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$$F(\theta)=\sin(\theta)\int_{-l}^{l} e^{-ikz\cos \theta} h(z)\,dz$$ We know that $F(\theta)$ is defined on $0\le \theta \le \pi$ and $h(z)$ is defined on $|z|\le l$
What is the period of $F(\theta)$?
Is $F(\theta)$ even or odd?
What if we change the dummy variable $z$ in the integral? does it affect the aspect of being even or odd of the function?
Is the function $e^{-ikz\cos \theta}$ standalone even or odd for variable $\theta$?
I know the even function is $F(\theta)=F(-\theta)$ and odd function is $F(\theta)=-F(\theta)$ but I am confused in this particular example!
Thanks in advance

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  • $\begingroup$ is z a real or a complex variable? $\endgroup$ – charlotte Dec 4 '14 at 7:36
  • $\begingroup$ $z$ is real, but it might be complex, don't ask such questions, suppose we're not sure, answer in both cases! $\endgroup$ – FreeMind Dec 4 '14 at 7:42
  • $\begingroup$ @FreeMind: "don't ask such questions"? It seems reasonable to assume that $z$ is real, but prudent to ask (since $z$ is often used for a complex variable). However, if $z$ is complex, then issues of the path of integration arise that will need to be answered. $\endgroup$ – robjohn Dec 4 '14 at 14:30
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    $\begingroup$ @charlotte: please ask such questions. $\endgroup$ – robjohn Dec 4 '14 at 14:30
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    $\begingroup$ @FreeMind: For $F(\theta)$ to be even or odd, it must be defined on a domain symmetric about $0$ since questions about oddness and evenness depend on comparing $F(\theta)$ and $F(-\theta)$. This cannot be done on $[0,\pi]$. $\endgroup$ – robjohn Dec 4 '14 at 14:32
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The integral, being a function of $\cos(\theta)$, is an even function of $\theta$. Multiplying an even function by an odd function, $\sin(\theta)$, gives an odd function. Therefore, if the domain were extended to $[-\pi,\pi]$, then the function would be odd.

Since $\sin(\theta)$ and $\cos(\theta)$ have a period of $2\pi$, $F(\theta)$ will have a period which is an integral divisor of $2\pi$.

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