I am trying to compute the following complex-valued integral $(1.3.10)$ by using contour integration and the residue theorem. The substitution is shown right below $(1.3.10)$ and the answer is shown in $(1.3.11)$. By goal is to reproduce $(1.3.11)$.
$$
Since $-\pi < k < \pi$ the contour integration should be clockwise around the unit circle. I will be using the residue theorem which is based on counterclockwise contours. We see
$$
P(x,z) = \frac{1}{2\pi} \int_{-\pi}^{\pi} \frac{e^{-ikx} dk}{1 – z(pe^{ik} + qe^{-ik})} = \frac{1}{2\pi} \int_{\text{clockwise}} \frac{w^x idw/w}{1 – z(pw^* + qw)} = \frac{i}{2\pi}\int_{\text{clockwise}} \frac{w^{x-1}dw}{1 – z(pw^* + qw)}
$$
$$
= -\frac{i}{2\pi}\int_{\text{counter clockwise}} \frac{w^{x-1}dw}{1 – z(pw^* + qw)} = \text{sum of residues}
$$
The above seems to suggest that for $x < 1$ there are additional poles at $w = 0$, and so it seemed that $P(x,z)$ would have two forms, one for $|x| < 1$ and $|x| \ge 1$, but the answer seems to give one form for $P(x,z)$.
Does that mean $w = 0$ is never a pole regardless of the value of x?
How do I get the final answer?
Best Answer
As you have already stated, the change of variable $w = e^{-ik}$ leads to $$ P(x,z) = \frac{1}{2\pi} \int_{-\pi}^{\pi} \frac{e^{-ikx}}{1 - z(pe^{ik} + qe^{-ik})} \mathrm{d}k = \frac{1}{2\pi i}\int_C \frac{w^{x-1}}{1 - z(pw^* + qw)} \mathrm{d}w $$ where $C$ is the unit circle. Now, we would like to apply the residue formula at some point; however, it is forbidden, because the integrand depends on the complex conjugate $w^*$. Fortunately, one has $w^* = w^{-1}$ on the unit circle, such that $$ P(x,z) = -\frac{1}{2\pi i}\int_C \frac{w^x}{qzw^2 - w + pz} \mathrm{d}w $$ Let's suppose $x \ge 0$ for the moment. Then, the integrand admits two first-order poles coming from the equation $qzw^2 - w + pz = 0$, hence $w_\pm = \frac{1\pm\sqrt{1-4pqz^2}}{2qz}$; their residues are given by $$ \begin{array}{rcl} \mathrm{Res}_{w=w_\pm}\left(\frac{w^x}{qzw^2 - w + pz}\right) &=& \displaystyle \lim_{w\rightarrow w_\pm} (w-w_\pm)\frac{w^x}{qzw^2 - w + pz} \\ &=& \displaystyle \frac{w_\pm^x}{qz(w_\pm-w_\mp)} \\ &=& \displaystyle \frac{\pm1}{\sqrt{1-4pqz^2}} \left(\frac{1\pm\sqrt{1-4pqz^2}}{2qz}\right)^x \end{array} $$ since $qzw^2 + w - pz = qz(w-w_+)(w-w_-)$ by factorization. Note that the poles can lie inside or outside the unit circle $C$ according to the range of the variables $p,q,z$ $-$ which you haven't specified here. In the present case, the authors seem to consider $w_-$ only, so that $$ \begin{array}{rcl} P(x,z) &=& \displaystyle -\frac{1}{2\pi i} \cdot 2\pi i \,\mathrm{Res}_{w=w_-}\left(\frac{w^x}{qzw^2 - w + pz}\right) \\ &=& \displaystyle \frac{1}{\sqrt{1-4pqz^2}} \left(\frac{1-\sqrt{1-4pqz^2}}{2qz}\right)^x \end{array} $$ Finally, in the case where $x \le 0$, consider the change of variable $w = e^{+ik}$ instead, so that $e^{-ikx} = w^{-x} = w^{|x|}$ cannot generate a singularity at the origin. Then, apply the same procedure as above and you will end up with the expression given by the authors.