Integration over delta distribution

[lavidaenrosa]

Homework Statement

a) ∫δ( x² cos x) dx with x from -∞ to +∞
b) Integral over C, with C = unit circle
∫_C δ(α-α₀) a ds
a =(1,1)

Homework Equations

a) δ(g(x)) =∑δ(x-x_i)/|g'(x_i)|
b)?

The Attempt at a Solution

a) g(x) = x² cosx
g'(x) = 2x cos x - x² sin x
x_i = 0 and (2n+1)π/2
g'(0)=0 then what is 1/|g'(0)|??

 

[andrewkirk]

For (a), why are you calculating $g'(x)$? The integral is simply equal to 1 if $x^2\cos x$ takes the value zero for any $x\in(-\infty,\infty)$ and 0 otherwise.

The same applies to (b). Assuming you are using $\alpha$ to represent the angle in a polar representation of the number plane ($\theta$ would be more conventional, by the way), is there any value of $\alpha$ in $[0,2\pi)$ that is equal to $\alpha_0$?

 

[lavidaenrosa]

Thx, but I don't get it! Is there a way to show this? Or at least an explanation? I know the relation ∫δ(x-a) dx= θ(x-a) but I've never seen

∫δ(g(x)-a) dx =

• 0 g(x) ≠a
  
• 1 g(x)=a

 

[Ray Vickson]

Thx, but I don't get it! Is there a way to show this? Or at least an explanation? I know the relation ∫δ(x-a) dx= θ(x-a) but I've never seen

∫δ(g(x)-a) dx =

• 0 g(x) ≠a
  
• 1 g(x)=a

There is a serious issue with (a), but first, let's simplify it. Around the positive zero $x_n = (2n+1)\pi/2$ of $f(x) = x^2 \cos(x)$ we have $f(x) \approx f'(x_n) (x-x_n)$, so integration of $\delta(f(x))$ near there gives just $1/f'(x_n)$, as you said. However, since f(x) is even, the corresponding term for $x = -x_n$ has $f'(-x_n) = -f'(x_n)$, so the contributions from $x = x_n$ and $x = -x_n$ cancel.

So, all that matters is the value of $\int_{-a}^a \delta(f(x)) \, dx$ for small $a > 0$, and this is where we run into trouble. For small $|x|$, $f(x)$ looks like just plain $x^2$, but the point is that its graph does not cross the x-axis; it grazes the x-axis from above, so one is essentially looking at $2\int_0^a \delta(f(x)) \, dx$. This integrates only half of the $\delta$ function, so can either be regarded as non-existent, or can be looked at through a limiting procedure that gives $\delta(\cdot)$ in the limit. If you regard $\delta(u)$ as a limit of even functions $f_n(u)$, then your "half-$\delta$" integral would be one half of the usual. However, one can equally well regard $\delta(u)$ as a limit of unsymmetrical (not even) functions $f_n(u)$, in which case the "half-$\delta$" integral can be a fraction other than 1/2---basically, you can make it equal anything from 0 to 1 if you want to.

However, let's not overlook the fact that f(0) = 0 in your case.

 

[lavidaenrosa]

ok, thx, and how do I take into account that f(0)=0?

And b)? I have still no clue

 

[mfb]

For (a), why are you calculating $g'(x)$? The integral is simply equal to 1 if $x^2\cos x$ takes the value zero for any $x\in(-\infty,\infty)$ and 0 otherwise.

This is not true, and I don't see how you got that.Another issue with (a) are the zeros of the cosine. We get a sum over 1/n which does not converge. You can find limits where the contributions cancel in a nice way, but that is an arbitrary limit procedure, and different ways would lead to different results

And b)? I have still no clue

You can convert it to a one-dimensional integral and solve that.
(b) doesn't have the issues (a) has.

 

[Ray Vickson]

This is not true, and I don't see how you got that.Another issue with (a) are the zeros of the cosine. We get a sum over 1/n which does not converge. You can find limits where the contributions cancel in a nice way, but that is an arbitrary limit procedure, and different ways would lead to different results

You can convert it to a one-dimensional integral and solve that.
(b) doesn't have the issues (a) has.

In (a) we get a convergent series. At $x = x_n = (2n+1)\pi/2$ we get the contribution $1/f'(x_n)$, where $f(x) = x^2 \cos(x)$. Since $f'(x_n) = -x_n^2 \sin(x_n) + 2 x_n \cos(x_n) = (-1)^n (\pi /2)^2 (2n+1)^2$ we get the convergent series $\sum c(-1)^n/(2n+1)^2$. Then, as I said in #4, the terms from positive $x_n$ are canceled by those at $-x_n$, so the whole thing comes out as 0 (except, perhaps, for the part from $x = 0$).

 

[mfb]

Ah right, the sine term contributes, not the cosine. Sorry, my mistake. That just leaves the weird x=0 thing.

 

[Ray Vickson]

ok, thx, and how do I take into account that f(0)=0?

And b)? I have still no clue

For (a): you tell me.

 

[andrewkirk]

Perhaps I have misunderstood your question. When you write:

a) ∫δ( x² cos x) dx with x from -∞ to +∞

do you mean the definite integral
$$\int_{-\infty}^\infty\delta(x^2\cos x)\,dx$$
or are you looking for an indefinite integral?

 

[lavidaenrosa]

Andrewkirk: Yes, right, that's the one.
Ray vickson, I get the same series but still have this x=0 problem.
I still don't know what is the correct answer to this cuestion a)

b) I tried to convert it in an one-dimensional integral with polar coord.:
$$\int_C \delta (\vec{\alpha_0}-\vec{\alpha}) (1,1) d\vec{s} = \int_0^1 \int_0^{2 \pi}\delta (\vec{\alpha_0}-\vec{\alpha}) d\alpha dr = \int_0^{2 \pi}\delta (\vec{\alpha_0}-\vec{\alpha}) d\alpha$$... and again without knowing if this is correct and how to proceed.