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

$1+widetilde{K}(p)=2 e^{-(c / 2)|p|} cosh left(frac{c}{2} pright)$

the decomposition

$$

1+tilde{K}(p)=frac{K_{+}left(frac{c}{2 pi} pright)}{K_{-}left(frac{c}{2 pi} pright)}

$$

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}}{Gammaleft(frac{1}{2}+i qright)} exp left(-i qleft(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.