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to prove that the delta function is symmetric, I need to show that $\delta(x) = \delta(-x)$ by employing a change in variables.

$$\delta(x) = {1\over 2\pi}\int_{-\infty}^\infty\exp(ikx)dk\tag{1}$$ $$\delta(-x) = {1\over 2\pi}\int_{-\infty}^\infty\exp(-ikx)dk$$ Let $k = -l$, $$\delta(-x) = {-1\over 2\pi}\int_\infty^{-\infty}\exp(ilx)dl ={1\over 2\pi}\int_{-\infty}^\infty\exp(ilx)dl\tag2 $$ $$=\delta(x)$$

Am I right to say that it doesn't matter whether it is k or l, so long as they produce the same expression, the 2 integrals (1) & (2) are equal? But why is it the case, when k is not equal to l?

mrf
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user91820
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  • Your method is Ok. You are using Fourier transform technique. – Mhenni Benghorbal Aug 28 '13 at 10:36
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    As it is presented, your method is outside of the circle of sound mathematical proofs (which does not mean there are not some ways to transform these considerations into a proof). – Did Aug 28 '13 at 14:05
  • The answer given is much cleaner than the Fourier transform technique, and its rigor is more apparent. – Stefan Smith Aug 29 '13 at 00:56
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    In fact, It depends on what course you are taking and your back ground. As I said "your method is ok". – Mhenni Benghorbal Aug 29 '13 at 02:21
  • "Dirac's delta" is a distribution ( http://en.wikipedia.org/wiki/Distribution_(mathematics) ). – Felix Marin Sep 02 '13 at 00:50
  • "Dirac's delta" is a distribution ( http://en.wikipedia.org/wiki/Distribution_(mathematics) ). However, it was introduced "phenomenologically" by P. A. M. Dirac in the 30's. It was considered as a generalized function by Sobolev. In the 40's, Schwartz put them in a rigorous foundation with his Theory of Distributions. You can think that Dirac delta is a Kronecker delta "continous version" BUT be careful any way. – Felix Marin Sep 02 '13 at 00:56

2 Answers2

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Beware that $\delta$ is not a function but a distribution and you need to be careful with what you mean by expressions such as "$\delta(-x)$".

To do this more rigorously, lets start by defining an operation on smooth functions: if $\phi$ is smooth, let $\check\phi$ be the function defined by $\check\phi(x) = \phi(-x)$. If $f$ is a distribution that comes from a smooth (or even $L^1_{loc}$ function), then for any test function $\phi$

$$ \langle \check f, \phi \rangle = \int_{-\infty}^\infty f(-x)\phi(x)\,dx = \int_{-\infty}^\infty f(x)\phi(-x)\,dx = \langle f, \check \phi \rangle $$ by a change of variables. Hence, it's reasonable to define $\check U$ for any distribution by $\langle \check U, \phi \rangle = \langle U, \check \phi \rangle$. For $\delta$, we get

$$ \langle \check\delta, \phi \rangle = \langle \delta, \check\phi \rangle = \check \phi(0) = \phi(0) = \langle \delta, \phi \rangle. $$

Hence $\check \delta = \delta$, which is the distribution theory version of even.

mrf
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2

Instead of using the Fourier transform to prove symmetry, I used the definition of the delta function to prove symmetry

$$f(a) = \int_{-\infty}^{\infty}f(x)\delta (x-a) \; dx$$

and

$$f(a) = \int_{-\infty}^{\infty}f(x)\delta (a-x) \; dx$$

Therefore

$$\int_{-\infty}^{\infty}f(x)\delta (x-a) \; dx = \int_{-\infty}^{\infty}f(x)\delta (a-x) \; dx$$

Finally

$$\delta (x-a) = \delta (a-x)$$

or

$$\delta (x) = \delta (-x)$$