$$\int_{0}^{1}dx\int_0^1dy \delta (\frac{1}{2}-xy)y^2$$
Solution:
$$\int_{0}^{1}dx\int_0^1dy \delta (\frac{1}{2}-xy)y^2=\int_{0}^{1}dx\int_0^1dy \delta (x(y-\frac{1}{2x})y^2=\int_{0}^{1}\frac{1}{x}dx\int_0^1dy \delta (y-\frac{1}{2x})y^2$$
Up to this point my solution is the same as the official solution, however,
$$\int_{0}^{1}\frac{1}{x}dx \int_{0}^{1} dy \delta(y-\frac{1}{2x})y^2=\int_{1/2}^1 dx\frac{1}{x}(\frac{1}{(2x)^2})$$
Here in the official solution, the lower limit of the integral is $1/2$ and not $0$, which is what i had....Can someone tell me why?
$y$ goes from 0 to 1, so if $y=\frac{1}{2x}$, $x$ must be $\geq \frac{1}{2}$.