# 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 $\cos\theta_0$.

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...

• Did you try Wolfram Integrator ? – 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

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

• 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. – Tomas Jan 16 '15 at 15:58
• @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). – Ron Gordon Jan 16 '15 at 16:01
• @Gordon, thanks for the comment. I will have a try. – Tomas Jan 16 '15 at 16:04
