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\subsubsection*{Hyperbolic Functions}

(Hyperbolic sine and cosine)

\begin{definition*}
    $\cosh z = \half (e^z + e^{-z})$, $\sinh z =
    \half (e^z - e^{-z})$. Alternatively, $\cosh z
    = \cos(iz)$, $\sinh z = -i\sin(iz)$.
\end{definition*}

\noindent
One can also prove that $(\cosh z)' = \sinh z$ and
$(\sinh z)' = \cosh z$. (This is left as an
exercise). We also have
\[ \cosh^2 z - \sinh^2 z = 1 \]
The rest of the trigonometric functions ($\tan,
\cot, \sec, \cosec$) are defined in the usual way.

\newpage
\section{Integration}

$f : [a, b] \to \RR$ \emph{bounded}. (i.e. there
exists $K$ such that $|f(x)| \le K\,\,\forall\,\,
x \in [a, b]$)

\begin{definition*}
    A dissection (or partition) $\mathcal{D}$ of
    $[a, b]$ is a finite subset of $[a, b]$
    containing the endpoints $a$ and $b$. We write
    \[ \mathcal{D} = \{x_0, x_1, \dots, x_4\} \]
    with $a = x_0 < x_1 < \cdots < x_{n - 1} < x_
    = b$.
\end{definition*}

\begin{definition*}
    We define the upper sum and lower sum
    associated with $\mathcal{D}$ by
    \[ S(f, \mathcal{D}) = \sum_{j = 1}^n (x_j -
    x_{j - 1}) \sup_{x \in [x_{j - 1}, x_j]} f(x)
    \tag{upper} \]
    \[ s(f, \mathcal{D}) = \sum_{j = 1}^n (x_j -
    x_{j - 1}) \inf_{x \in [x_{j - 1}, x_j]} f(x)
    \tag{lower} \]
    Clearly $s(f, \mathcal{D}) \le S(f,
    \mathcal{D})$ for all $\mathcal{D}$.
\end{definition*}

\begin{lemma}
    If $\mathcal{D}$ and $\mathcal{D}'$ are
    dissections with $\mathcal{D}' \supseteq
    \mathcal{D}$, then
    \[ S(f, \mathcal{D}) \ge S(f, \mathcal{D}') \ge
    s(f, \mathcal{D}') \ge s(f, \mathcal{D}) \]
\end{lemma}

\begin{proof}
    \[ S(f, \mathcal{D}') \ge s(f, \mathcal{D}') \]
    is obvious. Suppose $\mathcal{D}'$ contains an
    extra point than $\mathcal{D}$, let's say $y
    \in (x_{r - 1}, x_r)$. Then clearly
    \[ \sup_{x \in [x_{r - 1}, y]} f(x), \sup_{x
    \in [y, x_r]} \le \sup_{x \in [x_{r - 1},
    x_r]} f(x) \]
    \[ \implies (x_r - x_{r - 1}) \sup_{x \in
    [x_{r - 1}, x_r]} f(x) \ge (y - x_{r - 1})
    \sup_{x \in [x_{r - 1}, y]} f(x) + (x_r - y)
    \sup_{x \in [y, x_r]} f(x) \]
    \[ \implies S(f, \mathcal{D}) \ge S(f,
    \mathcal{D}') \]
    The same for $s$ and the same if
    $\mathcal{D}'$ has more extra points than
    $\mathcal{D}$.
\end{proof}

\begin{lemma}
    $\mathcal{D}_1, \mathcal{D}_2$ two arbitrary
    dissections. Then
    \[ S(f, \mathcal{D}_1) \ge S(f, \mathcal{D}_1
    \cup \mathcal{D}_2) \ge s(f, \mathcal{D}_1
    \cup \mathcal{D}_2) \ge s(f, \mathcal{D}_2) \]
    and in particular
    \[ S(f, \mathcal{D}_1) \ge s(f, \mathcal{D}_2) \]
\end{lemma}

\begin{proof}
    Take $\mathcal{D}' = \mathcal{D}_1 \cup
    \mathcal{D}_2 \supseteq \mathcal{D}_1,
    \mathcal{D}_2$ in the previous lemma.
\end{proof}

\begin{definition*}
    The \emph{upper integral} of $f$ is
    \[ I^*(f) = \inf_{\mathcal{D}} S(f,
    \mathcal{D}) \]
    (always exists!) The \emph{lower integral} of
    $f$ is
    \[ I_*(f) = \sup_{\mathcal{D}} s(f,
    \mathcal{D}) \]
    By Lemma 5.2,
    \[ I^*(f) \ge I_*(f) \]
    because
    \[ S(f, \mathcal{D}_1) \ge s(f, \mathcal{D}_2) \]
    \[ I^*(f) = \inf_{\mathcal{D}_1} S(f,
    \mathcal{D}_1) \ge s(f, \mathcal{D}_2) \]
    \[ I^*(f) \ge \sup_{\mathcal{D}_2} s(f,
    \mathcal{D}_2) = I_*(f) \]
\end{definition*}

\begin{definition*}
    A bounded function $f : [a, b] \to \RR$ is
    said to be \emph{Riemann integrable} (or just
    integrable) if
    \[ I^*(f) = I_*(f) \]
    and we set
    \[ \int_a^b f(x) \dd x = I^*(f) = I_*(f) =
    \int_a^b f \]
\end{definition*}

\begin{example*}
    \[ f(x) = \begin{cases}
        1 & x \in \QQ \cap [0, 1] \\
        0 & x \not\in \QQ \cap [0, 1]
    \end{cases}
    \]
    $f : [0, 1] \to \RR$; $f$ is \emph{not}
    Riemann integrable:
    \[ \sup_{x \in [x_{j - 1}, x_j]} f(x) = 1,
    \qquad \inf_{x \in [x_{j - 1}, x_j]} f(x) = 0 \]
    Hence $s(f, \mathcal{D}) = 1$ and $s(f,
    \mathcal{D}) = 0$ for all $\mathcal{D}$. Hence
    $I^*(f) = 1$, but $I_*(f) = 0$.
\end{example*}
