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I have a question related to "FourierCosTransform" command in Mathematica. I have tested the following codes:

g0[Z_] = Z^2
g1[t_] = InverseFourierCosTransform[g0[Z], Z, t]
g2[Z_] = FourierCosTransform[g1[t], t, Z]
g3[Z_] = Sqrt[2/Pi]*Integrate[g1[t]*Cos[Z*t], {t, 0, Infinity}]

My question is since both g2 and g3 represent the "FourierCosTransform" operation, g2 uses Mathematica command directly, g3 uses the definition of "FourierCosTransform" as explained in HELP document. Why these two can not produce the exactly same result. The result shown by Mathematica 12.0 on my laptop is:

Z^2
-Sqrt[2 \[Pi]] (DiracDelta^\[Prime]\[Prime])[t]
Z^2
-2 Z^2 (-1+HeavisideTheta[0])

Thank you very much for your help!

J. M.'s missing motivation
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Fredrich
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1 Answers1

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Try this for g3:

g3[Z_] = Sqrt[2/Pi]*Integrate[g1[t]*Cos[Z*t], {t, -Infinity, Infinity}]

The second derivative of Dirac is not properly defined mathematically.

Edit:

The reason both g2 and g3 do not produce the same result is because of how Mathematica handles integration of DiracDelta and its derivatives. So for g3, you can define the Fourier Cosine Transform using the integral by hand as an alternative representation.

Alternatively:

You could rewrite g1[t] to use the second derivative of Heaviside step function, and then redefine g3 with the new g1 representation:

g1[t_] = -2*(-1 + HeavisideTheta[t])
g3[Z_] = Sqrt[2/Pi]*Integrate[g1[t]*Cos[Z*t], {t, 0, Infinity}]

Because HeavisideTheta and its derivatives are distributions, their integrals should be interpreted as distributional pairings.

More Senne
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    Could you explain exactly how this answers the question? – MarcoB May 02 '23 at 13:49
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    The second derivative is defined mathematically. What is not defined is how to integrate against it when the singular point (zero in this case) coincides with an endpoint of the integral. – Daniel Lichtblau May 02 '23 at 13:52
  • That is correct thanks @DanielLichtblau. – More Senne May 02 '23 at 14:06
  • Distributions as real line limits of complex functions, analytic in the upper half plane always are defined as elements of the dual space wrt. to integration over R, the space of "test functions" on a) smooth functions with compact support, b) functions with decay at infinity faster than any polynomial. Their Fourier transforms are entier functions. In order to include polynomials, one has to introduce Gaussian cutoffs at infinity and taking the limit ->oo. Thats not trival. So distributions with support on the boundary are not defined. Its better to work out such problems without tables. – Roland F May 02 '23 at 16:24
  • After change the integration domain, g3[Z_] = Sqrt[2/Pi]Integrate[g1[t]Cos[Z*t], {t, -Infinity, Infinity}] gives 2Z^2, twice of Z^2 as expected. For this example, could you please help to construct g1[t] with the aid of Heaviside step function and/or its derivatievs? Many thanks! – Fredrich May 03 '23 at 10:27