so i have the same problem as here, but the answers are not sufficient.

How to integrate


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 $\delta(x-y)$: $$\delta(x-y) = \delta(v\sqrt2) = \frac{1}{\sqrt2} \delta(v).$$

  • $\begingroup$ Can you explain this? The diagonal is x = y, why then has the delta function this factor? $\endgroup$ – cs912381 Jun 21 '17 at 19:39
  • $\begingroup$ @cs912381, I've expanded my answer. $\endgroup$ – md2perpe Jun 21 '17 at 20:20
  • $\begingroup$ I think you answered my question in some way. I asked myself why this geometrical approach is not the same as the integral (i maybe was not precise enough). The integral is only the projection of the diagonal in one dimension. If we choose the coordinate system along the diagonal the projection on the axis (along the diagonal) is the length. $\endgroup$ – cs912381 Jun 22 '17 at 13:00
  • $\begingroup$ I've at least tried to answer your question. I can try to explain it more in words: Consider $\delta(x-y)$ as a diagonal line with some width $d$. The width it has in the $x$-direction is then not $d$ but $d\sqrt2$. And as you say, the projection of it on the $y$-axis is $a$. In your "geometric approach" you used the same width although in that case you should use the normal width $d$ and the length $a\sqrt2$. $\endgroup$ – md2perpe Jun 22 '17 at 13:53

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