2D Integrating With Quadratic Arg. of Delta Function

  • #1
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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?
 

Answers and Replies

  • #2
Ray Vickson
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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
18
1
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
Ray Vickson
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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
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1
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
Ray Vickson
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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?)
 

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