[?] Simplify the following expression under Integral sign?

Questions
1

How does one simplify the following input

Integrate[E^(I*p3*y)*hDag[p3]*Subscript[C, p3], p3]*
  Integrate[(h[p4]*Subscript[C, p4])/E^(I*p4*y), p4]

to produce a simplification in the exponent

E^(I (p3 - p4) y)
2

Several observations:

  • C is a reserved symbol, try using c
  • What is h?
  • What is hDag?
  • Subscript should only be used a typographical construct; sooner or later you’ll get into trouble… call it cp3 or so…
  • Add some Assumptions on p4, p3, y (real?, positive?) to Integrate
3

Thank you for the response.
h is an operator, can treat it as a variable here. “hDag” is the conjugate, i.e. h^{\dagger}, but can be treated as another variable in this specific instance. I wouldn’t like to use h^{\dagger} though, since Mathematica starts evaluating stuff that it understands as conjugate in steps when I don’t want it to.

C_p3 is supposed to be treated as C[p3], and Mathematica hasn’t given any problems thus far with C_{p3}. In future, will take note of your suggestion and make it c[p3]

p3, p4, y are all real (actually 4-dimensional vectors), can be positive or negative. In this specific example, they can be treated as just real numbers.

Not much luck with using c[p3] instead of C_{p3} though,

Simplify[Integrate[(c[p4]*h[p4])/E^(I*p4*y), p4]*
   Integrate[E^(I*p3*y)*c[p3]*hDag[p3], p3]]

returns just the input.

4

You can’t simplify it if c is a ‘general’ function…, same holds for h and hdag… You need to know what h and hdag are in order to do the integral…

5

Don’t want Mathematica to do any integrals whatsoever. Just want it to simply add the exponents inside a double integral.

For instance, doing this even separately

`Collect[(-I)*p3*y + I*p4*y + I*t*Subscript[\[Omega], p3] - 
   I*t*Subscript[\[Omega], p4], {I}]`

is a royal pain, because the output is

(-I)*p3*y + I*p4*y + I*t*Subscript[\[Omega], p3] - I*t*Subscript[\[Omega], p4]

Simplifying just gets one step closer

Simplify[I*((-p3)*y + t*Subscript[\[Omega], p3]) + 
     I*(p4*y - t*Subscript[\[Omega], p4])]

And yet one more step

Collect[(p3 - p4)*y - t*Subscript[\[Omega], p3] + 
  t*Subscript[\[Omega], p4], 
   {t}]

And then one can get back to substituting them in the original expressions. Tedious cut and paste of exponent part, and chance of making silly mistakes.

Just can’t figure out why it can’t simplify this to

E^(I*((-p3)*y + t*Subscript[\[Omega], p3]) + 
      I*(p4*y - t*Subscript[\[Omega], p4]))

under two Integrals?

6

I wonder if this is what you mean:

Integrate[E^(I*p3*y)*hDag[p3]*Subscript[C, p3], p3]*
  Integrate[(h[p4]*Subscript[C, p4])/E^(I*p4*y), p4] /. 
 Integrate[a_, b_]*Integrate[c_, d_] :> Integrate[a*c, b, d]
7

Thanks very much!

Is there a way to extend the formula above so Mathematica can add the exponents, this time in a triple integral, and group the terms in the exponents just as you did, but not do the Integral? Currently, Mathematica will force the ExpIntegralEi output, but I just want it to collect the exponents in x and y separately, this time under the triple integral.

Integrate[E^(I*p3*y)*hDag[p3]*Subscript[C, p3], p3]*
   Integrate[(h[p4]*Subscript[C, p4])/E^(I*p4*y), p4]*
   Integrate[E^(I*k*(x - y))/(\[Mu]^2 - k^2), k] /. 
  Integrate[a_, b_]*Integrate[c_, d_] :> Integrate[a*c, b, d]

which results in

(((-E^(2*I*(x - y)*\[Mu]))*ExpIntegralEi[I*(x - y)*(k - \[Mu])] + 
    ExpIntegralEi[I*(x - y)*(k + \[Mu])])*
   Integrate[E^(I*(p3 - p4)*y)*h[p4]*hDag[p3]*Subscript[C, p3]*
     Subscript[C, p4], p3, p4])/(E^(I*(x - y)*\[Mu])*(2*\[Mu]))
8

You can Inactivate the integrals and do repeated replacements:

Inactivate[
 Integrate[E^(I*p3*y)*hDag[p3]*Subscript[C, p3], p3]*
   Integrate[(h[p4]*Subscript[C, p4])/E^(I*p4*y), p4]*
   Integrate[E^(I*k*(x - y))/(\[Mu]^2 - k^2), k] //. 
  Integrate[a_, b_]*Integrate[c_, d__] :> 
   Integrate[Simplify[a*c], b, d],
 Integrate]
9

This is wizardry, Thanks!
Which part of your code is adding the (p3-p4)y, so I can apply it to get the output within the overall triple integral grouped for x, y, as

E^(I*k*x + I*(p3 - p4 - k)*y)
10

Thank you