I want to ask how to integrate $\int_{B_r(0)} \dfrac{1}{\sqrt{|x|}|x-a|^2}dx$ where $x \in \mathbb{R}^2$, does it has a explicit solution? or some estimate?
Thank you!!
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There should be 2 answers for $a$ in or not in the disk – mnmn1993 Jun 15 '16 at 14:36
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Thank a lot! I think I need to change to polar coordinte to apply your hint , but how can I deal with the term $|x-a|^2$? – mnmn1993 Jun 15 '16 at 14:56
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OH....So sorry for my poor writing, $|x-a|$ means the 2- norm of the vector $x-a$ , both of $x$ and $a$ are in $\mathbb{R}^2$ – mnmn1993 Jun 15 '16 at 15:03
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I think there is no square root in the bottom , and how to integrate this ? – mnmn1993 Jun 16 '16 at 01:41
1 Answers
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Let $I$ be the integral given by
$$I=\int_{-\pi}^\pi \int_0^r \frac{\sqrt{\rho}}{\rho^2-2|\vec a|\rho \cos(\phi)+|\vec a|^2}\,d\rho \,d\phi$$
Now, we note that the integral diverges for $|\vec a|<r$. Therefore, we assume that $|\vec a|>r$.
We can evaluate the iterated integral over $\phi$ by either using the Tangent Half-Angle Substitution or contour integration. Proceeding, we find
$$\begin{align} \int_{-\pi}^\pi \frac{1}{\rho^2-2|\vec a|\rho \cos(\phi)+|\vec a|^2} \,d\phi&=2\int_{0}^\pi \frac{1}{\rho^2-2|\vec a|\rho \cos(\phi)+|\vec a|^2} \,d\phi\\\\ &=4\int_0^\infty \frac{1}{(\rho^2+|\vec a|^2)(1+t^2)-2|\vec a|\rho (1-t^2)}\,dt\\\\ &=4\int_0^\infty \frac{1}{(|\vec a|+\rho)^2t^2+(|\vec a|-\rho)^2}\,dt\\\\ &=\frac{2\pi}{|\vec a|^2-\rho^2} \end{align}$$
Then, the integral of interest, $I$, reduces to
$$\begin{align} I&=2\pi \int_0^r \frac{\sqrt{\rho}}{|\vec a|^2-\rho^2}\,d\rho\\\\ &=\bbox[5px,border:2px solid #C0A000]{\frac{2\pi}{\sqrt{|\vec a|}} \left(\text{arctanh}\left(\sqrt{\frac{r}{|\vec a|}}\right)-\arctan\left(\sqrt{\frac{r}{|\vec a|}}\right)\right) } \tag 2 \end{align}$$
where we arrived at $(2)$ by enforcing the substitution $\rho \to \rho^2$, and following that with partial fraction expansion.
Mark Viola
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