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I am trying to find the characteristic function for Johnson's SU distribution by integrating the probability density function with Exp[I*t*x] but Mathematica is returning the input itself.

As the characteristic function always exists, I'm not able to understand why Mathematica is not finding the integral.

Here's the code:

expr1[x_] := PDF[JohnsonDistribution["SU", γ, δ, ξ, λ], 
  x]

Integrate[expr1[x]*Exp[I*t*x], {x, -Infinity, Infinity}]
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    $\begingroup$ I'm voting to close this question as off-topic because the OP is asking for functionality that is not supported given the constraints the OP is putting on the solution. $\endgroup$ – m_goldberg Jun 19 at 15:28
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    $\begingroup$ @m_goldberg Closing a question from a first timer abruptly may alienate the person from this forum and worse: from the Wolfram community. $\endgroup$ – nilo de roock Jun 20 at 7:55
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Please look at the characteristic function of the lognormal distribution.
In some cases there is no closed form.

One way to address this issue is using empirical' s one.

emCF = With[{dist = #}, 
    RandomVariate[dist, 100] // Exp[I t #] & /@ # & // Mean // 
     Set[approx, #] &;
    ReIm[approx]] &;

approximated characteristic function of NormalDistribution[0,1]

Plot[Evaluate[emCF[NormalDistribution[0, 1]]], {t, -1, 1}]

Mathematica graphics

exact characteristic function of NormalDistribution[0,1]

Plot[ReIm@CharacteristicFunction[NormalDistribution[0, 1], t], {t, -1,
   1}]

Mathematica graphics

approximated characteristic function of JohnsonDistribution["SU", a, b, c, d]

Manipulate[
 Plot[Evaluate[emCF[JohnsonDistribution["SU", a, b, c, d]]], {t, -1, 
   1}], {a, -5, 5}, {b, 0.01, 5}, {c, -5, 5}, {d, 0.01, 5}]

Mathematica graphics

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The first assignment is incorrect, resp. I don't know what it should mean from a Wolfram Mathematica syntax perspective. It can mean

expression: a correct form is

 expr1 = PDF[JohnsonDistribution["SU", γ, δ, ξ, λ], x]

or function definition: a correct form is

 expr1[x_] := PDF[JohnsonDistribution["SU", γ, δ, ξ, λ], x]

However, I think that this is not a major issue. It is that apparently the characteristic function for Johnson's distribution cannot be expressed in the form of a closed expression. Neither wikipedia nor other source know the formula for general characteristic function. The Mathematica answer is "I'm not able to calculate it", i.e.

 CharacteristicFunction[JohnsonDistribution["SU", γ, δ, ξ, λ], t]

CharacteristicFunction[JohnsonDistribution["SU", γ, δ, ξ, λ], t]

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    $\begingroup$ There's nothing wrong with the function definition as supplied. It simply uses immediate evaluation instead of delayed. The real answer is your second portion to this. $\endgroup$ – b3m2a1 Jun 20 at 7:58
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    $\begingroup$ @b3m2a1 Exactly right. Function assignments of the form f[x_] = ... are simply equivalent to f[x_] := Evaluate[...]. $\endgroup$ – Sjoerd Smit Jun 20 at 8:26

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