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It is known that:

$$\int_{-\infty}^\infty f(x) \, \delta(g(x)) \, dx = \sum_{i}\frac{f(x_i)}{|g'(x_i)|}$$

Where $x_i$ are the roots of $g(x)$.

My question is, what happens when $g'(x_i)$ is zero, but $f(x_i)$ is not? It seems that the integral on the left shoul exist irrespective of the value of $g'(x)$, so:

Is there a different formula for the integral one should use in this case, or conversely, is this indeed an indicator that the integral diverges?

Bounty Edit

As user88595 has explained this may be a consequence of $g(x)$ not having a simple root. I'm looking for a proof or a counterexample that for any $g(x)$ which does not have a simple root at $x_i$, the integral diverges.

Edit:

I thought I'd give a concrete example. Let's first look at: $$\int_{-\pi/2}^{\pi/2}e^x \delta\left(\sin(x)\right)dx$$ Since $\sin x$ only has one zero in the interval, and since the derivative at zero is one, the integral is equal to $e^0=1$. Now look at: $$\int_{-\pi/2}^{\pi/2}e^x \delta\left(\sin^3(x)\right)dx$$ Which leads to infinity by the above rule (since $g'(x) = 0$), and indeed diverges. Is this the case in general?

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  • $\begingroup$ You can think of delta as a functional: $\delta[f] = f(0)$. So, it's acting in a space of functions which are defined at $x = 0$. Indeed, you are integrating $\int_{-\pi/2}^{\pi/2}{{\rm e}^{x} \over 3x^{2}}\,\delta\left(x\right)\,{\rm d}x$ but ${{\rm e}^{x} \over 3x^{2}}$ is not defined at $x = 0$. $\endgroup$ – Felix Marin Jan 12 '14 at 9:59
  • $\begingroup$ @Felix - so you are saying that anytime the functions derivative is zero at $x_i$ the integral is divergent? I'd happily accept a proof this is always so as an answer. $\endgroup$ – nbubis Jan 13 '14 at 5:39
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    $\begingroup$ No. I say that the integral is defined for functions which are defined at $x = 0$. The rule the OP is discussing is, in some way, an 'operational trick'. The answer is: 'It's not defined in that case'. In Physics it happens sometimes but for other reasons. For example, some approximation was not correct or the model is neglecting some important fature. In that case, we go back to check it or modify the model. It happens in my PhD thesis and my advisor told me: "Go and check everything because the experimental people won't believe that". $\endgroup$ – Felix Marin Jan 13 '14 at 6:47
  • $\begingroup$ @FelixMarin - I realize the rule "is not correct", hence the question: What is the correct rule in this case, or is the integral (perhaps) always divergent in this case. $\endgroup$ – nbubis Jan 13 '14 at 6:50
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    $\begingroup$ The way I see it is that, if g' is zero for one of g's roots, then g stays zero for a longer time because that root is a stationary point, so the delta function is infinite for a longer time, thus making the integral diverge. I know this isn't rigorous, but I find it a little more intuitive. $\endgroup$ – matias Jan 21 '14 at 0:21
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The statement is true.

Given $g(x)$ everywhere differentiable and $f(x)$ continuous real to real functions defined for all $x$ real numbers. If $f(x_i)\neq 0$, $g'(x_i)=0$ for any $x_i$ that satisfies $g(x_i)=0$ then $\int_{-\infty}^{\infty} f(x) \delta(g(x)) dx $ diverges. The converse is true in a restricted sense: if $f(x_i)\neq 0$, $g'(x_i)\neq 0$ for all $x_i$ that satisfies $g(x_i)=0$ and if there are only finitely many such $x_i$ points then $\int_{-\infty}^{\infty} f(x) \delta(g(x)) dx $ is finite.

The support of the delta function is the infinitesimal neighborhood around $0$, thus we can restrict the integral to the open sets around $x_i$, label them $U_i$.

Consider first the case where $g(.)$ is monotonic around $x_i$. Then there exists an inverse function of $g$ for which $G_i(g(x)) = x$ for $x\in U_i$. Furthermore the image of the open set by a continuous function is an open set, label it $V_i = g[U_i]$.

Then you can write $$\int_{-\infty}^{\infty} f(x) \delta(g(x)) dx =\sum_i \int_{U_i} f(x) \delta(g(x)) dx =\sum_i \int_{V_i} f(G_i(g)) \delta(g) \frac{dG_i}{dg} dg\,.$$ Let us denote $G'_i = \frac{dG_i}{dg}$. From the definition of the delta function $$\sum_i \int_{V_i} f(G_i(g)) G'_i(g) \delta(g) dg = \sum_i f(G_i(0)) G'_i(0)$$.

Thus if $f(x_i) = f(G_i(0)) \neq 0$ then this is finite iff $G'_i(0)$ is finite. However $G'_i(0) = 1/g'(x_i)$ from the inverse function theorem, therefore the expression is finite exactly if $g'(x_i)\neq 0$.

The case where $g(.)$ has a local maximum or minimum at some $x_i$ is a bit more subtle. In this case, there is no inverse, and the previous argument is not applicable. Let us consider the case where $g(.)$ is at least $n$ times differentiable, such that if it has a minimum, $g(x) \rightarrow a x^{2n}$ in an infinitesimal neighborhood $U_i$, where $a> 0$ real and $n\geq 1$ integer. The delta function can be represented formally as the limit of the series $\phi(x) = (2\pi \sigma^2)^{-1/2}\exp(-x^2/2\sigma^2)$ where $\sigma\rightarrow 0$. Then

$$\int_{U_i} f(x) \delta(g(x)) dx = \lim_{\sigma\rightarrow 0}\int_{U_i-x_i}\frac{1}{\sqrt{2\pi \sigma^2}} f(x_i + z)\exp(-z^{4n}/2\sigma^2) dz\,. $$

Since $f(.)$ is continuous you can always find a $c$ in a sufficiently small neighborhood that $f(x_i)+c>f(x)> f(x_i)-c$. Thus we can give a upper/lower bound on the integral by replacing $f(x_i+z)$ with $f(x_i)\pm c$, which you can take out in front of the integral. Now you can show that $$\frac{1}{\sqrt{2\pi \sigma^2}} \int_{-\infty}^{\infty} \exp(-x^{2n}/2\sigma^{2}) dx =\sigma^{(1-n)/n}\times\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \exp(-y^{2n}/2) dy $$ The second term is finite in the limit $\sigma \rightarrow 0$, so this is asymptotically proportional to $\sigma^{(1-n)/n}$. In the limit $\sigma\rightarrow 0$, this goes to infinity. Since, the lower bound of the integral goes to infinity, the integral is divergent.

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  • $\begingroup$ Exactly what I was looking for! I'll award the bounty when I can :) $\endgroup$ – nbubis Jan 21 '14 at 0:30
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You can tame these $\delta$'s by simple subsitution. Your first example:

$$\int_{-\pi/2}^{\pi/2}e^x \delta\left(\sin(x)\right)dx$$

Put $y=\sin x$, so $dy=\cos x dx$ and $dx = dy/\sqrt{1-y^2}$. The integral becomes:

$$\int_{-1}^{1}\frac{e^{sin^{-1}y}}{\sqrt{1-y^2}} \delta(y)dy$$

This is a bit messy, but it doesn't matter, because it is just equal to the value of the integrand at $y=0$, which is $1$.

Your second example:

$$\int_{-\pi/2}^{\pi/2}e^x \delta\left(\sin^3(x)\right)dx$$

Put $y=\sin^3 x$, so $dy=3\sin^2 x \cos x dx$ and $dx = dy/(3\sin^2 x \cos x) = dy/\left(3y^{\frac23}\sqrt{1-y^{\frac23}}\right)$. The integral becomes:

$$\int_{-1}^{1}\frac{e^{sin^{-1}(y^{\frac13})}}{3y^{\frac23}\sqrt{1-y^{\frac23}}} \delta(y)dy$$

But now the value of the integrand at $y=0$ is infinite, because of that $y^{\frac23}$ term. So the integral diverges.

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  • $\begingroup$ I know that, I'm looking for a general proof, that will show the integral diverges for all cases which are not simple poles. $\endgroup$ – nbubis Jan 20 '14 at 17:00
  • $\begingroup$ @nbubis: If you know that, why did you say that the two integrals "intuitively should give the same result"? $\endgroup$ – TonyK Jan 20 '14 at 17:01
  • $\begingroup$ see the bounty text. $\endgroup$ – nbubis Jan 20 '14 at 17:03
  • $\begingroup$ @nbubis: You should edit that into your question. $\endgroup$ – TonyK Jan 20 '14 at 17:15
  • $\begingroup$ Done. Thanks anyway for the answer, it is a helpful pointer in the right direction. $\endgroup$ – nbubis Jan 20 '14 at 17:23
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Indeed this formula doesn't hold if $g'(x_i) = 0$. In fact this formula is only true if $g$ is a continuously differentiable function with $g'$ nowhere zero.

You could also say that $g$ has simple zeros which implies that the function converges to zero linearly.

If $g'(x_i)$ is zero then $g(x_i)$ is not a simple zero and therefore converges to zero quicker than linearly.

My guess (although I can't prove it) is that the formula is finite because the delta converges to infinity "as quickly" as $dx$ converges to $0$ creating something finite. From my reasoning if $g'(x_i) = 0$ then the integral will diverge at that point as you said.

Please tell me if anything doesn't make sense. From my point of view the dirac delta function is not well enough defined argue correctly but then there might be things I don't know about it. It's a function from the dark arts.

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  • $\begingroup$ +1 for "the dark arts" :) but seriously, I think you've hit the nail on the head by stating that $g(x_i)$ is not a simple zero. That being said, I really need the "corrected" formula. $\endgroup$ – nbubis Jan 12 '14 at 9:00
  • $\begingroup$ I'm not sure there is any, everywhere I looked stated $g$ to have simple zeros only. What makes you think it will converge? $\endgroup$ – user88595 Jan 12 '14 at 9:36
  • $\begingroup$ It's a physical multidimensional integral - so I can choose which variable to use. One variable gives a simple pole and another variable gives a non simple pole. Given that the order of integration shouldn't make a difference, I suspect both are actually convergent. $\endgroup$ – nbubis Jan 12 '14 at 9:58
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    $\begingroup$ "this formula is only true if g is a continuously differentiable function with g′ nowhere zero": That's not right. $g'(x)$ is allowed to be zero whenever $g(x)$ is non-zero. $\endgroup$ – TonyK Jan 20 '14 at 16:36

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