2
$\begingroup$

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?

$\endgroup$
0
2
$\begingroup$

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 $c$.

$\endgroup$
0
2
$\begingroup$

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 $K/\mu$.

$\endgroup$
1
  • $\begingroup$ Thank you so much for the proof! $\endgroup$
    – Rex Lee
    Oct 22 '20 at 15:41
1
$\begingroup$

Why would this be true ? In $[0,t]$, $K(t)\ne K$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.