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I Delta function in 2D

  1. Jan 27, 2017 #1
    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!!!
    Last edited by a moderator: Jan 27, 2017
  2. jcsd
  3. Jan 27, 2017 #2


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    Use "##" (without quotes) before your expression and after that, in order for the integral to display correctly.
  4. Jan 27, 2017 #3


    Staff: Mentor

    You can also use "$$". The hashtags gives an inline LaTeX and the dollar signs gives non inline LaTeX. I have edited the post
  5. Jan 27, 2017 #4


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    What do you know about delta function so far?
  6. Jan 27, 2017 #5


    Staff: Mentor

    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.
  7. Jan 27, 2017 #6


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    Mathematically, an integral such as this is undefined. You can't work with the delta function like this.
  8. Jan 27, 2017 #7


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    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.

    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

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