Is it true that $\int_0^\infty \delta(x-\alpha)d\alpha = \int_0^\infty \delta(\alpha - x)d\alpha$? Note that x is NOT the variable here. It's said that the delta function is even so the expression stands. But isn't those translated versions of the original delta function? Does the even property still hold?
2026-03-30 00:15:14.1774829714
Improper integral of Dirac delta function
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