Theorem 2 of de Bruijn's paper titled "The roots of trigonometric integrals" (Duke Math. J., 17 (1950)) is given by:
What does it mean by "the function $q(x)$ be regular in the sector...?
Does it only mean that there is no poles for $q(x)$ in this sector?
ADDED: If we set $N=1,p_1=p_{-1}=\frac{1}{2}a (a>0)$, $q(x)=\left( \frac{1}{2}(x+x^{-1})\right)^{-m} (m=1,2,3...)$, $x=e^t$, then the integral becomes the one represented in another question here.
Is this $q(x)$ function regular in the sector $-\frac{\pi}{2}<\arg x<\frac{\pi}{2}$ and on its boundary?
I think that this $q(x)$ function is regular in the sector but not on its boundary, because when $-\frac{\pi}{2}=\arg x =\arg (e^t)=-i t$ we have $\cosh(t)=\cos(it)=0$. Therefore $Q(t)=q(e^t)=(\cosh t)^{-m} $ has a simple pole and does not have Taylor expansion at $t=-i\frac{\pi}{2}$.
Am I right?
Thanks- mike