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Why is the area under the Dirac delta function equal to one and not zero? Shouldn't it be zero since the function is symmetric and the area under of each side cancels out the other?

mvw
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Jack
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    Areas cancel out for ANTI-symmetric functions. Integral of $x\delta(x)$ is indeed $0$. – A.S. Mar 18 '16 at 14:01
  • Dirac delta function is not a function. Intuitively, you can think of it as an equivalent classes of sequences of functions. Each function has unit area under its curve and as a sequence, the support of functions tends to the singleton ${0}$. So by definition, the area under a Dirac delta function has to be one. Otherwise, there is not point to use it. – achille hui Mar 18 '16 at 14:08
  • @ A.S Isn't the area x times y. If so, the write side has an area of integral of xδ(x) and the left side -xδ(x), and then the summation of these integral is zero. – Jack Mar 18 '16 at 14:08
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    Yes, but $\delta(x)$ and $x\delta(x)$ are two different objects. One is symmetric, one is anti-symmetric. Integral of one is $1$, integral of the other is $0$. – A.S. Mar 18 '16 at 14:25
  • Isn't ∫δ(x)dx=1, the same as the integral of dxδ(x) or in your case xδ(x) ? – Jack Mar 18 '16 at 14:34

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The Dirac delta is no function, but a distribution.

The desire is to have $$ \int\limits_{-\infty}^{\infty} \delta(x) f(x) \, dx = f(0) \quad (*) $$ It is clear, that no function $\delta$ can achieve this.

Interpreted as linear functional $$ \langle \delta, f \rangle = f(0) $$ where $\langle . , . \rangle$ is a scalar product, typically involving an integration. This works in the theory of distributions. It can be approximated by sequences, as achile pointed out in the comments.

So it is kind of fishy to speculate about the symmetry of $\delta$, which has to vanish for any point but the origin, and can not have a finite value at the origin. One might look at the symmetry of the approximating functions, but that will not necessarily allow a transfer to the limit function I believe.

If you apply the definition $(*)$ to $f(x) = 1$, you end up with $$ \int\limits_{-\infty}^{\infty} \delta(x)\, dx = 1 $$

mvw
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    an illustrative aside: $2\delta$ has the same "graph" as $\delta$, but the "area under" $2\delta$ is $2$. It just makes no sense to image $\delta$ as a classical function (on the reals). – user251257 Mar 18 '16 at 16:12