Delta Function - Gauge-institute.org

The highly restrictive conditions for the Fourier Integral Theorem,in the Calculus of Limits, are irrelevant to the simplest functions,such as constants, and useless for singular functions.In particular, the singularδ( x)violates these conditions the Hyper-real Deltaδ( x)is not defined in the Calculus ofLimits, and is not Piecewise Continuous.δ'( x)is not defined, and is not Piecewise Continuous in anybounded interval.But in Infinitesimal Calculus,Fourier integral Theoremδ( x)satisfies the Hyper-realk =∞ ⎛ ξ=∞⎞1−ikξikxδ( x) = δ( ξ)e2π∫ ∫ dξ e dk.k =−∞ ⎜⎝ξ=−∞ ⎠⎟Also, the constant function f ( x) ≡ 1 violates the sufficientx =∞conditions’ requirement of absolute integrability, ∫ 1 dx =∞.x =−∞But in Infinitesimal Calculus, f ( x) ≡ 1 satisfies the Hyper-realFourier integral Theorem1k =∞ ⎛ ξ=∞⎞1−ikξikx= e dξe d2πk =−∞ ⎜⎝ξ=−∞⎠⎟∫ ∫ k.