Knowing the limit of function, how to caculate the limit of its integral? Know that $K(t) \sim K $ (i.e., $K(t) \rightarrow K$ as $t \rightarrow \infty)$, I want to compute the limit of the following integration
$C(t) = \int_0^t K(u) e^{-\mu(t-u)} \mathrm{d}u$.
I am guessing that
$\lim_{t \rightarrow \infty} c(t) = K \lim_{t \rightarrow \infty} \int_0^t e^{-\mu(t-u)} \mathrm{d}u = K \lim_{t \rightarrow \infty} \frac{1}{\mu} (1-e^{-\mu t}) = \frac{K}{\mu}$.
Is there any theorem I could use to support or verify my guessing?
 A: This may not answer your question but it may be helpful.
According to first mean value theorem for integration, if $f : [a, b] → R$ is continuous and $g$ is an integrable function that does not change sign on $[a, b]$, then there exists $c$ in $(a, b)$ such that
$$\int_a^b f(x) g(x) \mathrm{d} x = f(c)\int_a^b g(x) \mathrm{d} x.$$
In your case, since $e^{-\mu(t-u)}$ doen't change sign and if you let $K$ to be continuous,then
$$\lim_{t\to \infty} c(t) =\lim_{t\to \infty} \int_0^t K(u) e^{-\mu(t-u)} \mathrm{d} u\\
=K(c) \lim_{t\to \infty} \int_0^t e^{-\mu(t-u)} \mathrm{d} u$$ for some $c$.
A: I'm assuming $K$ is integrable on finite intervals (so the integrals exist), and $\mu > 0$.  Given $\epsilon > 0$, take $N$ so $|K(u)-K| < \epsilon$ for $u \ge N$.  If $\int_0^N |K(u)| \; du = L$,
$$ \left|\int_0^N K(u) e^{-\mu(t-u)}\; du\right| \le e^{-\mu(t-N)} L \to 0 \ \text{as}\ t \to \infty$$
On the other hand
$$ \int_{N}^t  e^{-\mu(t-u)} \; du = \frac{1-\exp(N\mu - t \mu)}{\mu} $$
so
$$ (K-\epsilon) \frac{1-\exp(N\mu-t\mu)}{\mu} < \int_N^t K(u) e^{-\mu(t-u)}\; du < (K+\epsilon) \frac{1-\exp(N\mu-t\mu)}{\mu}$$
where the left and right bounds go to $(K-\epsilon)/\mu$ and $(K+\epsilon)/\mu$ as $t \to \infty$.
Thus the lim inf of your integral as $t \to \infty$ is at least $(K-\epsilon)/\mu$ and the lim sup is at most $(K+\epsilon)/\mu$.
Taking $\epsilon \to 0+$, we conclude that the limit is indeed $K/\mu$.
A: Why would this be true ? In $[0,t]$, $K(t)\ne K$.
