Dirac delta's definition says it is zero except at $x=0$, and its Riemann integral over the real line equals $1$. Why do we define and use things in mathematics than contain contradictions in its definition?

These two properties in its definition contradict each other, because the integral of a function that has only one argument where its value isn't zero cannot be positive. You might say, ok, but mathematicians know it very well and that's why Dirac delta is not a function. It's a distribution. Fine, but is it really sufficient to define something that IS NOT a function, call it 'distribution' and use it as if it were a function? How do we know the reuslts we get are reliable if the tool we chose contains a logical contradiciton? Why can we even expect correct results?

Maybe I don't know much about distributions, but just giving Dirac delta a label 'distribution' doesn't solve the problem connected with using it as if it was a usual function.

For example, I could define a function $f(x)=0$ and say $\int_{-\infty}^{\infty}f(x)dx=1$, and then use it in my calculations, but should I rely on the results I get? Obviously not. Defining a notion is not enough and we should first see if it's correct.

Could anyone help me understand why is it perfectly okay to use Dirac delta, and show that it's different from the example I gave?

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    $\begingroup$ The definition you give is a completely heuristic definition. It is not a rigorous definition. The rigorous definition is that $\delta$ is a map from continuous functions to the real numbers which satisfies $\delta(f) = f(0)$. This is a perfectly well-defined object. The reason we talk about the "pointwise" characteristics of the Dirac delta is that the sequences which converge to it (in some sense of the word converge) have that behavior: they integrate to $1$ and as the sequence goes on, the functions go to zero everywhere but at the origin. $\endgroup$ – Cameron Williams Aug 18 '15 at 17:04
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    $\begingroup$ You're right. The Dirac Delta is not a function. And the "integral" is not an integral. Rather, there is a functional such that $\langle \delta,\phi \rangle =\phi(0)$ for any test function $\phi$. $\endgroup$ – Mark Viola Aug 18 '15 at 17:04
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    $\begingroup$ Rigorously, we never treat it as a function but instead as a linear functional. The whole mess about the Dirac delta is a result of the fact that the idea sprang out of physics (as did a lot of early functional analysis), and the lack of rigor has pervaded the colloquial explanations as a result. $\endgroup$ – Cameron Williams Aug 18 '15 at 17:04
  • $\begingroup$ @CameronWilliams Exactly ... just as I wrote, the so-called integral isn't! By the way, I live in The Woodlands. So, we are geographically close. $\endgroup$ – Mark Viola Aug 18 '15 at 17:06
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    $\begingroup$ The integral of f(x)gx) where f is the delta function is just symbolic. Consider a sequence f1,f2,f3,... where the sequence of integrals of f1(x)g(x), f2(x)g(x),... converges to g(0). But the symbolic notation has uses also. $\endgroup$ – DanielWainfleet Aug 18 '15 at 18:28

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