Exercise 8.4.15

Question

\begin{enumerate}[label=(\alph*)] 
\item Show that the improper integral $\int_0^\infty e^{-xt} dt$ converges uniformly to $1/x$ on the set $[1/2, \infty)$.
\item Is the convergence uniform on $0, \infty$?
 \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
\item From our earlier work, we have that
$$\int_0^\infty e^{-xt} dt - \int_0^d e^{-xt} = \frac{e^{-xd}}{x} \leq 2e^{-d/2}$$
Notably this is not dependent on $x$, and since $\lim_{d \to \infty} 2e^{-d/2} = 0$, we have that $\int_0^\infty e^{-xt} dt$ converges uniformly.
\item For any fixed $d$, $$\lim_{x \to 0^+} \frac{e^{-xd}}{x} = \infty$$
implying that the convergence cannot be uniform on $(0, \infty)$.
 \end{enumerate}

Exercise 8.4.15

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Exercise 8.4.15

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