Evaluating a basic integral of the exponent.

Question

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?

Answer 1

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

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

Answer 2

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 $$\lim_{R \rightarrow \infty}{\int_{0}^{R}e^{-v}dv} = 1$$