Thanks to everybody in advance. I have this integral:
$$ e^{-at} \int_{0}^{t} e^{au}f(u) du $$ where f(t) is something I can't integrate.
Is there a way to put the exponential function back in the integral in order to have something like this:
$$ e^{-at} \int_{0}^{t} e^{au}f(u) du = \int_{0}^{t} g(t) dt $$
Thanks again for any help.
Ale
If we consider $\int e^{at}f(t)dt$, the integral can be written as a family of function $F(t)+C_1$, where $C_1$ is an arbitrary constant. Therefore $e^{-at}\int e^{at}f(t)dt = e^{-at}F(t) + C_1e^{-at}$.
However, if there exist such $g(t)$ in general, then the integral is in the form of $G(t)+C_2$. The arbitrary constant appears as coefficient of another function, which means the operation you wish to do is impossible, unless $a=0$.
For example, consider a simple case, where $f(t)=a$.
$$e^{-at}\int e^{at}adt = e^{-at}\left(e^{at}+C\right) = 1+Ce^{-at}$$
But if we differentiate that with respect to $t$,
$$\frac{d}{dt}\left(1+Ce^{-at}\right)=-Cae^{-at}$$
which means $g(t)$ is not a function independent of $C$.