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I would like Mathematica to pull constants outside of an integral: e.g., $\int_0^t f[t] g[s] dt \to g[s] \int_0^t f[t] dt$ This has previously been discussed at

replacement rule to pull independent expression outside of Integrate

but the solution does not seem to work for me. My code starts as follows:

ϕ[t_, n_, λ_] = -a[t, n] (λ^2 + I λ);

The next line I write in LaTeX because the mathematica code would not be readable

$ \text{$\Phi $op}[\text{n$\_$},\text{s$\_$},\text{t$\_$},\text{x$\_$},\lambda \_,\text{f$\_$}]\text{:=}\text{FullSimplify}\left[\frac{\phi [s,n,\lambda ]D\left[f \,\text{Exp}\left[i \lambda x +\int_s^t \phi [u,0,\lambda ] \, du\right],\{\lambda ,n\}\right]}{\text{Exp}\left[i \lambda x + \int_s^t \phi [u,0,\lambda ] \, du\right]}\right] $

If I type

Φop[1, s, t, x, λ, 1]

and shift + Enter then Mathematica returns

$ (1-i \lambda ) \lambda a[s,1] \left(x-i \int_s^t -(i+2 \lambda ) a[u,0] \, du\right) $

which is correct, but I want to pull out everything in the integral that does not depend on the variable of integration. So I try

Φop[1, s, t, x, λ, 1] /. 
Integrate[q_*r_, {v_, l_, h_}] /; FreeQ[r, v] :> r*Integrate[q, {v, l, h}]

which returns

$ (1-i \lambda ) \lambda a[s,1] \left(x-i (-i-2 \lambda ) \int_s^t a[u,0] \, du\right) $

Okay, good. This is what I want.

But, now is where I have difficulties. I need to compute

Integrate[Φop[1, s, t, x, λ, 1], {s, 0, t}]

But, I want to get pull everying out of the iterated integral that does not depend on the vaiables of integration, so I try

Integrate[Φop[1, s, t, x, λ, 1] (...) , {s, 0, t}] (...)

where "(...)" I put

/. Integrate[q_*r_, {v_, l_, h_}] /; FreeQ[r, v] :> r*Integrate[q, {v, l, h}]

and Mathematica returns

$ (1-i \lambda ) \lambda \int_0^t a[s,1] \left(x-i (-i-2 \lambda ) \int_s^t a[u,0] \, du\right) \, ds $

I cannot figure out how to get Mathematica to pull the factors $i (-i-2 \lambda )$ and $x$ outside of the outer integral.

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Could you give good Mathematica code for a perhaps simpler example? –  David Park Aug 5 '13 at 23:42
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1 Answer

OK, here is an example that is close to yours. Again I am going to use the Presentations Application, which I sell, because it has the ability to manipulate separable integrals before evaluation.

step0 = Integrate[\[Lambda]1 a[s, 
     1] (x + Integrate[\[Lambda]2 a[u, 0], {u, s, t}]), {s, 0, 
    t}] //. Integrate[q_*r_, {v_, l_, h_}] /; FreeQ[r, v] :> 
   r*Integrate[q, {v, l, h}]

The repeated use of the rule does not completely breakout the integral.

enter image description here

Here are the steps using Presentations. The integrate command is like Integrate but does not evaluate.

<< Presentations`

step1 = integrate[\[Lambda]1 a[s, 
    1] (x + integrate[\[Lambda]2 a[u, 0], {u, s, t}]), {s, 0, t}]
step2 = step1 // OperateIntegrand[Expand]
step3 = step2 // BreakoutIntegral // OperateIntegrand[Expand]
step4 = step3 // BreakoutIntegral 

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That completely breaks out the integral as you wished. Now, just for fun, let's substitute some explicit values for the a functions and evaluate. UseIntegrate[] invokes the regular Mathematica Integrate command (one could also use a specialized table of integrals). Assumption can be placed in the UseIntegrate[assumptions] statement so you don't have to put them in at the beginning and destroy the formatting of Integrate.

step5 = step4 /. {a[s, 1] -> s^2, a[u, 0] -> u}
step6 = step5 // UseIntegrate[] // UseIntegrate[]

enter image description here

enter image description here

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