A double integral defined by $$\iint\frac xy dA$$ is integrated over the region of $1<x<3, x<y<2x$. Using different orders of integration produces different results as demonstrated below: $$\begin{align*} \int_1^3\int_x^{2x}\frac xy \ dy \ dx&=\int_1^3[x\ln y]_x^{2x} dx\\ &=\int_1^3x\ln2x - x\ln x\,dx\\ &=\int_1^3x\ln\frac{2x}x\,dx\\ &=\int_1^3x\ln2\,dx\\ &=\ln2\left[\frac {x^2}2\right]_1^3\\ &=\ln2\cdot\frac{3^2-1^2}2\\ &=4\ln2. \end{align*}$$
On the other hand, $$\begin{align*} \int_1^6\int_\frac y2^{y}\frac xy \ dx \ dy &=\int_1^6\left[\frac {x^2}{2y}\right]_{\frac y2}^{y} \ dy\\ &=\int_1^6 \frac y2 - \frac y8 \ dy\\ &=\left(\frac 12 - \frac 18\right)\left[\frac {y^2}2\right]_1^6\\ &=\left(\frac 12 - \frac 18\right)\left(\frac {6^2}2-\frac {1^2}2\right)\\ & =\frac{105}{16}. \end{align*}$$
What is the mistake here? Is this integral somehow divergent over its domain? Or have I interpreted the bounds incorrectly? Thanks in advance!
As Josh and GEdgar pointed out, you should calculate the following three double integrals and sum them.
$\begin{align}\displaystyle\int_1^2\!\mathrm dy\!\int_1^y\frac xy\,\mathrm dx&=\int_1^2\left[\frac{x^2}{2y}\right]_1^y\mathrm dy=\!\int_1^2\left(\frac y2-\frac1{2y}\right)\mathrm dy=\\[3pt]&=\left[\frac{y^2}4-\frac12\ln y\right]_1^2=\frac34-\frac12\ln2\end{align}$
$\displaystyle\int_2^3\mathrm dy\int_{\frac y2}^y\frac xy\,\mathrm dx=\int_2^3\left[\frac{x^2}{2y}\right]_{\frac y2}^y\mathrm dy=\int_2^3\left(\frac y2-\frac y8\right)\mathrm dy=\frac{15}{16}$
$\begin{align}\displaystyle\int_3^6\mathrm dy\int_{\frac y2}^3\frac xy\,\mathrm dx&=\int_3^6\left[\frac{x^2}{2y}\right]_{\frac y2}^3\mathrm dy=\int_3^6\left(\frac9{2y}-\frac y8\right)\mathrm dy=\\[3pt]&=\left[\frac92\ln y-\frac{y^2}{16}\right]_3^6=\frac92\ln2-\frac{27}{16}\end{align}$
If you sum the results of the previous three double integrals, you get $\,4\ln2\,$ which is what you obtained by using the other order of integration.