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I want to solve the following integral:

Integrate[( E^(-ω/Λ) Sin[((ω + Ω) t)/2] Sin[(ω t)/ 2])/((ω + Ω)/4) , {ω, 0, ∞}]

But unfortunately Mathematica cannot solve it. Is there any way to solve this integral analytically? If not, what is the most innocent approximation to solve it?

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Assuming a little about domains lets it find the integral

Assuming[Λ > 0 && Ω > 0 && t > 0, 
  Integrate[(E^(-ω/Λ) Sin[((ω+Ω) t)/2] Sin[(ω t)/2])/((ω+Ω)/4), {ω, 0, ∞}]]

Λ being positive is required for the integral to converge. If t==0 the integral is simply zero. Assuming t>0 gives the same result as t<0, but assuming t!=0 doesn't seem to be quite enough. If you assume Ω<0 you get a very different result.

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  • $\begingroup$ The assumptions are physically reasonable: Λ and Ω are frequencies and t is time, so they are positive. Thanks man. $\endgroup$ – Farhad Jan 11 '16 at 19:00
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You can get a simpler form by converting the trig functions to exponentials (TrigToExp), expanding, integrating each term separately, and simplifying.

expr = (E^(-ω/Λ) Sin[((ω + Ω) t)/
         2] Sin[(ω t)/2])/((ω + Ω)/4) // 
    TrigToExp // Expand;

Assuming[{Λ > 0, Element[t, Reals], Ω > 0},
 Simplify[Integrate[#, {ω, 0, ∞}] & /@ expr]]

(*  E^(-(1/2) I t Ω + Ω/Λ) (-(1 + E^(
       I t Ω)) \
ExpIntegralEi[-(Ω/Λ)] - 
   Gamma[0, (-I t + 1/Λ) Ω] - 
   E^(I t Ω)
     Gamma[0, (I t + 1/Λ) Ω])  *)

Note that you need only assume that t is real.

For the case t == 0

FullSimplify[%, {Λ > 0, t == 0, Ω > 0}]

(*  0  *)
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  • $\begingroup$ Interesting that the initial solution has some discontinuities when it is plotted, but the second solution does not. I don't know whether this is due to the simpler function or the assumption that t is a real number. $\endgroup$ – Jack LaVigne Mar 7 '16 at 17:02

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