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\subsection{Improper Integrals (infinite integrals)}

\begin{definition*}
    Suppose $f : [a, \infty) \to \RR$ integrable
    (and bounded) on every interval $[a, R]$ and
    that as $R \to \infty$
    \[ \int_a^R f(x) \dd x \to l \]
    Then we say that $\int_a^\infty f(x) \dd x$
    \emph{exists} or \emph{converges} and that its
    value is $l$. If $\int_a^R f(x) \dd x$ does
    not tend to a limit, we say that
    $\int_a^\infty f(x) \dd x$ \emph{diverges}. A
    similar definition applies to
    $\int_{-\infty}^a f(x) \dd x$. If
    \[ \int_a^\infty f = l_1 \qquad \text{and}
    \qquad \int_{-\infty}^a f = l_2 \]
    we write
    \[ \int_{-\infty}^\infty f = l_1 + l_2 \]
    (independent of the particular value of $a$).
\end{definition*}

\begin{note*}
    This last bit is \emph{not} the same as saying
    that
    \[ \lim_{R \to \infty} \int_{-R}^R f(x) \dd x \]
    exists. It is stronger: for example
    \[ \int_{-R}^R x \dd x = 0 \]
\end{note*}

\begin{example*}
    $\int_1^\infty \frac{\dd x}{x^k}$ converges if
    and only if $k > 1$. Indeed, if $k \neq 1$
    then
    \[ \int_1^R \frac{\dd x}{x^k} = \left.
    \frac{x^{1 - k}}{1 - k} \right|_1^R =
    \frac{R^{1 - k} - 1}{1 - k} \]
    and as $R \to \infty$ this limit is finite if
    and only if $k > 1$. If $k = 1$,
    \[ \int_1^R \frac{\dd x}{x} = \log R \to
    \infty \]
\end{example*}

\subsubsection*{Remarks}

\begin{enumerate}[(1)]
    \item $\frac{1}{\sqrt{x}}$ continuous on
        $[\delta, 1]$ for any $\delta > 0$, and
        \[ \int_\delta^1 \frac{1}{\sqrt{x}} \dd x
        = \left. 2\sqrt{x} \right|_\delta^1 = 2 -
        2\sqrt{\delta} \to 2 \]
        as $\delta \to 0$.
        \begin{center}
            \includegraphics[width=0.6\linewidth]
            {images/179d97f2c4e011ec.png}
        \end{center}
        $\frac{1}{\sqrt{x}}$ is unbounded on $(0,
        1]$.
        \[ \int_0^1 \frac{\dd x}{\sqrt{x}} =
        \lim_{\delta \to 0} \int_\delta^1
        \frac{\dd x}{\sqrt{x}} = 2 \]
        Exercise: give a general definition for
        cases like this.
        \begin{align*}
            \int_0^1 \frac{\dd x}{x}
            &= \lim_{\delta \to 0} \int_\delta^1
            \frac{\dd x}{x} \\
            &= \lim_{\delta \to 0} \left( \left.
            \log x \right|_\delta^1 \right) \\
            &= \lim_{\delta \to 0} (\log 1 - \log
            \delta)
        \end{align*}
        does \emph{not exist}.
    \item If $f \ge 0$ and $g \ge 0$, for $x \ge a$
        and
        \[ f(x) \le Kg(x) \qquad \forall\,\, x \ge
        a \]
        with $K$ a constant, then
        \[ \int_a^\infty g \text{ converges}
        \implies \int_a^\infty f \text{ converges} \]
        and
        \[ \int_a^\infty f \le K \int_a^\infty g \]
        Just note that
        \[ \int_a^R f \le K \int_a^R g \]
        The function $R \to \int_a^R f$ is
        increasing ($f \ge 0$) and bounded above
        (since $\int_a^\infty g$ converges). Take
        $l = \sup_{R \ge a} \int_a^R f < \infty$,
        and check that
        \[ \lim_{R \to \infty} \int_a^R f = l .\]
        Given $\eps > 0$, $\exists\,\, R_0$ such
        that
        \[ \int_a^{R_0} f \ge l - \eps \]
        Thus if $R \ge R_0$,
        \[ \int_a^R f \ge \int_a^{R_0} \ge l -
        \eps \]
        \[ \implies 0 \le l - \int_a^R f \le \eps \]
        \begin{example*}
            $\int_0^\infty e^{-\frac{x^2}{2}} \dd
            x$. Note $e^{-\frac{x^2}{2}} \le
            e^{-\frac{x}{2}}$ for $x \ge 1$. Note that
            \[ \int_1^R e^{-\frac{x}{2}} \dd x =
            \half [e^{-\half} - e^{-\frac{R}{2}}]
            \to \frac{e^{-\half}}{2} \]
            hence $\int_0^\infty
            e^{-\frac{x^2}{2}}$ converges.
        \end{example*}
    \item We know that if $\sum a_n$ converges,
        then $a_n \to 0$. $\int_a^\infty f$
        converges may \emph{not} imply that $f \to
        0$.
        \begin{example*}
            \begin{center}
                \includegraphics[width=0.6\linewidth]
                {images/8f20573cc4e111ec.png}
            \end{center}
            \[ \text{Area}(\triangle) =
            \frac{2}{(n + 1)^2} \]
            so since $\sum \frac{2}{(n + 1)^2}$
            converges, $\int_0^\infty f$
            converges. But $f(n) = 1$, so $f
            \not\to 0$.
        \end{example*}
\end{enumerate}

\subsection{The Integral Test}

\begin{theorem}[integral test]
    Let $f(x)$ be a positive \emph{decreasing}
    function for $x \ge 1$. Then
    \begin{enumerate}[(1)]
        \item The integral $\int_1^\infty f(x) \dd
            x$ and the series $\sum_1^\infty f(n)$
            both converge or diverge.
        \item As $n \to \infty$,
            \[ \sum_{r = 1}^n f(r) - \int_1^n f(X)
            \dd x \]
            tends to a limit $l$ such that $0 \le
            l \le f(1)$.
    \end{enumerate}
\end{theorem}

\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/18e426bac4e211ec.png}
\end{center}

\begin{note*}
    $f$ decreasing $\implies$ $f$ integrable on
    every bounded subinterval by Theorem 5.4.
\end{note*}

\begin{proof}
    If $n - 1 \le x \le n$, then
    \[ f(n - 1) \ge f(x) \ge f(n) \]
    hence
    \[ f(n - 1) \ge \int_{n - 1}^n f(x) \dd x \ge
    f(n) \tag{$*$} \]
    Adding
    \[ \sum_1^{n - 1} f(r) \ge \int_1^n f(x) \dd x
    \ge \sum_2^n f(r) \tag{$**$} \]
    From this claim (1) is \emph{clear}. For the
    proof of (2) set
    \[ \phi(n) = \sum_1^n f(r) - \int_1^n f(x) \dd
    x \]
    Then
    \[ \phi(n) - \phi(n - 1) = f(n) - \int_{n -
    1}^n f(x) \dd x \le 0 \]
    (using ($*$)) From ($**$)
    \[ 0 \le \phi(n) \le f(1) \]
    Thus $\phi(n)$ is decreasing and tends to a
    limit $l$ such that
    \[ 0 \le l \le f(1) .\]
\end{proof}