And so this is what I meant by pseudoinfinity, because if I have 2 times the Dirac delta function, and if I'm taking the area under the curve of that, of 2 times the Dirac delta function t minus c dt, this should be equal to 2 times-- the area of just under the Dirac delta function 2 times from minus infinity to infinity of the delta function shifted by c dt, which is just 2 times-- we already showed you, I just said, by definition, this is 1, so this will be equal to 2.