calculate integral on a domain $D$

Question

I want to calculate $\int_D {{1}\over{x}} \log y $ on $D=\{(x,y) \in R^2: 0<\sqrt y<x<1\}$. Is it an improper integral? $\lim_{\epsilon \rightarrow0} \int_{\epsilon}^{1} \int_{\epsilon}^{x^2} {{1} \over {x}} \log y dy dx$

Answer 1

You're gonna have a hard time if you try to evaluate the integral that way. Try reversing the order of integration to get: $$\lim_{\epsilon \to 0} \int_{\epsilon}^1 \int_{\sqrt{y}}^1 \frac{1}{x} \cdot \log(y)\ dxdy $$ $$\lim_{\epsilon \to 0} \int_{\epsilon}^1 (\log(1) - \log(\sqrt{y})) \cdot \log(y)\ dy $$ $$\lim_{\epsilon \to 0} \int_{\epsilon}^1 -\log(\sqrt{y}) \log(y)\ dy $$ $$-\lim_{\epsilon \to 0} [\frac{1}{2}y(\log^2(y) - 2\log(y) + 2)|_{\epsilon}^1 $$ $$-\lim_{\epsilon \to 0} (1 - \frac{1}{2}\epsilon \log^2(\epsilon) + \epsilon \log(\epsilon) - 2\epsilon) $$

You can verify that $\lim_{\epsilon \to 0} \epsilon\log(\epsilon) = \lim_{\epsilon \to 0} \frac{\log(\epsilon)}{1/\epsilon}= 0$ and $\lim_{\epsilon \to 0} \epsilon\log^2(\epsilon) = \lim_{\epsilon \to 0} \frac{\log^2(\epsilon)}{1/\epsilon}= 0$ using L'Hopital's rule. Hence, $$\int_D \frac{1}{x} \log(y)\ dA = -1 $$