so i have the same problem as here, but the answers are not sufficient.
How to integrate
$$\int_0^a\int_0^a\delta(x-y)dxdy$$
My first solution would be
$$\int_0^a\int_0^a\delta(x-y)dxdy = \int_0^adx = a = \int_0^ady$$
but then the geometrical approach would result in a path integral over the diagonal from the square with edge length a, so
$$\int_0^{\sqrt{2}a}dx = \sqrt{2}a$$
Where is my error? Thanks!
(I didn't know if i should comment on the old post or make a new one, what would be the right thing to do in the future?)
Your geometric approach actually uses $\sqrt2 \delta(x-y)$ and therefore you get the extra factor.
Consider the following approximation to $\delta(x-y)$: $$D_\epsilon(x,y) = \begin{cases} \frac{1}{2\epsilon} & \text{if $-\epsilon \leq x-y \leq +\epsilon$} \\ 0 & \text{otherwise} \end{cases} $$ This function satisfies $$\int_{-\infty}^{\infty} D_\epsilon(x,y) \, dx = 1.$$
Now we change coordinates using $x = \frac{1}{\sqrt 2}(u+v)$, $y = \frac{1}{\sqrt 2}(u-v)$: $$ \hat D_\epsilon(u,v) = D_\epsilon(x(u,v), y(u,v)) = \begin{cases} \frac{1}{2\epsilon} & \text{if $-\epsilon \leq v\sqrt2 \leq +\epsilon$ i.e. if $-\frac{\epsilon}{\sqrt2} \leq v \leq +\frac{\epsilon}{\sqrt2}$} \\ 0 & \text{otherwise} \end{cases} $$
Notice that in the new coordinates the approximation is thinner by a factor $\sqrt2$, which makes $$\int_{-\infty}^{\infty} \hat D_\epsilon(u,v) \, dv = \frac{1}{\sqrt2}.$$
This carries over to $\delta(x-y)$: $$\delta(x-y) = \delta(v\sqrt2) = \frac{1}{\sqrt2} \delta(v).$$