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integrable in the Calculus of Limits, and is not Lebesgueintegrable in Measure Theory.In fact, in the Calculus of Limits, only the Cauchy Principal ValueIntegral of the Delta Function exists, and it equals zero.Only in Infinitesimal Calculus, can the Delta Function be defined,differentiated, and integrated.Infinitesimal Calculus allows us to resolve open problems such asWhat is δ(0) ?How is xδ( x)defined at x = 0 ?How is the Delta Function the derivative of a Step Function?How do we integrate the Delta Function?What is δ (x )?What is δ2 ( x)?What is δ(x )?What is δ3 ( x)?23The Delta Function enables us to define the Fourier Transformwith minimal requirements on the transformed function.Delta Function, the Fourier Transform, and FourierIntegral TheoremH. Vic DannonAbstract: The Fourier Integral Theorem guarantees that theFourier Transform and its Inverse are well defined operations, sothat the inversed transform is the originally transformed function.
It is believed to hold in the Calculus of Limits under some highlyrestrictive sufficient conditions. In fact,The Theorem does not hold in the Calculus of Limitsunder any conditions,because evaluating the Fourier Integral requires the integration ofthat diverges at x = ξ .k =∞ik( x−ξ)∫ e dk,k =−∞Only in Infinitesimal Calculus, the integral is the Hyper-realDelta Functionk =∞1 ik( x−ξ)δ( x − ξ)= e2π∫ dk,k =−∞and the Fourier Integral Theorem states the sifting property forthe Delta Functionξ=∞f ( x) = ∫ f( ξδξ ) ( −x)d ξ.ξ=−∞In infinitesimal Calculus we can integrate over singularities, andthe Fourier Integral Theorem holdsk =∞ ⎛ ξ=∞⎞1−ikξf ( x) = f( ξ)e ξ2πk =−∞ ⎜⎝ξ=−∞⎠⎟where the Integrals are Hyper-real.ikx∫ ∫ d e dk,
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