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$$\int^{\infty}_{0}dx\int^{\infty}_{0}dy \; \delta(\sqrt{y^2-x^2}-a)$$ Here, $$ a>0 $$ and using the Jacobian $$x=x , \quad y=\sqrt{z+x^2}$$ Then, $$\int^{\infty}_{0}dx\int^{\infty}_{0}dy \; \delta(\sqrt{y^2-x^2}-a)$$ $$=\int^{\infty}_{0}dx \int^{\infty}_{-x^2}\frac{dz}{2\sqrt{z+x^2}} \delta(\sqrt{z}-a)$$ $$=\int^{\infty}_{0}dx \left[\int^{\infty}_{0} \frac{dz}{2\sqrt{z+x^2}} \delta(\sqrt{z}-a) +\int^{x^2}_{0} \frac{dz}{2\sqrt{x^2-z}} \delta(i\sqrt{z}-a) \right] $$ Why did this problem come to me??

$$\int^{x^2}_{0} \frac{dz}{2\sqrt{x^2-z}} \delta(i\sqrt{z}-a) = ?? $$

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    $\begingroup$ Why would you want to play with $\delta$ ? It doesn't make sense, until you study the theory of distribution. $\endgroup$
    – reuns
    Dec 21, 2017 at 10:38
  • $\begingroup$ i should solve that calculation, if you can solve that, please teach me... $\endgroup$
    – 정재훈
    Dec 21, 2017 at 11:00
  • $\begingroup$ So first you will need to know what is meant to have a delta function with complex argument. (I don't know what it means, but maybe you do?) $\endgroup$
    – GEdgar
    Dec 21, 2017 at 11:44
  • $\begingroup$ I am sorry to disagree with previous comments, but the issue here has literally nothing to do with the delta function (or distribution). The issue has to do with the fact that we have a square root in the integrand. The question is as ill-posed as asking: 'What is the integral $\int_{-5}^{-3}\sqrt{x}dx$?' ... the integration bounds are inconsistent with the domain of the integrand. And this irrespective of whether we have or not a Dirac delta in the integrand (this is completely irrelevant). $\endgroup$
    – Pierpaolo Vivo
    Dec 21, 2017 at 11:53
  • $\begingroup$ @정재훈 We can't write a course on the theory of distribution just for you. By definition $\int_{-\infty}^\infty \delta(x) \varphi(x)dx = \lim_{n \to \infty} \int_{-\infty}^\infty 2n 1_{|x| < 1/n} \varphi(x)dx$ which converges and $=\varphi(0)$ whenever $\varphi$ is continuous. $\endgroup$
    – reuns
    Dec 21, 2017 at 12:11

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Your initial problem is ill-defined. Since the radicand is $y^2-x^2$, this means that (for $x,y$ positive), $y$ should be larger than $x$. The integral you are after is in all likelihood $$ \int^{\infty}_{0}dx\int^{\infty}_{x}dy \; \delta\left(\sqrt{y^2-x^2}-a\right)\ . $$

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  • $\begingroup$ why? there is no boundary about z. for integration range of x and y, z can be navative sign. $\endgroup$
    – 정재훈
    Dec 21, 2017 at 11:28
  • $\begingroup$ As far as I know, a square root is only real if its radicand is non-negative. Forget your $z$. I am talking about your original integral. $\endgroup$
    – Pierpaolo Vivo
    Dec 21, 2017 at 11:30

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