Integration of delta function over two variables

1. Dec 20, 2016

binbagsss

1. The problem statement, all variables and given/known data

I have
$\int dx \int dy \delta (x^{2}+y^{2}-E)$ [1]

I have only seen expressions integrating over $\delta$ where the $x$ or the $y$ appear seperately as well as in the delta function and so you can just replace e.g $y^2 = - x^{2} +E$ then integrate over $\int dy$, $\int dx$.

2. Relevant equations

see above

3. The attempt at a solution

I am unsure how to do this, here is my working so far:

$x^{2}=E-y^{2}$

=>

$2x dx = E - 2y dy$
$2(E-y^{2})^{1/2} dx = E-2y dy$
$dx = \frac{E}{ 2(E-y^{2})^{1/2}} - \frac{y}{ (E-y^{2})^{1/2}} dy$
$dx = \frac{E}{ 2(E-y^{2})^{1/2}} - (E-y^{2})^{1/2}$

So can this step in affect replace the integrating over $x$ part of the delta function, to replace $dx$ so that

[1] reduces to $\int dy \frac{E}{ 2(E-y^{2})^{1/2})} - (E-y^{2})^{1/2}$

Am i on the right track?

2. Dec 20, 2016

PeroK

Hint: forget about the second $dx$ integral for a moment. Just focus on the $dy$ integral, where $x$ is effectively a constant.

3. Dec 20, 2016

binbagsss

but i thought this is what i have done by saying replacing$y^{2}=E - x^{2}$ , this is the effect of integrating over $y$ treating $x$ as constant ?

4. Dec 20, 2016

PeroK

If $E$ and $x$ are constants, then settting $y^2 = E - x^2$ effectively sets $y$ as a constant. That makes no sense.

You could try $u = y^2$. That would be a variable substitution.

5. Dec 20, 2016

binbagsss

okay I have no idea then if I have $\int dy y^{2} \delta (y^{2} + k - E)$, the integrating over $y$ part tells me to replace $y^{2}$ with $E-k$, the $k$ playing the role of treating $x$ as a constant, I thought this is all I have done here by saying this is setting $y^{2}=E-x^2?$ (I am then stuck because $y$ does not appear elsewhere in the integral as it does here (or x)

could you give more of a hint please? thanks

6. Dec 20, 2016

PeroK

Is this a definite or an indefinite integral?

7. Dec 20, 2016

binbagsss

indefinite

8. Dec 20, 2016

PeroK

First, it's not clear what effect having $y^2$ inside the delta function has. So, you can't jump to any conclusions about replacing $y^2$ with whatever.

Second, if you had $y$, that approach would work for a definite integral. But, for an indefinite integral you'll get a Heaviside function. The question is what happens to the Heaviside function when you have $y^2$?

9. Dec 20, 2016

nrqed

You know that for a general function of x, we have

$$\delta(f(x)) = \frac{\delta(x-x_0)}{|f'(x_0)|}$$ \
where $x_0$ is the zero of the function f(x). This assumes that f(x) has only one zero and that $f'(x_0)$ is not zero. If there are several zeroes, one must add similar terms all all the zeroes.

10. Dec 20, 2016

PeroK

I found this, which looks useful!

$\delta(y^2 - a^2) = \frac{1}{2|a|} (\delta(y+a) + \delta (y-a))$

11. Dec 20, 2016

binbagsss

to define a delta identity dont you have to define it within an integral, so Im guessing this holds over $dy$ and not $dy^{2}$ ?

12. Dec 20, 2016

binbagsss

how would I approach using this since I have $\delta (f(x,y))$?

13. Dec 20, 2016

Ray Vickson

If your integral is over all of 2-D space--or at least, over a region containing a circle of radius $\sqrt{E}$,centered at the origin--then you can get an answer immediately by switching to polar coordinates: for $x = r \cos \theta, y = r \sin \theta,$ with $r \geq 0$ and $0 \leq \theta \leq 2 \pi$ your integral becomes
$$\int_{r=0}^{\infty} \int_{\theta=0}^{2 \pi} r \delta(r^2 - E) \, dr \, d \theta.$$
We have $r\, dr = \frac{1}{2} d(r^2)$ and $r \geq 0$, so only the positive square root applies.

14. Dec 20, 2016

binbagsss

so it is not true that $\int f(x) \delta (x-y) dx = f(y)$ ? unless I add limits?

Okay, or if I had $dy^{2}$ it would be valid to replace?

I know that the delta function is the derivative of the heaviside function, which is defined to be $0$ if $x<0$ and $1$ if $x>0$ - $\theta(x)$ is.

15. Dec 20, 2016

binbagsss

again I#m unsure how to use this in the case of $delta (x^{2},y^{2})$

16. Dec 20, 2016

vela

Staff Emeritus
Are you sure? I think generally the integrals are definite integrals from $-\infty$ to $\infty$. It's just kind of tedious to write the limits in each time, so it's understood that your integrating across all space. Are you sure the integrals in this problem are indefinite integrals?

17. Dec 21, 2016

PeroK

The delta function is tricky, but not that tricky. This is an identity for the delta function, so you can replace one with the other anywhere (including under an integral sign).

However, as others have pointed out, this is probably a definite integral and by far the best approach is to use polar coordinates as suggested in post #13.

18. Dec 22, 2016

binbagsss

treating the $y$ as a constant, say? to obtain a linear expression in just one of the variables , $x$ here, and then this is enough to apply the delta to 'replace' the $x$ ?

19. Dec 22, 2016

PeroK

It's not clear why you are unwilling to consider a normal integration substitution, such as $u=y^2$ to get the integrand into a usable form.

20. Dec 22, 2016

Ray Vickson

What, exactly do you have against the use of polar coordinates in this problem?