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

Posted 11 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

POSTED BY: Mike Luntz
6 Replies
Posted 11 years ago

The reply from Andreas gave me a few hints on how to proceed. Thanks.

POSTED BY: Mike Luntz
Posted 11 years ago
POSTED BY: Mike Luntz

If I understand you correctly, you want an answer that involves the indefinite integral of f[y]?

POSTED BY: Frank Kampas
Posted 11 years ago

Thanks for the reply Frank. It does seem to work when the only function being integrated is the exponential.

In[3]:= Integrate[Exp[-x^2 - y^2], y]

Out[3]= 1/2 E^-x^2 Sqrt[\[Pi]] Erf[y]

But when there is another function involved it no longer produces the desired result

In[4]:= Integrate[Exp[-x^2 - y^2] f[y], y]

Out[4]= \[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 BY: Mike Luntz

I think Mathematica can handle it automatically.

In[1]:= Integrate[Exp[-x^2 - y^2], {x, -\[Infinity], \[Infinity]}]

Out[1]= E^-y^2 Sqrt[\[Pi]]
POSTED BY: Frank Kampas
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