Doubt in expansion of this perturbation equation

Question

I have a query that hopefully someone can help with...

Given a small perturbation:

$$ x(t) = x_0(t) + \epsilon x_1(t) + \epsilon ^2 x_2(t) + O\left(\epsilon ^3 \right) $$

I want to substitute this into the following:

$$ \epsilon x(t) \int_0^\infty f(z)x(s) ds $$

Now, what I have done is as follows:

$$ \epsilon \left(x_0(t) + \epsilon x_1(t) + \epsilon ^2 x_2(t) \right) \int_0^\infty f(z)\left(x_0(s) + \epsilon x_1(s) + \epsilon ^2 x_2(s) \right) ds $$

Now, where I'm getting a little confused is, for the First Order do we have:

$$ \epsilon x_0(t) \int_0^\infty f(z)\epsilon x_1(s) ds $$

Or, am I correct in saying that due to $ \epsilon $ being constant, this is brought out of the integral and therefore it becomes second order and therefore, the First Order is actually:

$$ \epsilon x_0(t) \int_0^\infty f(z)x_0(s) ds $$

If I am correct and it is the second option, is there a way of simplifying the $ x(t) $ and the $ x(s) $ into two integrals, one over the range 0 to t and one over the range 0 to infinity ?

Any Pointers are appreciated.

Ok, I solved this on my own. The second option is correct and I have split into seperate integrals. Not sure how to close the question, but no further help required.

Answer 1

The second option above is correct as $ \epsilon $ is basically a constant that is taken outside of the Integral.

The second query I had is a simple splitting of the integral into two ranges, $ 0 $ to $ u $ and $ u $ to $ \infty $.