Delta Function - Gauge-institute.org

Gauge Institute Journal, Volume 8, No.2, May 2012H. Vic Dannon⎛⎞ ⎟= ⎜ ⎟⎝⎠ξ=∞ k =∞1 ik( x ξ)f () ξ ⎜−e dk∫ ξξ 2 π∫ d .=−∞ ⎜ k =−∞ ⎟However,xik( x−ξ)= ξ ⇒ e = 1,and the integral12πk =∞−ik( ξ−x)∫ e dk,k =−∞diverges.That is, the Fourier Integral Theorem cannot be written in theCalculus of Limits.Avoiding the singularity atξ = xdoes not recover the Theorem,because without the singularity the integral equals zero.Furthermore,5.2 Calculus of Limits Conditions are not sufficientfor the Fourier Integral TheoremProof: The Calculus of Limits Conditions are1. Piecewise Continuity of f ( x ), and f '( x ) in any boundedinterval.x =∞2. convergence of f ( x)dx∫x =−∞19