Gaussian integration with delta function

dirac deltagaussian-integralintegrationphysics

An integration is given:

\begin{equation}
I = \int d^2 \vec{x}_1 \, d^2 \vec{x}_2 \, \delta^2 (\vec{x}_1 + \vec{x}_2 – \vec{x}) f_1(\vec{x}_1) f_2(\vec{x}_2)
\tag{1}
\end{equation}

Where we can assume that; the average of the $\langle x_1^2 \rangle = \langle x_2^2 \rangle$, where the functions are given:

\begin{align*}
f_1(\vec{x}_1) &= \frac{1}{\langle x_1^2 \rangle} e^{- x_1^2 / \langle x_1^2 \rangle} \\
f_2(\vec{x}_2) &= \frac{1}{\langle x_2^2 \rangle} e^{- x_2^2 / \langle x_2^2 \rangle}
\end{align*}

If we multiply the two functions, we get:

$$f_1(\vec{x}_1) f_2(\vec{x}_2) = \frac{1}{\pi^2 \langle x_1^2 \rangle\langle x_2^2 \rangle} e^{-(x_1^2 + x_2^2)/\langle x_1^2 \rangle} $$

Inserting this in the equation $\text{(1)}$ will give, with considering $(\vec{x}_1 + \vec{x}_2 = \vec x )$, which means: $x_1^2 + x_2^2 = x^2+2x_2^2-2x\cdot x_2$ [just vector addition formula]

\begin{equation}
I = \int d^2 \vec{x}_1 \, d^2 \vec{x}_2 \, \delta^2 (\vec{x}_1 + \vec{x}_2 – \vec{x}) \frac{1}{\pi^2 \langle x_1^2 \rangle\langle x_2^2 \rangle} e^{-(x_1^2 + x_2^2)/\langle x_1^2 \rangle}
\tag{2}
\end{equation}

If we integrate with respect to the $d^2 \vec{x_1}$, we get setting $\vec x_1 = \vec x – \vec x_2 $ in equation $\text{(2)}$

\begin{equation}
I = \int d^2 \vec{x}_2 \, \frac{1}{\pi^2 \langle x_1^2 \rangle\langle x_2^2 \rangle} e^{-(x^2+2x_2^2-2x\cdot x_2+ x_2^2)}
\tag{2}
\end{equation}

Am I doing in the right way?

However the result needs to be

$$I = \frac{e^{-\frac{-x^2}{\langle x_1^2 \rangle + \langle x_2^2 \rangle}}}{{\pi(\langle x_1^2 \rangle + \langle x_2^2 \rangle)}}$$

The equation I'm trying to get is 6 in this article: which follows from equation 4.

Best Answer

Given the form of the desired answer, I will drop the assumption $\langle x_1^2\rangle=\langle x_2^2\rangle$ for generality. It looks like you meant to include a $\frac{1}{\pi}$ factor in your definitions, so I'll take $f_j(x_j):=\frac{1}{\pi\langle x_j^2\rangle}\exp\frac{-x_j^2}{\langle x_j^2\rangle}$. Integrating out $x_1$,$$\begin{align}I&=\frac{1}{\pi^2\langle x_1^2\rangle\langle x_2^2\rangle}\int d^2x_2\exp\left(-\frac{(x-x_2)^2}{\langle x_1^2\rangle}-\frac{x_2^2}{\langle x_2^2\rangle}\right)\\&=\frac{1}{\pi^2\langle x_1^2\rangle\langle x_2^2\rangle}\int d^2x_2\exp\left[-\frac{\langle x_1^2\rangle+\langle x_2^2\rangle}{\langle x_1^2\rangle\langle x_2^2\rangle}\left(x_2^2-2\frac{\langle x_2^2\rangle}{\langle x_1^2\rangle+\langle x_2^2\rangle}x\cdot x_2+\frac{\langle x_2^2\rangle}{\langle x_1^2\rangle+\langle x_2^2\rangle}x^2\right) \right].\end{align}$$Next we complete the square with $y:=x_2-\frac{\langle x_2^2\rangle}{\langle x_1^2\rangle+\langle x_2^2\rangle}x$, so$$\begin{align}I&=\frac{1}{\pi^2\langle x_1^2\rangle\langle x_2^2\rangle}\int d^2y\exp\left[-\frac{\langle x_1^2\rangle+\langle x_2^2\rangle}{\langle x_1^2\rangle\langle x_2^2\rangle}\left(y^2+\frac{\langle x_1^2\rangle\langle x_2^2\rangle}{(\langle x_1^2\rangle+\langle x_2^2\rangle)^2}x^2\right)\right]\\&=\frac{1}{\pi(\langle x_1^2\rangle+\langle x_2^2\rangle)}\exp\frac{-x^2}{\langle x_1^2\rangle+\langle x_2^2\rangle},\end{align}$$as required.

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[Math] Gaussian integral with Dirac delta function

I have a Gaussian noise source $V(x,t)=\int d y\, e^{-A (x-y)^2} \xi(x,t)$, where $\xi$ is a Brownian noise, possessing an average of zero $\langle \langle \xi(x,t)\rangle \rangle=0$, and an autocorrelation function $\langle \langle \xi(x_1,t),\xi(x_2,t_2)\rangle \rangle=\gamma\, \delta(x_1-x_2)\delta(t_1-t_2)$.

You must be mistaken, still.

You can rearrange that integral as $V(x, t) = \xi(x, t)\int dy~e^{-A (x - y)^2};$ it cannot fundamentally be a different sort of noise than $\xi$ is, except that it might have a different mean/variance.

Brownian noise does not have a $\delta$ autocorrelation; the (discrete) definition of Brownian noise is (in JavaScript ES6; Python is similar)

function* white_noise() {
    var r, a;
    while (true) { // Box-Muller method
        r = Math.sqrt(-2 * Math.log(Math.random()));
        a = 2 * Math.PI * Math.random();
        yield r * Math.sin(a);
        yield r * Math.cos(a);
    }
}
function* brown_noise(r) { 
    var current = 0, p = Math.sqrt(1 - r*r), white = white_noise();
    while (true) {
        current = r * current + p * white.next().value;
        yield current;
    }
}

and as you can see, it explicitly takes an autocorrelation argument and produces a stream of values which depend on prior history.

[Math] Bessel function representation of Dirac-delta

In response to the questioner's comment: one standard viewpoint is that distributions (or tempered distributions, similarly) are "weak-dual-topology" limits of test functions (or of Schwartz functions), where the weak-dual topology (also called "weak-star topology) is given by semi-norms $\mu_\varphi(u)=|u(\varphi)|$ for test functions (or Schwartz) $\varphi$, at first for other test functions (or Schwartz $\varphi$) and then on the (quasi-) completion, by continuity. Thus, in this viewpoint, the integral representation of $\delta$ is not a literal integral, but is an extension-by-continuity of literal integrals giving approximations to $\delta$. I think this viewpoint, of having integrals not being literal, but extensions-by-continuity, may be most consonant with a common physics-y viewpoint... and does (to my mind) convey a more natural sense of "generalized functions/distributions", not as "functionals", but as an extended notion of function.

Of course, the second viewpoint, perhaps more standard in mathematics per se, is that distributions are continuous linear functionals, and that the integrals written about don't make sense... again, at least not as literal integrals, but perhaps more... haughtily said? That is, the same expressions are obtained, but if/because one is insisting on literal integrals, implicitly disallowing writing integral signs for extensions-by-continuity, some new symbols must be made up. (As you can tell, I am not deeply moved by this viewpoint.)

Either way, the underlying mathematics and mechanisms are the same. That is, e.g., for each Schwartz function $\varphi$, the integrals against the truncated Fourier inversion expression for $\delta$ go to $\varphi(0)$ as the cut-off goes to infinity.

The non-uniformity issue/feature is that there is (provably) no estimate for this limit-approach uniform in $\varphi$. It depends, as it turns out, on $\varphi$.

Is this to the point?

EDIT: I forgot to make explicit the point that, whether or not the integrals are pointwise/numerically valued, literal-or-not, etc., the operator/functional indicated by the integral (as extended by continuity, say) has certain expected properties, so that the usual manipulations which would be correct for numerical/literal integrals are also correct for the non-literal (extended-by-continuity) maps written as integrals... because they extend those integrals, and have the same properties... apart, perhaps, from non-critical numerical/pointwise evaluability.

This is why many physics-y heuristics succeed, after all: what is written is not what is literally meant (from the most exclusionary viewpoint), and it is "still ok" because things extend by continuity, with the same correct transformation properties of whatever-it-is operator denoted by "integral" (and extended from a literal integral).

Yes, on good days, we have our cake and eat it, too. This is (to my mind) the marvelous-ness of mathematics, both internally and in applications.

True, on a bad day bad things happen which will be inexplicable to the (naive?) optimist. Yet, crazily enough, even those bad things can often be repurposed to interesting ends. That's the second miracle in this business, I think.

Edit-edit: following up on the edit to the original question, yes, in that example, for $f$ represented by its Taylor-Maclaurin series in a fixed interval $[-\varepsilon_o,\varepsilon_o]$, as $\varepsilon\to 0$ we can estimate the rate of approach to $f(0)$ uniformly.

But if we look at the aggregate of the classes of such functions over all $\varepsilon_o>0$, this uniform-ness is lost.

Also, although I'd not advocate making this one's first worry, there is an old theorem of E. Borel that says that an arbitrary sequence of real (or complex) numbers can be the Taylor-Maclaurin coefficients at $0$ of a smooth function (obviously, then, not necessarily represented by the infinite series, but the finite-series-with-error-term still works...). This shows, strikingly, that there's no way to hope for true uniformity.

Also, as a sort of paraphrase of the Borel resuls: in the classification of distributions supported at $0$ (essentially proven by Taylor-Maclaurin expansions), we find that they are only the finite linear combinations of $\delta$ and its derivatives. Thus, there is no sequence $c_n$ such that $\sum_{n\ge 1} c_n\cdot \delta^{(n)}$ converges as a distribution.

Best Answer

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