2D Integrating With Quadratic Arg. of Delta Function

In summary: I am trying to say is that I suspect we are chasing a red herring.)In summary, there is difficulty in integrating the given 2D integral due to the presence of poles from the delta function. Various approaches, such as polar coordinates or hyperbolic coordinates, have been tried but all encounter problems. It is not clear if the integral is well-defined and if a Cauchy Principal value interpretation would be justified.
  • #1
junt
18
1

Homework Statement


I have a 2D integral that contains a delta function:

##\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\exp{-((x_2-x_1)^2)+(a x_2^2+b x_1^2-c x_2+d x_1+e))}\delta(p x_1^2-q x_2^2) dx_1 dx_2##,

where ##x_1## and ##x_2## are variables, and a,b,c,d,e,p and q are some real constants.

Homework Equations


How can one integrate this without resulting in poles. Because integral over, for example, ##x_1## leads to a pole, given the property of delta function.

The Attempt at a Solution


If inside the delta function, the argument would be ##x_1^2+x_2^2##, and not ##x_1^2-x_2^2##, then one could go to polar coordinates, and get rid of the poles coming from the delta function integral. Sadly, this is not the case here.
Further one can get rid of pole by going into hyperbolic coordinates ##x_1=RCosh(\theta)## and ##x_2=RSinh(\theta)##. The R integral can be done exactly because the poles cancel out due to Jacobian. However, the integral over ##\theta## then becomes unbounded. So this method also has a problem. Is there a smarter way to perform integral with quadratic argument of delta function without encountering poles?
 
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  • #2
junt said:

Homework Statement


I have a 2D integral that contains a delta function:

##\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\exp{-((x_2-x_1)^2)+(a x_2^2+b x_1^2-c x_2+d x_1+e))}\delta(p x_1^2-q x_2^2) dx_1 dx_2##,

where ##x_1## and ##x_2## are variables, and a,b,c,d,e,p and q are some real constants.

Homework Equations


How can one integrate this without resulting in poles. Because integral over, for example, ##x_1## leads to a pole, given the property of delta function.

The Attempt at a Solution


If inside the delta function, the argument would be ##x_1^2+x_2^2##, and not ##x_1^2-x_2^2##, then one could go to polar coordinates, and get rid of the poles coming from the delta function integral. Sadly, this is not the case here.
Further one can get rid of pole by going into hyperbolic coordinates ##x_1=RCosh(\theta)## and ##x_2=RSinh(\theta)##. The R 2 integral can be done exactly because the poles cancel out due to Jacobian. However, the integral over ##\theta## then becomes unbounded. So this method also has a problem. Is there a smarter way to perform integral with quadratic argument of delta function without encountering poles?

If ##p,q > 0## we have ##px_1^2 - q x_2^2 = ( \sqrt{p} x_1 -\sqrt{q} x_2)(\sqrt{p} x_1 + \sqrt{q} x_2)##, so near ##\sqrt{p} x_1 = \sqrt{q} x_2## the ##\delta##-function looks like ##\delta( (\sqrt{p} x_1 - \sqrt{q} x_2) 2 \sqrt{q} x_2)##, but near ##\sqrt{p}x_1 = -\sqrt{q} x_2## it looks like ##\delta((\sqrt{p} x_1 + \sqrt{q} x_2) (-2 \sqrt{q} x_2)##. That makes the ##x_1##-integration easy.
 
  • #3
Ray Vickson said:
If ##p,q > 0## we have ##px_1^2 - q x_2^2 = ( \sqrt{p} x_1 -\sqrt{q} x_2)(\sqrt{p} x_1 + \sqrt{q} x_2)##, so near ##\sqrt{p} x_1 = \sqrt{q} x_2## the ##\delta##-function looks like ##\delta( (\sqrt{p} x_1 - \sqrt{q} x_2) 2 \sqrt{q} x_2)##, but near ##\sqrt{p}x_1 = -\sqrt{q} x_2## it looks like ##\delta((\sqrt{p} x_1 + \sqrt{q} x_2) (-2 \sqrt{q} x_2)##. That makes the ##x_1##-integration easy.
What if, I modify my argument of the delta function a little bit, which looks as follows:

##\delta(px_1^2 - q x_2^2 +r x_1-s x_2+\epsilon)##, where r, s and ##\epsilon## are again some constants.

Does your argument still hold?
In fact, I just noticed that, your arrangement of the argument still doesn't solve the pole problem. Since ##\frac{d}{dx_1}((\sqrt{p} x_1 + \sqrt{q} x_2) (-2 \sqrt{q} x_2)## is ##x_2##, which basically becomes the denominator of the integrand.
 
  • #4
junt said:
What if, I modify my argument of the delta function a little bit, which looks as follows:

##\delta(px_1^2 - q x_2^2 +r x_1-s x_2+\epsilon)##, where r, s and ##\epsilon## are again some constants.

Does your argument still hold?
In fact, I just noticed that, your arrangement of the argument still doesn't solve the pole problem. Since ##\frac{d}{dx_1}((\sqrt{p} x_1 + \sqrt{q} x_2) (-2 \sqrt{q} x_2)## is ##x_2##, which basically becomes the denominator of the integrand.

You rare asking a new question, so it should go in a new thread. Anyway, you can certainly try to answer it for yourself, perhaps using the old response as aI have given a hint, and that is just about all the PF rules allow me to do.
 
  • #5
Ray Vickson said:
You rare asking a new question, so it should go in a new thread. Anyway, you can certainly try to answer it for yourself, perhaps using the old response as aI have given a hint, and that is just about all the PF rules allow me to do.
Thanks. But I think it is safe to say that your argument still leads to pole, which is linear in ##x_2##. Thanks though for replying
 
  • #6
junt said:
Thanks. But I think it is safe to say that your argument still leads to pole, which is linear in ##x_2##. Thanks though for replying

I am not sure the integral is well-defined. To simplify the discussion, let's change variables to ##u = \sqrt{p} x_1 - \sqrt{q} x_2## and ##v = \sqrt{p} x_1 + \sqrt{q} x_2##. Changing variables gives
$$\int \int F(u,v) \delta(uv) \frac{du dv}{2 \sqrt{pq}}$$
Along the line ##u = 0## we have ##\delta(uv) = (1/v) \delta(u)##, which is OK as long as ##v \neq 0##. Similarly, along the line ##v = 0## we get ##(1/u) \delta(v)##, and that is OK as long as ##u \neq 0##. The trouble happens where the two lines ##u = 0## and ##v = 0## intersect. I am not sure the ##\delta## even makes sense at such a point. As you say, you get poles or order 1, but the ##F##-part of the integrand is not zero at the pole locations, so you do not get cancellation. As an ordinary integral the result may not exist, but perhaps it does exist a some type of Cauchy Principal value. (But what would be the justification for such a re-interpretation if it did happen to work?)
 

1. What is a 2D integrating with quadratic arg. of delta function?

A 2D integrating with quadratic arg. of delta function is a mathematical tool used in physics and engineering to analyze and solve problems involving two-dimensional systems. It involves integrating a function over a two-dimensional space with a quadratic argument of the delta function, which is a mathematical representation of an infinitely narrow and infinitely tall spike at a specific point.

2. What is the purpose of using a 2D integral with quadratic arg. of delta function?

The purpose of using a 2D integrating with quadratic arg. of delta function is to simplify and solve complex problems involving two-dimensional systems. It allows for the manipulation of functions and equations to obtain useful information and insights about the system being studied.

3. How is a 2D integral with quadratic arg. of delta function calculated?

A 2D integral with quadratic arg. of delta function is calculated by first expressing the function being integrated in terms of the delta function. Then, the delta function is expanded using its quadratic argument, and the integral is evaluated using standard integration techniques.

4. What are some applications of 2D integrating with quadratic arg. of delta function?

2D integrating with quadratic arg. of delta function has various applications in physics and engineering, such as in solving problems involving electric and magnetic fields, fluid dynamics, and heat transfer. It is also used in signal processing and image analysis.

5. Are there any limitations to using 2D integrating with quadratic arg. of delta function?

Like any mathematical tool, 2D integrating with quadratic arg. of delta function has its limitations. It may not be applicable to all problems, and the results obtained may not always be physically meaningful. It is important to understand the assumptions and constraints of using this method and to use it appropriately in the appropriate context.

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