A limit of sequence of improper integral

Question

I was baffled to deal with such problem. The result of this problem is obvious but I don’t how to solve it with a precise method because too many parameters in this integral.

Answer 1

Since $f$ is bounded the improper integral is same as Lebesgue integral. The answer is $k/\alpha$. Make the substitution $y=a_nx$ to get $\frac 1 {na_n} \int_0^{\infty} f(\frac y {a_n}) e^{-y}\, dy$. It is enough to show that $\int_0^{\infty} f(\frac y {a_n}) e^{-y}\, dy \to k$. Since $a_n \to 0$ we have $f(\frac y {a_n}) \to k$. The proof can be completed using DCT.

Answer 2

Enforcing the substitution $x\mapsto nx$ reveals

$$\frac1n\int_0^\infty e^{-a_nx}f(x)\,dx=\int_0^\infty f(nx)e^{-na_nx}\,dx$$

Since $f$ is continuous, it is bounded. And since $na_n\to \alpha>0$ is also bounded, the Dominated Convergence Theorem (with dominating function $||f||_\infty e^{-\min(na_n)x}$) guarantees that

$$\begin{align} \lim_{n\to\infty}\int_0^\infty f(nx)e^{-na_nx}\,dx&=\int_0^\infty \lim_{n\to\infty}f(nx)e^{-na_nx}\,dx\\\\ &=\int_0^\infty ke^{-\alpha x}\,dx\\\\ &=k/\alpha \end{align}$$