I Evaluating 2D Delta Function Integral - Any Help Appreciated

[junt]

I am quite new here, and was wondering if anybody knows how this 2D integral is evaluated.
$$ \int_{-\infty}^\infty \int_{-\infty}^\infty \delta(k_1 x-k_2y)\,dx\,dy$$Any help is greatly appreciated! Thanks!

 

[QuantumQuest]

I am quite new here, and was wondering if anybody knows how this 2D integral is evaluated.
$ \int_{-\infty}^\infty \int_{-\infty}^\infty \delta(k_1 x-k_2y)\,dx\,dy$

Use "##" (without quotes) before your expression and after that, in order for the integral to display correctly.

 

[Dale]

You can also use "$$". The hashtags gives an inline LaTeX and the dollar signs gives non inline LaTeX. I have edited the post

 

[QuantumQuest]

...and was wondering if anybody knows how this 2D integral is evaluated.

What do you know about delta function so far?

 

[Dale]

I am quite new here, and was wondering if anybody knows how this 2D integral is evaluated.
$$ \int_{-\infty}^\infty \int_{-\infty}^\infty \delta(k_1 x-k_2y)\,dx\,dy$$

I think that integral diverges. You can evaluate the inside integral using the "sampling" property of the Delta function. Then you integrate that, and I don't think it converges.

 

[micromass]

I am quite new here, and was wondering if anybody knows how this 2D integral is evaluated.
$$ \int_{-\infty}^\infty \int_{-\infty}^\infty \delta(k_1 x-k_2y)\,dx\,dy$$Any help is greatly appreciated! Thanks!

Mathematically, an integral such as this is undefined. You can't work with the delta function like this.

 

[jasonRF]

I think that integral diverges. You can evaluate the inside integral using the "sampling" property of the Delta function. Then you integrate that, and I don't think it converges.

I Agree. For $k_1 \neq 0$ we have,
$$ \begin{eqnarray}
\int_{-\infty}^\infty \int_{-\infty}^\infty \delta(k_1 x - k_2 y) dx \, dy & = & \frac{1}{|k_1|}\int_{-\infty}^\infty \int_{-\infty}^\infty \delta(x - k_2 y/k_1) dx \, dy \\
& = & \frac{1}{|k_1|}\int_{-\infty}^\infty dy
\end{eqnarray} $$
which diverges.

Mathematically, an integral such as this is undefined. You can't work with the delta function like this.

Even some applied math treatments (eg. L. Schwartz's "mathematics for the physical sciences") echo you point, and even frown on writing the delta distribution with an argument, such as $\delta(x)$. However, most engineers and scientists write expressions like $\delta(x)$ or integrals with delta function integrands, and whether or not they realize it, these expressions are basically symbolic representations. One just needs to learn how the symbology works. In the above, the integral over $x$ can be viewed as the convolution of the delta function and the test-function $1$, which results in a constant. The second integral is then the integral of that constant over the real line, which diverges. As an engineer it doesn't bother me ... but I realize that some confusion and wrong results can probably be traced back to the integral notation. Just my 2 cents

Jason