Is This Week's POTW Centered on an Anti-Holomorphic Function?

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Here is this week's POTW:

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Show that the complex function

$$F(z) = \frac{1}{\pi}\int_0^1 \int_{-\pi}^\pi \frac{r}{re^{i\theta} + z}\, d\theta\, dr$$

is anti-holomorhpic (i.e., the conjugate $\bar{F}$ is holomorphic) in the open unit disc, $\Bbb D$, and holomoprhic in complement $\Bbb C \setminus \bar{\Bbb D}$ of the closed unit disc.
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No one answered this week's problem. You can read my solution below.

Spoiler

If $z = 0$, then by $F(z) = \frac{1}{\pi} \int_0^1 \int_{-\pi}^\pi e^{-i\theta}\, d\theta\, dr = 0$. For $0 < r < 1$ and $z\in \Bbb C\setminus \{0\}$,

$$\int_{-\pi}^\pi \frac{r}{re^{i\theta} + z}\, d\theta = \oint_{\lvert w\rvert = r} \frac{r}{w + z}\frac{dw}{iw} = \frac{r}{iz}\oint_{\lvert w\rvert = r} \left(\frac{1}{w} - \frac{1}{w + z}\right)\, dw = \frac{r}{iz}\cdot 2\pi i(1 - \delta_{\lvert z\rvert < r}) = \frac{2\pi r}{z}(1 - \delta_{\lvert z\rvert < r}), $$

where $\delta_{\lvert z\rvert < r}$ equals $1$ when $\lvert z\rvert < r$ and $0$ if $\lvert z\rvert > r$. Thus

$$F(z) = \frac{1}{\pi} \int_0^1 \frac{2\pi r}{z}(1 - \delta_{\lvert z \rvert < r})\, dr = \frac{1}{z} \int_0^1 2r(1 - \delta_{\lvert z \rvert < r})\, dr$$

If $0 < \lvert z \rvert < 1$, then

$$F(z) = \frac{1}{z}\int_0^{\lvert z\rvert} 2r(1 - \delta_{\lvert z\rvert < r})\, dr + \frac{1}{z}\int_{\lvert z\rvert}^1 2r(1 - \delta_{\lvert z\rvert < r})\, dr = \frac{1}{z}\int_0^{\lvert z\rvert} 2r\, dr + \frac{1}{z}\int_{\lvert z\rvert}^1 2r(0)\, dr = \frac{\lvert z\rvert^2}{z} = \bar{z}$$

If $\lvert z \rvert > 1$, then

$$F(z) = \frac{1}{z}\int_0^1 2r\, dr = \frac{1}{z}$$

In summary,

$$F(z) = \begin{cases} \bar{z}, & z\in \Bbb D\\\frac{1}{z}, & z\in \Bbb C\setminus \bar{\Bbb D}\end{cases}$$

In particular, $F(z)$ is holomorphic in $\Bbb C \setminus \bar{\Bbb D}$. Since the conjugate of $F(z)$ is $z$ in $\Bbb D$, $F$ is antiholomorphic in $\Bbb D$.