# Help with a limit of an integral: $\lim_{h\to \infty}h\int_{0}^\infty{{ {e}^{-hx}f(x)} dx}=f(0)$

I'm not sure how to handle limits and integral and I would like some help with the following one:

let $f:[0,\infty)\rightarrow \Bbb{R}$ be a continuous and bounded function, show that $$\lim_{h\to \infty}h\int_{0}^\infty{{ {e}^{-hx}f(x)} dx}=f(0)$$

I tried many things from The fundamental theorem of calculus and define $F$ such that $F'=f$ and use integration by parts to get $\int_{0}^\infty{{ {e}^{-hx}f(x)} dx}={{ F(x){e}^{-hx}} dx}|_{0}^{\infty}+{h}\int_{0}^\infty{{ {e}^{-hx}F(x)} dx}$ but it leads nowhere.

I think we can say $\int_{0}^\infty{{ {e}^{-hx}f(x)} dx}=L<\infty$ from Dirichlet test but now sure how to use it.

thx

• Are you working with Riemann or with Lebesgue integrals? Jul 19, 2015 at 22:21
• Riemann integrals
– MSm
Jul 19, 2015 at 22:22
• Either way, I think starting with the substitution $y = hx$ is the best method. Jul 19, 2015 at 22:22
• $y=hx$ or $y=-hx$ ? because I tried it already
– MSm
Jul 19, 2015 at 22:24
• $y = hx$. Have you thought of splitting the integral at some $a$? Jul 19, 2015 at 22:26

Note that $h\int_{0}^\infty{{ {e}^{-hx}} dx} = 1$. Fix $\epsilon> 0$, by continuity there is a $\delta>0$ such that $|x|< \delta \Rightarrow |f(x)-f(0)|\leq \epsilon$. Note also (by boundedness) that there is a $M>0$ such that for every $x \in \Bbb{R}$,$|f(x)| \leq M$. Therefore, \begin{align}\bigg|h\int_{0}^\infty{{ {e}^{-hx}f(x)} dx} - f(0)\bigg| &=\bigg| h\int_{0}^\infty{{ {e}^{-hx}(f(x) - f(0))} dx} \bigg|\\&=\bigg| h\int_0^\delta {{ {e}^{-hx}(f(x) - f(0))} dx} + h\int_\delta^\infty {{ {e}^{-hx}(f(x) - f(0))} dx} \bigg|\\ &\leq \bigg|h\int_\delta^\infty {{ {e}^{-hx}}}\epsilon\, dx \bigg|+ \bigg|2M h\int_\delta^\infty {{ {e}^{-hx}} dx}\bigg| \\ &\leq \epsilon + 2M (e^{-h\delta} ) \overset{h \to \infty}{\leq} 2 \epsilon \end{align}
Since $e^{-h\delta} \overset{h \to \infty}{\longrightarrow} 0$
• thanks!, I have a question about it, how the $f(0)$ got into the integral?,
• Since $h\int_{0}^\infty{{ {e}^{-hx}} dx} = 1$ then $f(0) = f(0)h\int_{0}^\infty{{ {e}^{-hx}} dx} = h\int_{0}^\infty{{ {e}^{-hx}}f(0) dx}$. Jul 19, 2015 at 22:57