Why did Feynman do this integral this way to calculate the field from a plane of oscillating charges? In Feynmen Lectures on Physics Vol I Feynman calculated the field for a plane of oscillating charges in the following way: (given in the photos, I am having trouble uploading them in the right place)
But why did he take the upper limit of the integral to be infinity?
At a certain time $t$ the charges for which $r>ct$ Won't be able to contribute to the field cause their field wouldn’t be able to reach the point $P$ in that given time. Because the field propagates with the speed $c$. So Won't the upper limit of this integral be $r=ct$?
Also why did he use the complex notation? The position of the charges were given by a cosine function which can be easily Integrated. So why did he need to use the complex notation and get into a mess? 
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 A: If you let the upper limit of integral be $ct$, you presuppose that the charges in plane start to oscillate at the initial time $t_0=0$. So if we let the initial time be $t_0= -\infty$, the upper limit of integral is $+\infty$.
I think the reason for using complex notatin, is that complex exponention function is natural setting in Linear ODE, say: $\lambda\in \mathbb C,\ (\frac{d}{dt})^k e^{\lambda t}=\lambda^k e^{\lambda t}$, so if $\lambda$ is a root of (real) polynomial $y^n+a_{n-1} y^{n-1}+\cdots+ a_0=0$, then $e^{\lambda t}$ is a solution of ODE $ x^{(n)}+a_{n-1} x^{(n-1)}+\cdots+a_0 x=0$; in particular, ODE $x''+px'+qx=0$ (harmonic oscillator without external force). Another reason is that, it's easy to calculate the derivative and anti-derivative of $e^{\lambda t}$.
To avoid get into a mess, note that if we identify $\mathbb C\cong \mathbb R^2$, the inclusion $x\hookrightarrow x+i0$ and projection $x+iy\twoheadrightarrow x$ and $x+iy\twoheadrightarrow y$ are linear map, so it communitcate with many other operation such as derivative, integral.
I guess the meaning of "in physics, $e^{-i\infty}=0$" or "then the coefficient
$\eta$ (charge/area density) in the exact integral would decrease toward zero" is that feynmann use $\eta=\eta_0 e^{-Vr},\ V\in \mathbb R_{>0}$. (By dominated convergence theorem, now it's integrable.) So we can calculate
$$\int_{z}^{\infty} e^{-Vr} e^{\frac{-i\omega}{c} r}\, dr=\int_{z}^{\infty} e^{(\frac{-i\omega}{c}-V)r}\, dr=\frac{1}{\frac{-i\omega}{c}-V} e^{(\frac{-i\omega}{c}-V)r}\Big|_{r=z}^{r=\infty}=\frac{-1}{\frac{-i\omega}{c}-V} e^{(\frac{-i\omega}{c}-V)z}$$
because $\lim_{r\to\infty} e^{-Vr}\cdot \frac{1}{\frac{-i\omega}{c}-V}\cdot e^{(\frac{-i\omega}{c})r}=0$. If we then let $V\to 0^+$, we get $\frac{c}{i\omega} e^{-i\omega z/c}$.
And the for "the graph of our integral would then become a curve which is a spiral. The spiral would eventually end up at the center of our original circle, as drawn in Fig. 30-12.", it corresponds to the graph of $r\mapsto(e^{-Vt}\cos\frac{\omega}{c} r, e^{-Vt}\sin\frac{\omega}{c} r)$ when $V>0$ is small (It's really a spiral curve).
Note: if we consider "There is also another reason ... we have omitted for the projection of the acceleration ..." and do not use $\eta=\eta_0 e^{-Vt}$, then it can prove that the integral become $\int_z^\infty \frac{1}{2}(1+\frac{z^2}{r^2})e^{\frac{-i\omega}{c} r}\, dr$, which still diverge. And unfortunately, the anti-derivative $\int \frac{e^{-at}\cos bx}{x^2}\, dx$ (or $\sin$) is not an elementary function.
