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I am trying to simplify the following expression

code:

Cos[(Sqrt[Integrate[(-1 - I)*dper*Ex[t1], {t1, 0, Infinity}]]*
    Sqrt[Integrate[(-1 + I)*dper*Ex[t1], {t1, 0, Infinity}]])/hbar

with previous assumption declared as

$Assumptions = Element[\[Alpha] | \[Beta] | ap | a0 | E0 | Ep | t | Ex[t1] | Ex[] |
 t1 | Nd | T | hbar | dper | dpara, Reals];

However, somehow it seems Mathematica is not able to understand (-1+i)dper does not depend on t1, and therefore, it cannot be simplified. Any thoughts on how I could improve that?

Thanks!

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6
  • 1
    $\begingroup$ What is Ex? Should it be Exp? $\endgroup$
    – Bob Hanlon
    Mar 8, 2021 at 18:49
  • 2
    $\begingroup$ Can you, please, include the code shown in the picture as easily copy-&-paste-able text in a code block here? $\endgroup$
    – CA Trevillian
    Mar 8, 2021 at 18:55
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    $\begingroup$ The code yo included is not WL. and what is dper? $\endgroup$
    – Daniel Huber
    Mar 8, 2021 at 19:31
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    $\begingroup$ To paraphrase the question: Why is Integrate[a*f[x], x] auto-simplified to a*Integrate[f[x], x] but Integrate[a*f[x], {x, x1, x2}] remains as-is without extracting the constant a from the integral? $\endgroup$
    – Roman
    Mar 8, 2021 at 19:36
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    $\begingroup$ @denis - prior to copy and paste of your code, convert it to Raw InputForm $\endgroup$
    – Bob Hanlon
    Mar 8, 2021 at 19:37

1 Answer 1

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This is your expression:

expr = Cos[(Sqrt[Integrate[(-1 - I)*dper*Ex[t1], {t1, 0, Infinity}]]*
      Sqrt[Integrate[(-1 + I)*dper*Ex[t1], {t1, 0, Infinity}]])/hbar];

Try this:

expr2 = Simplify[
  expr /. Sqrt[Integrate[a_*Ex[t1], {t1, 0, Infinity}]] :> 
     Sqrt[a]*Sqrt[(Integrate[Ex[t1], {t1, 0, Infinity}])] /. 
   Sqrt[a_]*Sqrt[b_] :> Sqrt[a*b], dper \[Element] Reals]

(*  Cos[(Sqrt[2] Abs[dper] \!\(
\*SubsuperscriptBox[\(\[Integral]\), \(0\), \(\[Infinity]\)]\(Ex[
     t1] \[DifferentialD]t1\)\))/hbar]  *)

enter image description here

Have fun!

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