# Wiener-Hopf Factorization: How are these contour integrals done?

$$\Phi_+(\alpha) = \frac{1}{2\pi i} \int_{C_1}\ln\left(2 e^{-\frac{c | z| }{2}} \cosh \left(\frac{c z}{2}\right) \right) \frac{dz}{z-\alpha}$$

and

$$\Phi_-(\alpha) = - \frac{1}{2\pi i} \int_{C_2} \ln\left(2 e^{-\frac{c | z| }{2}} \cosh \left(\frac{c z}{2}\right) \right) \frac{dz}{z-\alpha}$$

where the contours $$C_1$$ and $$C_2$$ are parallel to the real line, but pass above and below the point $$z=\alpha$$

Here $$c$$ is a complex parameter.

This is called Weiner-Hopf factorization. These came when I was trying to solve an integral equation of the form $$\phi(x)=f(x)+\int_0^\infty K(x-x')\phi(x)$$. Where one has to factorize the Kernel into the two parts which are analytic in upper and lower half plane. According to wikipedia https://en.wikipedia.org/wiki/Wiener%E2%80%93Hopf_method these integrals gives such factorization.

But I am not sure how are these integrals done.

For an example if I had to factorize $$\frac{1}{z^2+9}$$ into $$\phi_+(\alpha)+{\phi_-(\alpha)}$$ such that $$\phi_+$$ is analytic in upper half plane and $$\phi_-$$ is analytic in the lower half plane.

I could use see that the function is anaytic in the strip $$-3<\Im \alpha<3$$ and use

$$\phi_+=\frac{1}{2\pi i}\int_{-\infty+d'i}^{\infty+d'i} \frac{1}{(z^2+9)(z-\alpha)}=\frac{i}{6 (\alpha +3 i)}$$ where $$\Im \alpha >d'>3$$ and

$$\phi_-=-\frac{1}{2\pi i}\int_{-\infty+c'i}^{\infty+c'i} \frac{1}{(z^2+9)(z-\alpha)}=\frac{-i}{6 (\alpha -3 i)}$$

where $$\Im\alpha

But in my case, I am not sure how to carry out these integrals. As I could not use the residue theorem (as the function does not have any poles) like in the example I presented. Any hints or resources on these?

Summary:

How can $$\ln\left(2 e^{-\frac{c | z| }{2}} \cosh \left(\frac{c z}{2}\right) \right)$$ be written as $$\phi_+ -\phi_-$$ such that $$\phi_+$$ is analytic in upper half plane and $$\phi_-$$ is analytic in lower half plane?