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Upon reading some mathematical literature, I have encountered the following computation:

$x\in X$, a Banach space, $\alpha=\text{Re }(z)$ for $z\in\mathbb{C}$ and $\omega$ is the growth bound.

$$\int_{0}^{\infty}e^{-(\alpha-\omega)t}\|x\|dt=\frac{1}{\alpha-\omega}\|x\|$$

Could someone please explain this to me?

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    $\begingroup$ is ||x|| a constant? If yes, shame on you. If not the computation seems questionable $\endgroup$
    – tired
    Sep 4, 2015 at 17:15
  • $\begingroup$ @tired Yes, it is a constant. Yes, shame me if you must but I have my excuses (although I will save writing them on here). $\endgroup$
    – Jason Born
    Sep 4, 2015 at 17:18
  • $\begingroup$ that's good, then you have just to calculate the easiest integral of all... :-) $\endgroup$
    – tired
    Sep 4, 2015 at 17:20
  • $\begingroup$ @tired not really. you at the minimum have to specify relationships between the other parameters, $\alpha$ and $\omega$. $\endgroup$
    – ir7
    Sep 4, 2015 at 17:24
  • $\begingroup$ @ir7 $\omega$ is the growth bound and $\alpha=\text{Re }(z)$, where $z\in\mathbb{C}$. $\endgroup$
    – Jason Born
    Sep 4, 2015 at 17:26

2 Answers 2

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Supposing that $\|x\|$ doesn't depend on $t$ and $R \in \mathbb{R}$:

$$\int_{0}^{\infty}e^{-(\alpha-\omega)t}\|x\|dt=\lim_{R \to \infty}\|x\|\int_{0}^{R}e^{-(\alpha-\omega)t}dt=\lim_{R \to \infty}\|x\|\frac{e^{-(\alpha-\omega)t}}{\omega - \alpha}\Bigg|_{0}^{R}=\lim_{R \to \infty}\|x\|\frac{e^{-(\alpha-\omega)R}-1}{\omega - \alpha}=\frac{1}{\alpha-\omega}\|x\|$$

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  • $\begingroup$ Works only if $\alpha > \omega$. $\endgroup$
    – ir7
    Sep 4, 2015 at 17:57
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Note that the variable of integration is $t$, so, we can use the following steps.

$$\int_{0}^{\infty}e^{-(\alpha-\omega)t}\|x\|dt \Rightarrow\|x\|\int_{0}^{\infty}e^{-(\alpha-\omega)t}dt$$.

Here we do a little substitution and let $v = (\alpha - \omega)t$ and so we have $\frac{dv}{dt}=(\alpha - \omega) \Rightarrow \frac{dv}{(\alpha-\omega)} = dt$. Upon substituting, we see that

$$\|x\|\int_{0}^{\infty}e^{-(\alpha-\omega)t}dt \Rightarrow \|x\|\lim_{R \rightarrow \infty}{\frac{1}{(\alpha - \omega)}\int_{0}^{R}e^{-v}dv} \Rightarrow \|x\| \frac{1}{(\alpha - \omega)} \lim_{R \rightarrow \infty} {\int_{0}^{R}e^{-v}dv}$$

Which evaluates to $$\|x\| \frac{1}{(\alpha - \omega)}$$ Since $$\lim_{R \rightarrow \infty}{\int_{0}^{R}e^{-v}dv} = 1$$

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  • $\begingroup$ Works only if $\alpha > \omega$. $\endgroup$
    – ir7
    Sep 4, 2015 at 17:57
  • $\begingroup$ @ir7 yes. You are right. $\endgroup$
    – 9301293
    Sep 4, 2015 at 19:00

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