. For $0<\sigma_0<1$, one can take the limit without any issues numerically, but when $\sigma_0>1$, two new poles appear which give an oscillating behaviour (which is expected).
What I've done so far:
From what I've learnt previously, I tried to create a contour in the complex plane which encircles the two poles that satisfy $\Delta(z) = 0$ which corresponds to the solutions of $z_{1,2} = \sigma_0 \tanh(z_{1,2})$ when $\epsilon=0$, and avoid all poles along the imaginary axis. For this, I considered a rectangular contour with the following line segments; $\Gamma_1$ from $-R$ to $R$, $\Gamma_2$ along $(R,R+ih)$, $\Gamma_3$ along $(R+ih,-R+ih)$ and finally $\Gamma_4$ along $(-R+ih,R)$ where $h$ is a suitable constant. As $R\to \infty$, the line integrals along $\Gamma_2,\Gamma_4$ go to $0$ and so
\begin{align} \int_{-\infty}^{\infty}\frac{\hat{f}(z)}{\Delta(z)}e^{izx} &= \sum_{i=1}^2 \text{Res}\left(\frac{\hat{f}(z)e^{izx}}{\Delta(z)} ,z_i\right) + \int_{-\infty+ih}^{\infty+ih}\frac{\hat{f}(z)}{\Delta(z)}e^{izx} dz \\ &= \sum_{i=1}^2 \text{Res}\left(\frac{\hat{f}(z)}{\Delta(z)}e^{izx} ,z_i\right) + \int_{-\infty}^{\infty}\frac{\hat{f}(u-ih)}{\Delta(u-ih)}e^{i(u-ih)x}du \\ &= \sum_{i=1}^2 \text{Res}\left(\frac{\hat{f}(z)}{\Delta(z)}e^{izx} ,z_i\right) + \int_{-\infty}^{\infty}\frac{\hat{f}(u-ih)}{(u-ih)\cosh(u-ih)-\sigma_0\sinh(u-ih)} e^{i(u-ih)x} du \end{align}
where the limit $\epsilon\to 0$ is taken on the rhs in the last step. However, when I compare these two expressions, they are nowhere near.
Is it simply so that it is likely that I've done things wrong when implementing this numerically (does the above mentioned way seem reasonable)? Or could this be a case where the regularisation I'm using is not suitable? Or is there a way to handle this case?
Edit: The orignal problem comes from considering the following case. Note that the notation below is not the same as above, and the inverse Fourier transform above is a simplified version of the one obtained from the pde below (though I think considering the one above is sufficient to learn how to handle the one below).
\begin{align} \sigma_0 \partial_z\Phi + \partial_{xx}\Phi - \epsilon\partial_x\Phi &= 0, \quad z=0\\ \partial_z\Phi - V_0f'(x) &= 0, \quad z=-D \end{align}
which in frequency domain becomes
\begin{align} \sigma_0\partial_z\phi-k^2\phi-i\epsilon k\phi &=0, \quad z=0 \\ \partial_z\phi - V_0 ik\hat{f}(k) &=0, \quad z=-D \end{align}
where $V_0$ is a constant related though related to $\sigma_0$ and $f(k)$ is some function. I have from here assumed that the solution can be written on the form
$$ \phi(k,z) = Ae^{kz} + Be^{-kz} $$
and when solving for the constants $A,B$ i eventually end up with a similar looking integral as above, altough the integral in question is simplified.