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I have some long expressions like exp which can be divided into three specified part. I want to simplify each part to its simplest one use FullSimplify.

The specified parts should be:

  1. the first part is term1 = factor1 * E^(-(t/τe1)),
  2. the second part is term2 = factor2 * DiracDelta[t],
  3. the third part is term3 = factor3.

Here factor1, factor2, and factor3 shold not have E^(-(t/τe1)) or DiracDelta[t].

The one long expression is like this

exp = 1/(16 π (-1 + δ)) (1 + 
    2 (1/(9 η0^2 (λ0 + μ0)^4) E^(-(t/τe1)) μ0^2 (-(3 
λ0 + 2 μ0) (6 η0 (λ0 + μ0) (λ0 +2 μ0) - 
             t μ0^2 (3 λ0 + 2 μ0)) - 
          x (6 η0 λ0 (λ0 + μ0) + 
             t μ0^2 (3 λ0 + 2 μ0))) + ((λ0 + 
            2 μ0) (x + λ0 + 2 μ0) DiracDelta[
           t])/(λ0 + μ0)^2 - 
       3 (-((E^(-(t/τe1)) μ0^2 (3 λ0 + 
                 2 μ0))/(3 η0 (λ0 + μ0)^2)) + ((λ0 + 
               2 μ0) DiracDelta[t])/(λ0 + μ0))))

I know firstly I should Expand it into some terms, then divide these terms into three part, then use FullSimplify to simplify them.

But I don't know how to use MatchQ to get these three specified parts.

Please help me!

Thank you very much!

Edit 2017/05/27

The exp without synatix error is like this:

exp=(1 + 2 ((E^(-(
  t/\[Tau]e1)) \[Mu]0^2 ((-3 \[Lambda]0 - 
      2 \[Mu]0) (6 \[Eta]0 (\[Lambda]0 + \[Mu]0) (\[Lambda]0 + 
         2 \[Mu]0) - t \[Mu]0^2 (3 \[Lambda]0 + 2 \[Mu]0)) - 
   x (6 \[Eta]0 \[Lambda]0 (\[Lambda]0 + \[Mu]0) + 
      t \[Mu]0^2 (3 \[Lambda]0 + 2 \[Mu]0))))/(
9 \[Eta]0^2 (\[Lambda]0 + \[Mu]0)^4) + ((\[Lambda]0 + 
   2 \[Mu]0) (x + \[Lambda]0 + 2 \[Mu]0) DiracDelta[
  t])/(\[Lambda]0 + \[Mu]0)^2 - 
3 (-((E^(-(t/\[Tau]e1)) \[Mu]0^2 (3 \[Lambda]0 + 2 \[Mu]0))/(
    3 \[Eta]0 (\[Lambda]0 + \[Mu]0)^2)) + ((\[Lambda]0 + 
      2 \[Mu]0) DiracDelta[
     t])/(\[Lambda]0 + \[Mu]0))))/(16 \[Pi] (-1 + \[Delta]))

After using exp2=Collect[exp, {DiracDelta[t], Exp[-\[Tau]/\[Tau]e1]}, Simplify], I get

exp2=(E^(-(t/\[Tau]e1)) (9 E^(
 t/\[Tau]e1) \[Eta]0^2 (\[Lambda]0 + \[Mu]0)^4 - 
2 \[Mu]0^2 (x (6 \[Eta]0 \[Lambda]0 (\[Lambda]0 + \[Mu]0) + 
      t \[Mu]0^2 (3 \[Lambda]0 + 2 \[Mu]0)) - (3 \[Lambda]0 + 
      2 \[Mu]0) (t \[Mu]0^2 (3 \[Lambda]0 + 2 \[Mu]0) + 
      3 \[Eta]0 (\[Lambda]0^2 - \[Mu]0^2)))))/( 144 \[Pi] (-1 + \[Delta]) \[Eta]0^2 (\[Lambda]0 + \[Mu]0)^4) + ((x - 
2 \[Lambda]0 - \[Mu]0) (\[Lambda]0 + 2 \[Mu]0) DiracDelta[t])/( 8 \[Pi] (-1 + \[Delta]) (\[Lambda]0 + \[Mu]0)^2)

Its return is very close to the answer I want. But exp2 has two terms, not three. The first term of exp2 contains E^(-(t/\[Tau]e1)) (temp1*(E^(t/\[Tau]e1)+temp2).

If I try Collect[Part[%, 1], Exp[-\[Tau]/\[Tau]e1], Simplify] on the first term, then MMA just return it without change.

Could you help me to get three terms? I have about 50 long expressions like exp1, So I need a function to do this work. Thank you!

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  • $\begingroup$ Your expression was not posted correctly. I tried to fix the formatting but there are syntax errors so will not evaluate. Would you please try again? never mind, I think bbgodfrey fixed it. $\endgroup$ – Mr.Wizard♦ May 26 '17 at 15:48
  • $\begingroup$ Both @Mr.Wizard and I tried to fix the many formatting errors, and I think we got them all. In any case, use Collect[exp, {Exp[-(t/τe1)], DiracDelta[t]}, Simplify]. $\endgroup$ – bbgodfrey May 26 '17 at 15:50
  • $\begingroup$ @Mr.Wizard I have edited my question, thank you very much! $\endgroup$ – tanghe2014 May 27 '17 at 0:46
  • $\begingroup$ @bbgodfrey I try your code, then I get two terms, not three terms. I try to use Collect[Part[%,1], Exp[-(t/τe1)], Simplify ] on the first term to divide it into two, but it doesn't work. $\endgroup$ – tanghe2014 May 27 '17 at 0:50
  • $\begingroup$ @tanghe2014 I obtain three terms when applying Collect to exp in your edit. $\endgroup$ – bbgodfrey May 27 '17 at 0:52
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T0 make my comment more concrete, use

Collect[exp, {Exp[-t/τe1], DiracDelta[t]}, Simplify]

to obtain

1/(16 π (-1 + δ)) +
(E^(-(t/τe1)) μ0^2 (-x (6 η0 λ0 (λ0 + μ0) + t μ0^2 (3 λ0 + 2 μ0)) + (3 λ0 + 2 μ0) 
    (t μ0^2 (3 λ0 + 2 μ0) + 3 η0 (λ0^2 - μ0^2))))/(72 π (-1 + δ) η0^2 (λ0 + μ0)^4) + 
((x - 2 λ0 - μ0) (λ0 + 2 μ0) DiracDelta[t])/(8 π (-1 + δ) (λ0 + μ0)^2)

In general, Collect with Simplify can be used to gather together and simplify the coefficients of various subexpressions in larger expressions.

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  • $\begingroup$ Thanks, Collect[exp, {Exp[-t/τe1], DiracDelta[t]}, Simplify], then get (1/(144 \[Pi] (-1 + \[Delta]) \[Eta]0^2 (\[Lambda]0 + \[Mu]0)^4)) E^(-(t/\[Tau]e1)) (9 E^(t/\[Tau]e1) \[Eta]0^2 (\[Lambda]0 + \[Mu]0)^4 + 2 \[Mu]0^2 (-x (6 \[Eta]0 \[Lambda]0 (\[Lambda]0 + \[Mu]0) + t \[Mu]0^2 (3 \[Lambda]0 + 2 \[Mu]0)) + (3 \[Lambda]0 + 2 \[Mu]0) (3 \[Eta]0 (\[Lambda]0 - \[Mu]0) (\[Lambda]0 + \ \[Mu]0) + t \[Mu]0^2 (3 \[Lambda]0 + 2 \[Mu]0)))) + ((x - 2 \[Lambda]0 - \[Mu]0) (\[Lambda]0 + 2 \[Mu]0) DiracDelta[t])/( 8 \[Pi] (-1 + \[Delta]) (\[Lambda]0 + \[Mu]0)^2) $\endgroup$ – tanghe2014 May 27 '17 at 0:41
  • $\begingroup$ I get two terms not three terms. It is very close to the answer I want. How to get three terms? Thank you! $\endgroup$ – tanghe2014 May 27 '17 at 0:45
  • $\begingroup$ @tanghe2014 I cannot obtain the expression in your comment just above. $\endgroup$ – bbgodfrey May 27 '17 at 0:57
  • $\begingroup$ Sorry for my post. If add Expand on exp in your code, It will return what we want. $\endgroup$ – tanghe2014 May 27 '17 at 1:42
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Following the comment, I find this code works well!

intoThreeTerms[exp_]:=Collect[
                        exp // Expand,
                        {Exp[t/\[Tau]e1], DiracDelta[t]},
                        FullSimplify];
myexp2=intoThreeTerms[myexp1]

Thank your patient comments and some useful hints!

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