Need help in finding integral value of $\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} dx \int_{-1}^{1} dy\ \delta(\sin2x)\ \delta(x-y)$?

Question

I am trying to find value of the following integral:

$\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} dx \int_{-1}^{1} dy\ \delta(\sin2x)\ \delta(x-y)$

I have some experience in solving integrals that contain Dirac delta function in one variable, but here it is in two variable. I am completely clueless and don't even know how to approach this problem. I am not expecting the solution, need someone to guide me what approach will help me to evaluate this integral?

Parcly Taxel · Accepted Answer · 2019-02-27T17:42:45.623

1

To extend change-of-variables manipulations to the Dirac delta, the following composition is defined: $$\delta(\mathbf f(\mathbf x))=\sum_{\mathbf r_i}\frac{\delta(\mathbf x-\mathbf r_i)}{|\mathbf J_{\mathbf f}(\mathbf r_i)|}$$ where $\mathbf r_i$ are the roots of $\mathbf f$. For the given $\mathbf f(x,y)=(\sin2x,x-y)$, there is only one root in the domain of integration, $(0,0)$, and $$|\mathbf J_{\mathbf f}|=\begin{vmatrix}2\cos2x&1\\0&-1\end{vmatrix}=-2\cos2x$$ Evaluating this at $(0,0)$ and taking the absolute value gives 2, so (within the domain of integration) $$\delta(\mathbf f(\mathbf x))=\frac{\delta(\mathbf x-(0,0))}2$$ Thus the integral over the domain is $\frac12$.

edited Feb 27 '19 at 17:42

answered Feb 27 '19 at 16:50

Parcly Taxel

103,344

Thanks, @Parcly. I am afraid the book gives the answer as $\frac{1}{2}$. I am confused! Could you please elaborate? I am not able to follow. Apologies for naive understanding of calculus. – orionphy Feb 27 '19 at 17:15
@orion OK, fixed. – Parcly Taxel Feb 27 '19 at 17:42

Need help in finding integral value of $\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} dx \int_{-1}^{1} dy\ \delta(\sin2x)\ \delta(x-y)$?

1 Answers1