Are there explicit forms of these integrals Let $a>0$, and $z_0=r_0e^{i\theta_0}$, where $0<r_0<a$, $0<\theta<\gamma<\frac{\pi}{2}$, do we have a closed form of each of the following  integrals
$$
I_1(r_0,\theta_0)=\int_{0}^{a}{\frac{r}{|r-z_0|}}dr,
$$
$$
I_2(r_0,\theta_0,\gamma)=\int_{0}^{a}{\frac{r^{-\beta}}{|re^{i\gamma}-z_0|}}dr,~~~\beta>0,
$$
and
$$
I_3(r_0,\theta_0,\gamma)=\int_{0}^{\gamma}{\frac{(\cos\theta)^{-\beta}}{|ae^{i\theta}-z_0|}}d\theta ~~?
$$
The closed form (if existed) may look very "ugly", especially the second and the third one, what I really want is to obtain upper bound for these integrals by a relatively simple function, say an elementary function of $r_0$ and $\cos\theta_0$.
Thanks in advance.
Edit: The motivation is related to this question I asked earlier on this site, see estimating a particular analytic function on a bounded sector.. In the above, I want to use the Cauchy integral formula to get an estimate, but it seems that this is not enough to get the desired result...
 A: The first one, $I_1$, looks very innocent, but it gets a little messy.  Still, it is very doable.  Begin by noting that
$$\left |r-r_0 e^{i \theta_0} \right|^2 = r^2 - 2 r_0 r \cos{\theta_0} + r_0^2 $$
so the integral may be written as
$$\int_0^a dr \frac{r}{\sqrt{\left ( r-r_0 \cos{\theta_0} \right)^2 + r_0^2 \sin^2{\theta_0}}} $$
This may be rewritten in a form that allows us to recognize antiderivatives: sub $r=\rho + r_0 \cos{\theta_0}$ to get
$$\int_{-r_0 \cos{\theta_0}}^{a-r_0 \cos{\theta_0}} d\rho \frac{\rho+r_0 \cos{\theta_0}}{\sqrt{\rho^2+r_0^2 \sin^2{\theta_0}}} $$
Split the integral apart; each piece has a well-known antiderivative.  I trust you can find these.  The result I get after evaluation is
$$\frac12 \left (\sqrt{a^2+r_0^2-2 a r_0 \cos{\theta_0}}-r_0 \right ) + r_0 \cos{\theta_0} \log{\left [\frac{a-r_0 \cos{\theta_0} + \sqrt{a^2+r_0^2-2 a r_0 \cos{\theta_0}}}{r_0 (1-\cos{\theta_0})}\right ]}$$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\Li}[1]{\,{\rm Li}_{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
'For the time being' the best we can do is to reduce the $\ds{\,{\rm I_{3}}}$
integral to a series which involve Legendre Polynomial integrals. I 'guess' there is not any further reduction:
\begin{align}
\color{#66f}{\large\,{\rm I_{3}}\pars{r_{0},\theta_{0},\gamma}}
&=\int_{0}^{\gamma}
\frac{\cos^{-\beta}\pars{\theta}}{\verts{a\expo{\ic\theta} - z_{0}}}\,\dd\theta
\\[5mm]&=\int_{0}^{\gamma}
\frac{\cos^{-\beta}\pars{\theta}}
{\root{\bracks{a\cos\pars{\theta} - r_{0}\cos\pars{\theta_{0}}}^{\, 2} + \bracks{a\sin\pars{\theta} - r_{0}\sin\pars{\theta_{0}}}^{\, 2}}}\,\dd\theta
\\[5mm]&=\int_{0}^{\gamma}
\frac{\cos^{-\beta}\pars{\theta}}
{\root{a^{2} -2ar_{0}\cos\pars{\theta - \theta_{0}} + r_{0}^{2}}}\,\dd\theta
\\[5mm]&={1 \over \verts{a}}\int_{-\theta_{0}}^{\gamma - \theta_{0}}
\frac{\cos^{-\beta}\pars{\theta + \theta_{0}}}
{\root{1 -2\pars{r_{0}/a}\cos\pars{\theta} + \pars{r_{0}/a}^{2}}}\,\dd\theta
\end{align}
Now, we insert the Legendre Polynomial $\ds{\,{\rm P}_{\ell}}$ Generating  Function
$$\frac{1}{\root{1 - 2xh + h^{2}}}
=\sum_{\ell=0}^{\infty}h^{\ell}\,{\rm P}_{\ell}\pars{x}\,,\qquad\verts{h} < 1
$$
such that
\begin{align}
\color{#66f}{\large\,{\rm I_{3}}\pars{r_{0},\theta_{0},\gamma}}
&=\color{#66f}{\large%
{1 \over \verts{a}}\sum_{\ell=0}^{\infty}\pars{\frac{r_{0}}{a}}^{\ell}
\int_{-\theta_{0}}^{\gamma - \theta_{0}}\cos^{-\beta}\pars{\theta + \theta_{0}}
\,{\rm P}_{\ell}\pars{\cos\pars{\theta}}\,\dd\theta}
\end{align}
