Integration of delta function over two variables

[binbagsss]

Homework Statement

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 $\int dy$, $\int dx$.

Homework Equations

see above

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?

Many thanks in advance.

 

[PeroK]

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

 

[binbagsss]

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

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 ?

 

[PeroK]

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 ?

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.

 

[binbagsss]

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.

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

 

[PeroK]

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

Is this a definite or an indefinite integral?

 

[binbagsss]

Is this a definite or an indefinite integral?

indefinite

 

[PeroK]

indefinite

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 $y^2$?

 

[nrqed]

Homework Statement

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 $\int dy$, $\int dx$.

Homework Equations

see above

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?

Many thanks in advance.

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.

 

[PeroK]

I found this, which looks useful!

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

 

[binbagsss]

I found this, which looks useful!

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

to define a delta identity don't you have to define it within an integral, so I am guessing this holds over $dy$ and not $dy^{2}$ ?

 

[binbagsss]

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.

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

 

[Ray Vickson]

Homework Statement

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 $\int dy$, $\int dx$.

Homework Equations

see above

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?

Many thanks in advance.

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.

 

[binbagsss]

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 $y^2$?

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.

 

[binbagsss]

to define a delta identity don't you have to define it within an integral, so I am guessing this holds over $dy$ and not $dy^{2}$ ?

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

 

[vela]

indefinite

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?

 

[PeroK]

to define a delta identity don't you have to define it within an integral, so I am guessing this holds over $dy$ and not $dy^{2}$ ?

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.

 

[binbagsss]

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

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 $x$ ?

 

[PeroK]

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 $x$ ?

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.

 

[Ray Vickson]

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 $x$ ?

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

 

[binbagsss]

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

mate I don't.
Ive done it that way and got $\pi$, I am trying to check my understanding of things in general by doing it both ways...

 

[PeroK]

mate I don't.
Ive done it that way and got $\pi$, I am trying to check my understanding of things in general by doing it both ways...

Both the simple substitution $u = y^2$ or, better, the identity I posted in post #10 look easier alternatives. Especially given $f$ has two zeros in this case.

 

[Ray Vickson]

mate I don't.
Ive done it that way and got $\pi$, I am trying to check my understanding of things in general by doing it both ways...

It's easier if you write $\delta(x^2+y^2-r^2)$ instead of $\delta(x^2+y^2-E)$. So, in the $r^2$ version, if $|x| < r$ you can use the result of Perok in Post #10, to write (for example)
$$\delta(y^2 -(r^2-x^2)) = \frac{1}{2\sqrt{r^2-x^2}} \left[ \delta(y - \sqrt{r^2-x^2}) + \delta(y + \sqrt{r^2-x^2}) \right] $$.
If you are integrating over all of $\mathbb{R}^2$ your $y$-integration will go through both points $y = \pm \sqrt{r^2-x^2}$, (when $|x| < r$) so the $y$-integration produces
$$y-\text{integral} = \begin{cases} \frac{2}{2 \sqrt{r^2-x^2}},& \text{if} \; -r < x < r \\
0, & \text{if} \; |x| > r
\end{cases}
$$
You should think about why the second case happens.

Now do the $x$-integration.

 

[vela]

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 $x$ ?

Yes.