# 2D Integrating With Quadratic Arg. of Delta Function

## 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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Ray Vickson
Homework Helper
Dearly Missed

## 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.

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.

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.

Ray Vickson
Homework Helper
Dearly Missed
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.

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.

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

Ray Vickson
$$\int \int F(u,v) \delta(uv) \frac{du dv}{2 \sqrt{pq}}$$