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I'm trying to solve this integral and getting confused. Any help would be much appreciated.

$$\int_{-\infty}^{\infty}dy\int_{-\infty}^{\infty}dx \,e^{-A(x-y)^2}\delta(x-y)$$

Update! In responding to a comment asking to provide some context, I realise I've gotten the actual question wrong and it might not be unreasonable that it was coming back as infinite. So, to explain and update the question:

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 $\langle \langle \xi(x_1,t),\xi(x_2,t_2)\rangle \rangle=\gamma\, \delta(x_1-x_2)\delta(t_1-t_2)$.

I'm trying to work out what the autocorrelation function is for $V$, and so it looks like I should get

\begin{align*} \langle \langle V(x_1,t_1), V(x_2,t_2)\rangle \rangle &= \int \int dy_1 \, dy_2\, e^{-A(x_1-y_1)^2}e^{-A(x_2-y_2)^2}\langle \langle \xi(x_1,t),\xi(x_2,t_2)\rangle \rangle\\ &=\int \int dy_1 \, dy_2\, e^{-A(x_1-y_1)^2}e^{-A(x_2-y_2)^2}\gamma \,\delta(y_1-y_2)\delta(t_1-t_2) \end{align*}

So! This is in fact my problem. Any help very much appreciated.

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    $\begingroup$ It's infinite. Would Mathematics be a better home for this question? $\endgroup$
    – Qmechanic
    Commented Oct 6, 2015 at 14:31
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    $\begingroup$ Isn't it effectively $\int_{-\infty}^\infty 1\, dx$? $\endgroup$ Commented Oct 6, 2015 at 14:33
  • $\begingroup$ Dan, why not expand your question to give the physical background? We might be able to spot why you've ended up with an infinite definite integral. That would also bring your question back on topic. $\endgroup$ Commented Oct 6, 2015 at 14:36
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    $\begingroup$ If you are confused you could set $u=x-y$ and apply the definition of $\delta$: $\int_{\mathbb R} f(u)\delta(u) = f(0)$ (where $f$ is a locally integrable function). $\endgroup$
    – anderstood
    Commented Oct 6, 2015 at 15:01

2 Answers 2

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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.

  1. 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.

  2. 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.

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  • $\begingroup$ Thanks for the response. The disagreement here comes in that I was taking the average autocorrelation function (as implied by the double angle brackets) and you have taken the single value in a given instance, which is indeed a random value. I believe the answer below by anderstood is correct for average values. $\endgroup$
    – Dan Goldwater
    Commented Oct 7, 2015 at 10:06
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\begin{align}&\int \int dy_1 \, dy_2\, e^{-A(x_1-y_1)^2}e^{-A(x_2-y_2)^2}\gamma \,\delta(y_1-y_2)\delta(t_1-t_2) \\ =&\gamma \delta(t_1-t_2)\int e^{-A(x_1-y_1)^2}e^{-A(x_2-y_1)^2}\text{d}y_1 \\ =&\gamma \delta(t_1-t_2)e^{-A({x_1}^2+{x_2}^2)}\int e^{2Ay(2(x_1+x_2)-y)}\text{d}y \end{align}

And if $A>0$, $$\int e^{2Ay(x-y)}\text{d}y=\sqrt{\dfrac{\pi}{2A}}\exp{\dfrac{Ax^2}{2}}$$

so in the end:

$$\langle\langle V(x_1,t_1),V(x_2,t_2)\rangle\rangle=\sqrt{\dfrac{\pi}{2A}}\gamma\delta(t_1-t_2)e^{-A({x_1}^2+{x_2}^2)}e^{A(x_1+x_2)^2/2}$$

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