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)

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]))

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?

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…