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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

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  • $\begingroup$ Are you working with Riemann or with Lebesgue integrals? $\endgroup$ – Daniel Fischer Jul 19 '15 at 22:21
  • $\begingroup$ Riemann integrals $\endgroup$ – MSm Jul 19 '15 at 22:22
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    $\begingroup$ Either way, I think starting with the substitution $y = hx$ is the best method. $\endgroup$ – Daniel Fischer Jul 19 '15 at 22:22
  • $\begingroup$ $y=hx$ or $y=-hx$ ? because I tried it already $\endgroup$ – MSm Jul 19 '15 at 22:24
  • $\begingroup$ $y = hx$. Have you thought of splitting the integral at some $a$? $\endgroup$ – Daniel Fischer Jul 19 '15 at 22:26
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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 $e^{-h\delta} \overset{h \to \infty}{\longrightarrow} 0$

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  • $\begingroup$ thanks!, I have a question about it, how the $f(0)$ got into the integral?, $\endgroup$ – MSm Jul 19 '15 at 22:54
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    $\begingroup$ 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} $. $\endgroup$ – Conrado Costa Jul 19 '15 at 22:57
  • $\begingroup$ wow, thanks again. that's a very nice trick! $\endgroup$ – MSm Jul 19 '15 at 22:59

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