Consider this function $$\frac{k^{2}-\xi^{2}}{k^{2}+1}$$ which has singularities at $k=\pm i$, the strips where it is analytic are $$ -1<k^{\prime \prime}<0 \quad \text { or } \quad 0<k^{\prime \prime}<1 $$ Let us concentrate on the strip $-1<k^{\prime \prime}<0 .$ The decomposition in this strip yields $$ 1-\hat{K}(k)=\frac{k^{2}-\xi^{2}}{k^{2}+1}=\frac{A(k)}{B(k)}=\frac{k^{2}-\xi^{2}}{k-i} \frac{1}{k+i} $$ where $$ \begin{array}{c} A(k)=\frac{k^{2}-\xi^{2}}{k-i} \\ B(k)=k+i \end{array} $$ such that now $A(k)$ is analytic for $k^{\prime \prime}<0$ and $B(k)$ is analytic for $k^{\prime \prime}>-1$. I worked out it just based on trial and error(Or I would say it is easy to find out such factors in this case).

Now a paper that I am reading claims that for $1+\widetilde{K}(p)=2 e^{-(c / 2)|p|} \cosh \left(\frac{c}{2} p\right)$ the decomposition $$ 1+\tilde{K}(p)=\frac{K_{+}\left(\frac{c}{2 \pi} p\right)}{K_{-}\left(\frac{c}{2 \pi} p\right)} $$ in such a way that $K_{+}\left(K_{-}\right)$ has singularities only in the upper (lower) half plane of the complex $p$ plane is given by

\begin{aligned} K_{+}(q) &=K_{-}^{-1}(-q)=\frac{(2 \pi)^{1 / 2}}{\Gamma\left(\frac{1}{2}+i q\right)} \exp \left[-i q\left(1+\frac{i \pi}{2}-\ln (-q+i \varepsilon)\right]\right] \end{aligned}

Is there a general process by which they got it?

Also, I have another problem where $c$ is now a complex variable instead of real as in this decomposition. How does changing $c$ to $c_r+i c_i$ change the above factorization?

The necessity for such a factorization appears in study of integral equations that are solvable by Wiener-Hopf method.