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


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

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 $\Im\alpha<c'<3$

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?


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?


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