What is the Centroid of $z=\frac{1}{(1-i\tau)^{i+1}},\ \ \tau\in (-\infty,\infty)$ Some years ago I developed the closed curve (Ref. 1)
$$z=\frac{1}{(1-i\tau)^{i+1}}, \ \ \tau\in (-\infty,\infty)$$
I was able to calculate the arc length and area,
$$
s=\int |\dot z| d\tau=2\sqrt{2}\sinh\left(\frac{\pi}{2}\right)\\
A=\frac{1}{2}\int \mathfrak{Im}\{z^*\dot z\}d\tau=\frac{\sinh \pi}{4}
$$
However, the centroid has always eluded me,
$$z_c=\frac{1}{3A}\int z\ \mathfrak{Im}\{z^*\dot z\}d\tau$$
You may find it easier to work with the transform $\tau=\tan\theta$ and
$$z(\theta)=\cos^{1+i}\theta\ e^{i(1+i)\theta},\ \ \theta\in[-\pi/2,\pi/2]$$
References


*

*The Apple of My $i$ (and the linked PDF therein)

*Zwikker, C. (1968). The Advanced Geometry of Plane Curves and Their Applications, Dover Press.
 A: I can reproduce your results for $s$ and $A$, but I too am stuck on $z_c$.  Querying Wolfram Alpha with a modified integrand, yields an unrelated "closed" form solution with a hypergeometric function as part of the solution.
Documenting some intermediate expressions:
$\tau = \tan{\theta}$
$d\tau = \sec^{2}{\theta} d\theta$
$z = \dfrac{1+i\tau}{1+\tau^2}e^{\tan^{-1}{(\frac{-\tau}{1})}}e^{-i\frac{1}{2}\ln{(1+\tau^2)}} = \dfrac{1+i\tan{\theta}}{\sec^{2}\theta}e^{-\theta}e^{-i\ln{(\sec{\theta})}} = \cos{(\theta)} e^{-\theta}e^{i(\theta-\ln{(\sec{\theta})})}$
$\bar{z} = (1+i\tau)^{-(1-i)}$
$\dot{z} = (-1+i)(1-i\tau)^{-(2+i)}$
$|\dot{z}|=\dfrac{\sqrt{2}}{1+\tau^2}e^{\tan^{-1}(\frac{-\tau}{1})} = \dfrac{\sqrt{2}}{\sec^{2}{\theta}}e^{-\theta}$
${\rm Im}(\bar{z}\dot{z}) = \dfrac{(1-\tau)}{(1+\tau^2)^2}e^{2\tan^{-1}{(\frac{-\tau}{1})}} = \dfrac{\cos{\theta}(\cos{\theta}-\sin{\theta})e^{-2\theta}}{\sec^{2}{\theta}}$
I arrive at the following integral for the centroid:
$$z_c = \frac{1}{3A}\int_{-\pi/2}^{\pi/2} \cos^{2}\theta(\cos{\theta}-\sin{\theta})e^{-3\theta}e^{i(\theta-\ln{(\sec{\theta})})} d\theta$$
which can be manipulated into
$$z_c = \frac{1}{3A}\int_{-\pi/2}^{\pi/2} e^{-6\theta}(1-\tan{\theta})(e^{\theta}\cos{\theta})^{3+i} d\theta$$
and with an integration by parts using $u = \dfrac{1}{3A}e^{-6\theta}$ and $dv$ as the remaining portion of the above integrand, the integral can be converted to
$$z_c = \dfrac{3-i}{5A}\int_{-\pi/2}^{\pi/2} e^{-6\theta}(e^{\theta}\cos{\theta})^{3+i} d\theta$$
and then I'm stuck.
Update after making more progress in a different line:
Starting from
$z{\rm Im}(\bar{z}\dot{z}) = \dfrac{(1-\tau)(1+i\tau)}{(1+\tau^2)^3(1-i\tau)^i}\left[\dfrac{1+i\tau}{1-i\tau}\right]^{i}$
the integral
$$z_c = \dfrac{4}{3\sinh{\pi}} \int_{-\infty}^{\infty} \dfrac{(1-\tau)(1+i\tau)}{(1+\tau^2)^3(1-i\tau)^i}\left[\dfrac{1+i\tau}{1-i\tau}\right]^{i} d\tau$$
can't be solved using Residue Theory, because of the items raised to the power $i$ which introduces singularities in the complex plane from the complex Log() function.
Luckily, the integral does have a solution:
$$z_c = \dfrac{i(1+i\tau)^{-1+i}}{6\sinh(\pi)4^i}\left[{}_2F_1\left(-1+i,3+2i;i;\dfrac{1+i\tau}{2}\right)-(1+i\tau) {}_2F_1\left(i,3+2i;1+i;\dfrac{1+i\tau}{2}\right)\right]\biggr\rvert_{-\infty}^{\infty}$$
From a plot of this function, it appears that it converges rapidly in both directions.  By $\tau = 2$ the function is $\approx 0$.  By $\tau = -30$ the real and imaginary parts are relatively flat.
So it is very likely that the limit need only be evaluated at $-\infty$ to get a closed form expression
$$z_c = \lim_{\tau\to-\infty} -\dfrac{i(1+i\tau)^{-1+i}}{6\sinh(\pi)4^i}\left[{}_2F_1\left(-1+i,3+2i;i;\dfrac{1+i\tau}{2}\right)-(1+i\tau) {}_2F_1\left(i,3+2i;1+i;\dfrac{1+i\tau}{2}\right)\right]$$
So this has been reduced to a limit problem, which the substitution
$\tau = \tan\theta$, and hence $1+i\tau = \sec(\theta)e^{i\theta}$, may help.
I dug into evaluating the limit a bit, and I don't think L'Hospital's rule can be used here.  I could be wrong.
Update after applying hypergeometric transform and evaluating limits
After applying the Pfaff transform to the hypergeometric functions in the expression and evaluating the limits as $\tau$ goes to $\pm \infty$, I get a surprising cancellation of the $\sinh(\pi)$ term, and I arrive at:
$$z_c = \dfrac{i}{6 \cdot 2^i}\left[{}_2F_1(-1+i,-3-i;i;1) + 2_2F_1(i,-2-i;1+i;1)\right]$$
which after applying
$${}_2F_1(a,b;c;1) = \dfrac{\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)}$$
yields
$$z_c = \dfrac{i}{6 \cdot 2^i}\left[\dfrac{\Gamma(i)\Gamma(4+i)}{\Gamma(1)\Gamma(3+2i)} +2 \cdot \dfrac{\Gamma(1+i)\Gamma(3+i)}{\Gamma(1)\Gamma(3+2i)}\right]$$
or, after some manipulation:
$$z_c = \dfrac{\sqrt{2}}{2}\cdot e^{i\left(\frac{3\pi}{4}-\ln{2}\right)}\cdot\dfrac{\Gamma(i)\Gamma(3+i)}{\Gamma(3+2i)}$$
$$z_c = \dfrac{1}{\sqrt{2}}\cdot 2^{-i}\left(-\dfrac{\sqrt{2}}{2}+\dfrac{\sqrt{2}}{2}i\right)\cdot\dfrac{\Gamma(i)\Gamma(3+i)}{\Gamma(3+2i)}$$
$$z_c = \dfrac{1}{\sqrt{2}}\cdot 2^{-i}\left(-\dfrac{\sqrt{2}}{2}+\dfrac{\sqrt{2}}{2}i\right)\cdot \mathrm {B}(i, 3+i)$$
Simplifying the $\Gamma()$ functions a bit more:
$$z_c = \dfrac{1}{4\sqrt{2}}\cdot2^{-i}\left(\dfrac{3}{5}+\dfrac{4}{5}i\right)\left(\dfrac{\sqrt{2}}{2}+\dfrac{\sqrt{2}}{2}i\right) \cdot \dfrac{\Gamma(i)^2}{\Gamma(2i)}$$
(A radius, 3 rotations, and then a ratio of $\Gamma()$ functions with purely imaginary arguments.)
