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# How can I change exponential of sum to product of exponentials?

Posted 9 years ago
 I would like to be able to translate Exp[a+b]  to Exp[a] Exp[b]  but when I execute Exp[a+b]/.Exp[m_+n_]-> Exp[m] Exp[n]  Mathematica seems to re-combine the product. The reason for wanting to split the exponential is that I have an integral of the form Integrate[Exp[f[x]+g[y]] h[y],y]  that I would like to evaluate symbolically to yield Exp[f[x]] Integrate[Exp[g[y]] h[y],y]  I have determined that Mathematica can recognize when a factor that does not depend on the variable of integration can be pulled outside an integration so that Integrate[Exp[x] f[y],y]  is evaluated as Exp[f[x]] Integrate[f[y],y]  The above is the gist of my question, but a concrete example of what I am trying to get is to convert this Integrate[Exp[(I T(l(-lam+th)+n(mu-w)))] P[lam] P[th] P[-mu+th] P[lam-w],th,lam]  to this Exp[ I n(mu-w)] Integrate[Exp[(I T(l(-lam+th)))]P[lam] P[th] P[-mu+th] P[lam-w],th,lam]  I copied and pasted these equations from a notebook and tried to edit the result to make it more readable, so if there is a mismatched parenthesis the problem is my editing, not a fundamental error in the equation EDIT:: I tried using HoldForm which did maintain the product of exponentials but the integration did not move the exponential outside the integration  mykern = Exp[u + v] h[v] /. Exp[a_ + b_] -> HoldForm[Exp[a] Exp[b]] Integrate[mykern, v]  results in Integrate[h[v] (Exp[u] Exp[v]),v]  and releasing the hold just on that result just changes back to an exponential of the sum
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Posted 9 years ago
 The reply from Andreas gave me a few hints on how to proceed. Thanks.
Posted 9 years ago
Posted 9 years ago
 Yes. I desire that anything not depending on the variable of integration be moved outside the integral. In my problem there are several of these integrals that are summed together. Some will have the same function under the integral that depends on the variable of integration. For these I need to sum up that part of the function not depending on the variable of integration.
Posted 9 years ago
 Thanks for the reply Frank. It does seem to work when the only function being integrated is the exponential. In:= Integrate[Exp[-x^2 - y^2], y] Out= 1/2 E^-x^2 Sqrt[\[Pi]] Erf[y] But when there is another function involved it no longer produces the desired result In:= Integrate[Exp[-x^2 - y^2] f[y], y] Out= \[Integral]E^(-x^2 - y^2) f[y] \[DifferentialD]y In my problem the function f[y] is not specified so that the integral must be left as a general indefinite integral
Posted 9 years ago
 If I understand you correctly, you want an answer that involves the indefinite integral of f[y]?
Posted 9 years ago
 I think Mathematica can handle it automatically. In:= Integrate[Exp[-x^2 - y^2], {x, -\[Infinity], \[Infinity]}] Out= E^-y^2 Sqrt[\[Pi]]