My Blog: Anh-Tourguide: Several Interesting Limits

Recently, I have learnt from my friend, ZJ, the following result

Assume that $F:\mathbb R \to \mathbb R$ is absolutely integrable. Then
$\displaystyle\begin{gathered} \mathop {\lim }\limits_{t \to \pm \infty } {e^{2t}}\int_t^{ + \infty } {F(x){e^{ - 2x}}dx} = 0, \hfill \\ \mathop {\lim }\limits_{t \to \pm \infty } {e^{ - 2t}}\int_{ - \infty }^t {F(x){e^{ - 2x}}dx} = 0. \hfill \\ \end{gathered}$

The result seems reasonable by the following observation, for example, we consider the first identity when $t \to +\infty$ . Then the factor

$\displaystyle\int_t^{ + \infty } {F(x){e^{ - 2x}}dx}$

decays faster then the exponent function $\exp (2t)$ . This may be true, of course we need to prove mathematically, because the integrand contains the term $\exp (-2x)$ which turns out to be a good term since $x \geqslant t$ . So here is the trick in order to solve such a problem.

A proof of

$\displaystyle\mathop {\lim }\limits_{t \to+\infty } {e^{2t}}\int_t^{ + \infty } {F(x){e^{ - 2x}}dx} = 0.$

To prove this, we split the function under the limit sign into two parts as the following

$\displaystyle {e^{2t}}\int_t^{ + \infty } {F(x){e^{ - 2x}}dx} = {e^{2t}}\left[ {\int_t^{2t} {F(x){e^{ - 2x}}dx} + \int_{2t}^{ + \infty } {F(x){e^{ - 2x}}dx} } \right].$

The term

$\displaystyle {e^{2t}}\int_t^{2t} {F(x){e^{ - 2x}}dx}$

can be estimated as follows

$\displaystyle\begin{gathered} \left| {{e^{2t}}\int_t^{2t} {F(x){e^{ - 2x}}dx} } \right| \leqslant {e^{2t}}\int_t^{2t} {|F(x)|{e^{ - 2x}}dx} \hfill \\ \qquad\qquad\qquad\leqslant {e^{2t}}\int_t^{2t} {|F(x)|{e^{ - 2t}}dx}= \int_t^{2t} {|F(x)|dx} \to 0 \hfill \\ \end{gathered}$

as $t \to +\infty$ . The term

$\displaystyle {{e^{2t}}\int_{2t}^{ + \infty } {F(x){e^{ - 2x}}dx} }$

can be estimated as the following

$\displaystyle\begin{gathered} \left| {{e^{2t}}\int_{2t}^{ + \infty } {F(x){e^{ - 2x}}dx} } \right| \leqslant {e^{2t}}\int_{2t}^{ + \infty } {|F(x)|{e^{ - 2x}}dx} \hfill \\ \qquad\qquad\qquad\leqslant {e^{2t}}\int_{2t}^{ + \infty } {|F(x)|{e^{ - 4t}}dx} \hfill \\ \qquad\qquad\qquad= {e^{ - 2t}}\int_{2t}^{ + \infty } {|F(x)|dx}\leqslant {e^{ - 2t}}\int_{ - \infty }^{ + \infty } {|F(x)|dx} \to 0 \hfill \\ \end{gathered}$

as $t \to +\infty$ . Thus, it is clear now to see why the first identity holds.

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Several Interesting Limits

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