Integral evaluation of delta function with Gaussian

Question

I am trying to evaluate the integral of the form,

$$ \int_0^\infty d\tau \frac{e^{-x^2/(4\tau)}}{\sqrt{4 \pi \tau}} \delta(\tau) $$ Shouldn't this integral be equal to $\delta(x)$, as the integral essentially corresponds to something of the form $$ \lim_{\tau \rightarrow 0} \frac{e^{-x^2/(4\tau)}}{\sqrt{4 \pi \tau}} = \delta(x) $$ which is the definition of Dirac delta function through limit distributions.

But, when I evaluate this integral using Mathematica, I get the answer 0.

Is my above argument wrong or is Mathematica not giving the correct answer for this case.

Answer 1

The use of the symbol $\delta(x)$ by physicists is a calamity, since $ \delta$ is not a function but a measure which should be denoted by $\delta_a(dx)$ meaning $\int_R h(x)\delta_a(dx) =h(a)$ for a continuous function $h$ on $a$. We can indulge $\delta(dx)=\delta_0(dx)$ in your case $a=0.$

Now your integral $$I(x)=\int_0 ^{\infty}d\tau g(x,\tau )\delta (\tau),$$ while understanding that your $\delta(\tau)d\tau$ means the measure $\delta(d\tau)$, your $I(x)$ say, makes no sense since $g(x,0)$ is infinite. This explains craziness in the Mathematica answer.

To save the day, we have to use a sequence of positive functions $d_n(\tau)$ such that for any $h$ continuous we have $\int_R h(\tau)d_n(\tau)d\tau\to h(0)$. Now consider

$$I_n(x)=\int_0^{\infty}g(x,\tau)d_n(\tau)d\tau.$$ We claim that for any $h$ positive and continuous 0n zero we have $$\lim_{n}\int_Rh(x)I_n(x)dx=h(0)/2.\ (*)$$ To see this, we just exchange the integral signs (no problem, everything is positive) and we get

$$ \lim_{n}\int_Rh(x)I_n(x)dx=\lim_{n}\int_Rh(x)\left(\int_0^{\infty}g(x,\tau)d_n(\tau)d\tau\right)dx$$$$=\lim_{n}\int_0^{\infty}\left(\int_R h(x)g(x,\tau)dx\right)d_n(\tau)d\tau=\lim_{n}\int_0^{\infty}\left(\int_R h(z\sqrt{2\tau})g(z,1)dz\right)d_n(\tau)d\tau$$ Since the function $\tau \mapsto \int_R h(z\sqrt{2\tau})g(z,1)dz$ is continuous and since its value on $\tau=0$ is $h(0)\int_0^{\infty}g(z,1)dz=h(0)/2$ then (*) is proved. The 'result' of your nonsense integral $I(x)$ is $\delta(dx)/2.$

Answer 2

If you make the problem smoother and, since the Dirichlet kernel converges to a Dirac delta function, use $$\delta(\tau)=\frac 2 \pi\,\,\underset{a\to \infty }{\text{limit }}\frac{\sin (a \tau )}{\tau }$$ $$I=\frac{1}{\pi ^{3/2}}\int_0^\infty \frac{e^{-\frac{x^2}{4 \tau }} \sin (a \tau )}{\tau ^{3/2}}\,d\tau=\frac 2{\pi x}e^{-x \sqrt{\frac{a}{2}}} \sin \left(x\sqrt{\frac{a}{2}}\right)$$

Reintrodue now $\delta(x)$ from its approximation.

Answer 3

If $\delta$ is defined as a distribution, then the object (you wrote it as an integral) is ill-defined. This is because the product of distributions is not defined in general. Mathematica might be using the right-side limit of the Heaviside function to evaluate this ill-defined object.

Here, let $f_x(\tau)=\frac{e^{-x^2/4\tau}}{\sqrt{4\pi\tau}}$ and $H(\tau)$ denote the Heaviside function. The object written as $\int_0^\infty f_x(\tau)\delta(\tau)\,d\tau$ can be interpreted as the functional $\langle \delta,f_xH \rangle$. Inasmcuh as $f_x(\tau)H(\tau)$ is discontinuous at $\tau=0$, the functional is not defined. That is to say, $f_x(\tau)H(\tau)$ is not a test function.

In fact, the functional $\langle \delta,H\rangle$ is not defined. To see this, let $\delta_n(\tau)$ be a regularization of the Dirac Delta. For any test function $\phi$, we have

$$\lim_{n\to\infty}\int_{-\infty}^\infty \phi(\tau)\delta_n(\tau)\,d\tau=\phi(0)$$

Note that there is no requirement that $\delta_n(\tau)$ be symmetric about the origin. Therefore, the integral $\int_{-\infty}^\infty H(\tau)\delta_n(\tau)\,d\tau=\int_0^\infty \delta_n(\tau)\,d\tau$ need not be equal to $1/2$. In fact, the value of the integral can be any value between $0$ and $1$ (See THIS ANSWER).

Inasmuch as the limit, $\lim_{n\to\infty}\int_{-\infty}^\infty H(\tau)\delta_n(\tau)\,d\tau$ depends on the choice of regularization $\delta_n$, the functional $\langle \delta,H\rangle$ fails to exist.