# Double Dirac delta integral

Let $I=\int_{-1}^0\int_{0}^1 \delta(x-y)dxdy$ where $\delta(t)$ is defined as the limit of a symmetric Gaussian pdf. The ranges overlap only on the zero-length range $x=y=0$. Is the result here $0$ or is it something else? Thanks

Using the fourier integral representation, we can write $$\delta(x-y)=\frac{1}{2\pi}\int\limits_{-\infty}^{\infty} \cos(px-py)dp$$

therefore we have $$\frac{1}{2\pi}\int\limits_{-1}^{0} \int\limits_{0}^{1} \int\limits_{-\infty}^{\infty} \cos(px-py)dp dx dy$$ Changing the order of integration , integrate with respect to x first,then y $$\frac{1}{2\pi} \int\limits_{-\infty}^{\infty} \int\limits_{-1}^{0} \frac{\sin(yp)-\sin(yp-p)}{p} dy dp$$

$$\frac{2}{\pi} \int\limits_{-\infty}^{\infty} \frac{\sin^2(p/2)\cos(p)}{p^2} dp$$

For evaluating this integral, use contour integration by considering $$f(z)=\frac{\sin^2(z/2)\cos(z)}{z^2} dz$$

Using Cauchy's theorem $\int_C f(z)dz=0$ $$\int\limits_{-R}^{-r} \frac{\sin^2(x/2)\cos(x)}{x^2}dx+\int_{\gamma}f(z)dz+\int\limits_{r}^{R}\frac{\sin^2(x/2)\cos(x)}{x^2}dx+\int_{\Gamma}f(z)dz=0$$

Evaluation of $\int_{\gamma}f(z)dz$ $$\int_{\gamma}f(z)dz=\int\limits_{-\pi}^{0} \frac{\sin^2(re^{i\theta}/2)\cos(re^{i\theta})}{re^{i\theta}}id\theta$$

As $r \to 0$,the above integral also tends to $0$

Evaluation of $\int_{\Gamma}f(z)dz$

$$\int_{\Gamma}f(z)dz=\int\limits_{0}^{\pi} \frac{\sin^2(Re^{i\theta}/2)\cos(Re^{i\theta})}{Re^{i\theta}}id\theta$$

As $R \to \infty$ , the above integral also tends to $0$. Substituting these values as $r \to 0 and R \to \infty$ in Cauchy theorem ,, we are left with. $$\int\limits_{-\infty}^{0} \frac{\sin^2(x/2)\cos(x)}{x^2}dx+\int\limits_{0}^{\infty} \frac{\sin^2(x/2)\cos(x)}{x^2}dx =0$$

which gives,, $$\int\limits_{-\infty}^{\infty} \frac{\sin^2(x/2)\cos(x)}{x^2}dx=0$$

Thus, $$\boxed{\int\limits_{-1}^{0} \int\limits_{0}^{1} \delta(x-y)dx dy =0}$$

It is zero indeed. Denote $w_h(t)\to\delta(t)$, $h\to+0$, a symmetric Gaussian pdf. Making an orthogonal change of variables $(x,y)\to(u,v)$ in the square $[0,1]\times[-1,0]$ s.t. $u=x-y$ is increasing along the diagonal of the 4th quarter, $v=x+y$, we have $$\int_{-1}^0\int_{0}^1 w_h(x-y)dxdy= \frac12\int_0^2 w_h(u)\int_{|1-u|-1}^{1-|1-u|}\, dvdu=$$ $$\int_0^2(1-|1-u|) w_h(u)\,du\to0,\ h\to+0,$$ because the function $1-|1-u|$ vanishes at $u=0$.

Let us perform first the $y$-integral $$\int_{0}^1 \delta(x-y)dy =1\qquad\text{if }x\in (0,1)\qquad\text{and }0\text{ otherwise}\ .$$ Therefore $$\int_{-1}^0 dx{\bf 1}_{x\in (0,1)}=0\ ,$$ and the final result is zero.

• The $y$ bounds are from $-1$ to $0$, so actually it is $1$ iff $x \in (-1,0)$, which is good, given what the $dx$ limits are...
– Ian
Jun 28, 2016 at 18:39