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I am using the sign convention of FourierParameters->{0, -2 Pi} for calculating the inverse FT of $Aexp(-2\iota\pi f K)$, where $A$, $K$ are real numbers >0 and $f$ is the frequency.

S = InverseFourierTransform[A E^(-2 I f K \[Pi]), f, x, FourierParameters -> {0, -2 \[Pi]}, Assumptions -> {Element[A | K, Reals], K > 0, A > 0}]

The answer is A DiracDelta[K - x], however if we put a real value of K, say 5,

S = InverseFourierTransform[A E^(-2 I f 5 \[Pi]), f, x, FourierParameters -> {0, -2 \[Pi]}, Assumptions -> {Element[A, Reals], A > 0}]

The answer is A DiracDelta[-5 + x]

Why do we see a difference in the sign of x when we have already specified that K is a real number greater than 0? The DiracDelta is said to be even, but can we still get general answer A DiracDelta[-K + x] in Mathematica, without inserting a real positive constant in place of K. I have tried GenerateConditions but still the general answer is A DiracDelta[-K + x] instead of A DiracDelta[K - x].

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  • $\begingroup$ delta is "on" only when its argument is zero. Right? In the first case this happens when $x=k$. Correct? In the second case, it is "on" when $x=5$. This makes sense since you put $k=5$. So I do not know see any issue? There is no difference if you write $-k+x$ or $k-x$ in both cases you get $x=k$ in order for the delta argument to be zero. $\endgroup$
    – Nasser
    Apr 2, 2022 at 3:47
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    $\begingroup$ DiracDelta[a] = DiracDelta[-a], so DiracDelta[x - 5] = DiracDelta[5 - x]. Either is correct. Whichever form Mathematica chooses, it generally outputs variables in expressions in alphabetical order. $\endgroup$
    – Bill Watts
    Apr 2, 2022 at 7:20
  • $\begingroup$ I agree both answers are correct but if we replace K with another letter, c, it is still DiracDelts[-c+x]. I was wondering about the sign associated with x. $\endgroup$
    – AChem
    Apr 2, 2022 at 10:53
  • $\begingroup$ Both DiracDelta[x-5] and DiracDelta[5-x] mean the $\delta$-distribution with the support at $x=5$. $\endgroup$
    – user64494
    Apr 3, 2022 at 5:33

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I can't say I always understand Mathematica's formatting rules, but you do have some choices. You probably have your default output format set to StandardForm. Maybe you would like TraditionalForm better.

S = InverseFourierTransform[A E^(-2 I f K \[Pi]), f, x, 
  FourierParameters -> {0, -2 \[Pi]}, 
  Assumptions -> {Element[A | K, Reals], K > 0, A > 0}]

$A\ DiracDelta[K-x]$

S = InverseFourierTransform[A E^(-2 I f K \[Pi]), f, x, 
   FourierParameters -> {0, -2 \[Pi]}, 
   Assumptions -> {Element[A | K, Reals], K > 0, A > 0}] // 
  TraditionalForm

$A\ \delta (K-x)$

In the case of 5 in place of K

StandardForm

$A\ DiracDelta[-5+x]$

TraditionalForm

$A\ \delta (x-5)$

My earlier comment about variables in alphabetical order applies to StandardForm. TraditionalForm operates under different rules and I don't pretend to know all of them.

I would expect the variable c to behave the same as K because they are both ahead of x in the alphabet. Try y instead because it is after x.

StandardForm

$A\ DiracDelta[-x+y]$

TraditionalForm

$A\ \delta (y-x)$

As to which variable it makes negative, I cannot predict, but both ways are correct for DiracDelta. Outputting to TraditionalForm, unlike MatrixForm and other formatting options, does not ruin the ability to use the results in further calculations.

While I use StandardForm for input, since Version 3 I have used TraditionalForm as my default output format, but I realize that is not for everybody. The only problems I have seen is using the Notation package, since that requires both input and output formats to be the same.

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  • $\begingroup$ Okay thanks . I have also written to Mathematica. Let us see what they respond. $\endgroup$
    – AChem
    Apr 3, 2022 at 16:00

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