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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] 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


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 9 years ago

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

POSTED BY: Mike Luntz
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 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 9 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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