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I have the following code in mathematica

\[Alpha] = 3;
 f[s_] := Exp[-A*s^(2/\[Alpha])]; 
 F = Assuming[{A > 0, t > 0, {t, A} \[Element] Reals},Integrate[f[s]*Exp[s*t] /. s -> I*y, {y, 0, Infinity}]/Pi]; 
 Z= FullSimplify[ComplexExpand@Re[F], {t > 0, A > 0, {t, A} \[Element] Reals}]

We can see that $Z$ is a function of $t$.

Now I want to perform the following operation

$$S(x)=\int_0^{\infty}te^{-tx}Z(t)dt$$

For which I write the code

S = Integrate[t*Exp[-t*x]*Z, {t, 0, \Infinity}]

But, I get some error. How can I resolve this?

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    $\begingroup$ You know that LaplaceTransform[] and InverseLaplaceTransform[] are built-in, no? $\endgroup$ – J. M. will be back soon Feb 24 '16 at 2:56
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Two things.

  1. Your integration statement has a slash in front of infinity.

  2. Put the same Assumption about A in the final integral as you had in the integration that generated F.

Then you get an answer:

S = Assuming[{A > 0, A \[Element] Reals}, 
  Integrate[t*Exp[-t*x]*Z, {t, 0, Infinity}]]

Mathematica graphics

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  • $\begingroup$ Thanks for your answer. Yes, Now S is function of both t and x. I just want it as a function of x. Is it possible to find that? $\endgroup$ – Dimitrios Feb 24 '16 at 4:49

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