# 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?

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 $$c$$.

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 $$K/\mu$$.

• Thank you so much for the proof! Oct 22 '20 at 15:41

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