Exercise 8.4.10

Question

\begin{enumerate}[label=(\alph*)] 
\item Use the properties of $e^t$ previously discussed to show
$$\int_0^\infty e^{-t} dt = 1.$$
\item Show
$$\frac{1}{\alpha} = \int_0^\infty e^{-\alpha t} dt, \text{\quad for all } \alpha > 0.$$
 \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
\item The antiderivative of $e^{-t}$ is $-e^{-t}$, so
$$\lim_{b \to \infty} \int_0^b e^{-t} dt = \lim_{b \to \infty} = \lim_{b \to \infty} -e^{-b} + e^0 = 1$$
\item The antiderivative of $e^{-\alpha t}$ is $-e^{-\alpha t} / \alpha$, so
$$\lim_{b \to \infty} \int_0^b e^{-\alpha t} dt  = \lim_{b \to \infty} \frac{1}{\alpha }\left(1 -e^{-\alpha b}\right) = \frac{1}{\alpha}$$
 \end{enumerate}

Exercise 8.4.10

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Exercise 8.4.10

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