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...

  • $\begingroup$ Did you try Wolfram Integrator ? $\endgroup$ – Yves Daoust Jan 16 '15 at 11:45

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

$$\frac12 \left (\sqrt{a^2+r_0^2-2 a r_0 \cos{\theta_0}}-r_0 \right ) + r_0 \cos{\theta_0} \log{\left [\frac{a-r_0 \cos{\theta_0} + \sqrt{a^2+r_0^2-2 a r_0 \cos{\theta_0}}}{r_0 (1-\cos{\theta_0})}\right ]}$$

  • $\begingroup$ Thanks very much, I also computed $I_1$, and got the same result as you did. I think $I_3$ is the hardest one to estimate, which I have no idea how to do it. Do you have any idea to get a relatively good bound? Thanks again. $\endgroup$ – Tomas Jan 16 '15 at 15:58
  • 1
    $\begingroup$ @sun: nothing comes to mind. For $I_2$, for example, you can't split the integral as I did for $I_1$. For $I_3$, you;re effectively introducing a factor of $1/\sqrt{a^2-r^2}$ into the integral (roughly speaking). $\endgroup$ – Ron Gordon Jan 16 '15 at 16:01
  • $\begingroup$ @Gordon, thanks for the comment. I will have a try. $\endgroup$ – Tomas Jan 16 '15 at 16:04

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ '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}


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