# Integrating the function $\frac{e^{ipz}}{(\cos z)^{a}} \frac{1}{z- \xi}$ on the complex plane

In a paper titled General Theorems in Contour Integration with Some Applications, G.H. Hardy states the following:

Let us integrate

$$(i) \ \int \frac{e^{ipz}}{(\cos z)^{a}} \frac{\mathrm dz}{z- \xi} \, ,$$

$$(ii) \ \int \frac{e^{ipz}}{(\sin z)^{a}} \frac{\mathrm dz}{z- \xi} \, ,$$

round contours $$K'$$. For the present we suppose $$\xi$$ not real, and $$a <1 ,0 < p + a$$; $$a$$ may be negative. We begin with $$(i)$$, taking that value of $$(\cos z)^{a}$$ which is real at the origin.

As $$z$$ moves around any one of the points $$\left( n+\frac{1}{2} \right) \pi$$, the subject of integration acquires a factor $$e^{ ia \pi}$$; and so the contribution of $$\Xi$$ is, in the limit,

$$\sum_{n=-\infty}^{\infty} \int^{(n+\frac{1}{2})\pi}_{(n-\frac{1}{2}) \pi} \frac{e^{i(na \pi + p x)} }{|\cos x |^{a}} \frac{\mathrm dx}{x-\xi} \, ,$$

or $$\sum_{n=-\infty}^{\infty} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{e^{ipu}}{(\cos u)^{a}} \frac{e^{i n \pi(p+a)}}{u+n \pi - \xi} \, \mathrm du.$$

The contour $$K'$$ is a closed semicircle in the upper half-plane, and $$\Xi$$ is the part of the contour that lies just above the real axis.

Since $$\frac{1}{(\cos z)^{a}} = \frac{e^{-ia \arg (\cos z)}}{|\cos z |^{a}} \, ,$$ Hardy seems to be saying that $$\arg (\cos z) = - n \pi$$ on $$\Xi$$ if $$\Re(z) \in \Big( \left(n-\frac{1}{2}\right) \pi, \left(n+ \frac{1}{2} \right)\pi \Big)$$.

How exactly is $$\log (\cos z)$$ being defined here?

• I fixed the latex for $\pi$. Please check that I did not alter the meaning as well.
– robjohn
Jan 30, 2014 at 20:58
• That was supposed to be a p. I was just copying it verbatim from the book and didn't change the ordering of p and i. Jan 30, 2014 at 21:24
• My apologies. $piz$ looked as if it might have been intended to be $\pi z$ or $\pi iz$. Now it does not look ambiguous.
– robjohn
Jan 30, 2014 at 21:30
• $\log(\cos(z))$ can be well defined by integrating $\tan(z)$ as long as we don't circle the poles of $\tan(z)$. One domain would be $\mathbb{C}\setminus\{x\in\mathbb{R}:|x|\ge\pi/2\}$.
– robjohn
Jan 31, 2014 at 4:47
• Your cut on the intervals will not work. As Daniel Fischer pointed out in a comment to my now modified answer, the residue of $\tan(z)$ is $-1$ at each pole. They will never cancel. The author seems to be using the branch cut I suggested in my previous comment and their results follow.
– robjohn
Jan 31, 2014 at 12:39

The author seems to be using branch cuts along the real axis where $|x|\ge\pi/2$. With this branch cut, $\log(\cos(z))$ would be defined by integrating $-\tan(z)$ from $z=0$ (where $\log(\cos(0))=0$). Then $$\cos(z)^a=e^{a\log(\cos(z))}$$ As the point on $\Xi$ passes $\left(n+\tfrac12\right)\pi$, $\log(\cos(z))$ decreases by $\pi i$ (since $-\tan(z)$ has residue $1$ and the contour by passes in a clockwise semi-circle).
$\hspace{3cm}$
Thus, along $\Xi$, $$\cos(z)^a= \begin{array}{} |\cos(z)|^ae^{-ian\pi}&\text{if }\mathrm{Re}(z)\in\left[\left(n-\tfrac12\right)\pi,\left(n+\tfrac12\right)\pi\right] \end{array}$$