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I would like to know if it is possible to compute something like

$$\int_{-\infty}^{\infty}f\left(\delta(x-a)\right)dx,$$

where $f(x)$ is a function, or if it is even defined.

Thanks in advance!

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  • $\begingroup$ How would you define $f(\delta(x-a))$ when $x=a$? $\endgroup$ – Clive Newstead Feb 5 '19 at 18:49
  • $\begingroup$ I would naively define it to be to $f(1)$ if $x=a$ and $f(0)$ otherwise, but I guess that is not a Dirac delta. $\endgroup$ – Carl Feb 5 '19 at 20:37
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Strictly, it's not defined, but there might be cases (other than $f$ being an affine function) where it can be given a meaning.

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  • $\begingroup$ So, would it make sense to define something like $$\int_{-\infty}^{\infty}\delta(x-a)\ln[\delta(x-a)]dx=0$$, where $\delta$ is the dirac delta? The thing is I still need the property $$\int_{-\infty}^{\infty}\delta(x-a)dx=1$$. $\endgroup$ – Carl Feb 5 '19 at 20:46
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    $\begingroup$ @Carl. I cannot make anything meaningful out of that integral. The most natural would be $\ln \delta(a-a)$ with a value of $+\infty.$ $\endgroup$ – md2perpe Feb 5 '19 at 21:02
  • $\begingroup$ I see. Well, thank you for your answer :) $\endgroup$ – Carl Feb 5 '19 at 21:15
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$$\int_{-\infty}^{\infty}f(x)(\delta(x-a))dx = f(a)$$ This comes from the fact that: $$\int_{-\infty}^{\infty}\delta(x)dx = 1$$

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    $\begingroup$ Yes, but the question is what happens if the delta is the argument of the function, not if it multiplies the function. Maybe changing the parenthesis would be clearer: $$\int_{-\infty}^{\infty}f[\delta(x-a)]dx$$ $\endgroup$ – Carl Feb 5 '19 at 18:10

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