What is the dimensions of Dirac Delta Function and why is it so ?
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1Hint: if $f(x)=\int g(y)h(x,,y)d^ny$ then $[h]=\frac{[f]}{[g]\prod_{i=1}^n[y_i]}$. Now take $g(y)=f(y),,h(x,,y)=\delta(x-y)$. – J.G. Aug 07 '21 at 08:32
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Voting to close as a math question. – Aug 07 '21 at 08:38
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2Does this answer your question? https://physics.stackexchange.com/q/33760/ – justauser Aug 07 '21 at 09:10
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3Does this answer your question? What are the units or dimensions of the Dirac delta function? – jng224 Aug 07 '21 at 09:21
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$$1 = \int_{-\infty}^{+\infty} \delta(x-x_0) dx$$ Since the product $\delta(x-x_0) dx$ is adimentional, the dimension of $\delta$ must be the inverse of the dimension of $x$.
Prallax
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