Calculating $\int_{t=0}^{2 \pi} \int_{r=0}^{2 \rho} \log \left(\frac{r}{2 \rho} \right)r \, dr dt$

Question

I am having trouble calculating an integral from Ransford's Potential Theory in the Complex Plane (Corollary 3.2.4). The equality for $\rho >0$ \begin{align} \int_{t=0}^{2 \pi} \int_{r=0}^{2 \rho} \log \left(\frac{r}{2 \rho} \right)r \, dr dt =-2 \pi \rho ^2 \end{align} is stated. Can someone help me with this calculation?

Answer 1

The $r$ integral can be simplified to the following using a substitution $r=2\rho u$. We can then compute the integral by parts.$$\begin{align}\int_0^{1}u\log u\,\mathrm du&=\left[\frac12u^2\log u\right]_0^1-\int_0^1\frac12u\,\mathrm du\\&=0-\left[\frac14 u^2\right]_0^1=-\frac14\end{align}$$ So your answer is $$\int_0^{2\pi}\,\mathrm dt\int_0^1(2\rho)^2u\log u\,\mathrm du=(2\pi)(2\rho)^2(-1/4)=-2\pi\rho^2$$