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\centerline{MATH 136 -- Calculus 2}
\centerline{First Practice on Improper Integrals}
\centerline{February 25, 2020}
\vskip 10pt
\noindent
{\it Background}
\vskip 10pt
Whenever $a = -\infty$ or $b = +\infty$ or both, the integral 
$$\int_a^b f(x)\ dx$$
is said to be an \emph{improper integral}. Improper integrals are always
 handled by \emph{taking limits of
``ordinary'' integrals}.  
\begin{itemize}
\item We say $\int_a^\infty f(x)\ dx$ \emph{converges}
if the limit 
$$\lim_{b \to \infty} \int_a^b f(x)\ dx$$ 
exists and we say the integral \emph{diverges} otherwise.
\item We say $\int_{-\infty}^b f(x)\ dx$ \emph{converges}
if the limit 
$$\lim_{a \to -\infty} \int_a^b f(x)\ dx$$ 
exists and we say the integral \emph{diverges} otherwise.
\item We say $\int_{-\infty}^\infty f(x)\ dx$ \emph{converges}
if, splitting at any finite value $c$ (often at $c = 0$), \emph{both of the limits} here: 
$$\lim_{a \to -\infty} \int_a^c f(x)\ dx + \lim_{b \to +\infty} \int_c^b f(x)\ dx$$ 
exist and we say the integral \emph{diverges} otherwise.  (Note this says that if either one
of the two limits does not exist, then the integral diverges.)
\end{itemize}

When $[a,b]$ is a finite interval, the integral 
$$\int_a^b f(x)\ dx$$
is also said to be \emph{improper} whenever $f(x)$ has one or more \emph{discontinuities}
in the interval $[a,b]$.  These can happen at the endpoints $a$ or $b$, or at some
$c$ in $(a,b)$.  In all cases, improper integrals are handled by \emph{taking limits of
``ordinary'' integrals}.  For the purposes of this discussion, let us suppose that
$f(x)$ has only one point of discontinuity in the interval.
\begin{itemize}
\item If $f$ is discontinuous at $a$, then we say $\int_a^b f(x)\ dx$ \emph{converges}
if the limit 
$$\lim_{c \to a^+} \int_c^b f(x)\ dx$$ 
exists and we say the integral \emph{diverges} otherwise.
\item If $f$ is discontinuous at $b$, then we say $\int_a^b f(x)\ dx$ \emph{converges}
if the limit 
$$\lim_{c \to b^-} \int_a^c f(x)\ dx$$ 
exists and we say the integral \emph{diverges} otherwise.
\item If $f$ is discontinuous at $d$ in $(a,b)$, then we say $\int_a^b f(x)\ dx$ \emph{converges}
if \emph{both of the limits} here: 
$$\lim_{c \to d^-} \int_a^c f(x)\ dx + \lim_{c \to d^+} \int_c^b f(x)\ dx$$ 
exist and we say the integral \emph{diverges} otherwise.  (Note this says that if either one
of the two limits does not exist, then the integral diverges.)
\end{itemize}
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\noindent
{\it Questions}
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For each of the following,
\begin{enumerate}
\item[(i)] Explain why the integral is improper.
\item[(ii)] Set up the appropriate limit(s) and evaluate.
\item[(iii)]  Say whether the integral converges or diverges.
\end{enumerate}
\begin{enumerate}
\item $\displaystyle \int_{-\infty}^2 e^{10x}\ dx$
\item[] {\it Solution:} This is improper because of the infinite lower limit
of integration. We have 
$$\lim_{a\to -\infty} \int_a^2 e^{10x}\ dx = \lim_{a\to -\infty} \frac{1}{10}(e^{20} - e^a) = 
\frac{1}{10}e^{20}.$$
Hence the integral \emph{converges}.
\item $\displaystyle \int_4^{\infty} \frac{1}{x^2 + 16}\ dx$
\item[] {\it Solution:} For this and the next example, we use the indefinite integration formula.
$$\int \frac{1}{x^2 + 16}\ dx = \frac{1}{4} \tan^{-1}(x/4) + C$$
This integral is improper because of the infinite upper limit of integration.  We have 
\begin{align*}
\lim_{b\to \infty} \left. \frac{1}{4}\tan^{-1}(x/4)\right|_4^b 
&= \lim_{b\to\infty} \frac{1}{4}\left(\tan^{-1}(b/4) - \frac{\pi}{4}\right)\\
&=\frac{1}{4}\left(\frac{\pi}{2} - \frac{\pi}{4}\right)\\
&=\frac{\pi}{16}.
\end{align*}
The integral \emph{converges}.
\item $\displaystyle \int_{-\infty}^\infty \frac{1}{x^2 + 16}\ dx$
\item[] {\it Solution:}  This is improper because both limits of integration 
are infinite.  The integral equals
$$\lim_{a \to -\infty} \int_a^0 \frac{1}{x^2 + 16}\ dx + \lim_{b\to\infty} \int_0^b \frac{1}{x^2 + 16}\ dx.$$
Using the formula from the previous problem, this equals 
$$\lim_{a\to -\infty} -\frac{1}{4}\tan^{-1}(a/4) + \lim_{b\to\infty} \frac{1}{4}\tan^{-1}(b/4) = 
\frac{\pi}{8} + \frac{\pi}{8} = \frac{\pi}{4}.$$
The integral \emph{converges}.
\item $\displaystyle \int_1^2 \frac{1}{\sqrt[3]{x - 1}}\ dx$
\item[] {\it Solution:}  Improper because the function is discontinuous at $a = 1$ (an infinite
discontinuity).
Using a $u$-substitution, the integral is 
$$\lim_{a\to 1^+} \left.\frac{3}{2}(x - 1)^{2/3}\right|_a^2 = \lim_{a\to 1^+} \frac{3}{2}(1 - (a - 1)^{2/3}) = \frac{3}{2}.$$
Since the limit is finite, the integral converges.  (Note this would not be true if
the exponent of the $x - 1$ in the original integral was $\ge 1$.)
The original  
\item $\displaystyle \int_0^2 \frac{1}{x^2 - 5x + 6}\ dx$
\item[] {\it Solution:}  Improper because the function is discontinuous at $x = 2$.  
By partial fractions, the integral is $-\ln|x - 2| + \ln|x - 3|$.  Then
$$\lim{b \to 2^-} -\ln|b - 2| + \ln|b-3| + \ln|-2| - \ln|-3|$$
does not exist because $y=\ln|x|$ has a vertical asymptote at $x = 0$.
This integral \emph{diverges}.
\item $\displaystyle \int_1^3 \frac{1}{x^2 - 4x + 4}\ dx$
\item[] {\it Solution:}  This integral is improper because the function 
$$\frac{1}{x^2 - 4x + 4} = \frac{1}{(x-2)^2}$$
is discontinuous at $x = 2$ in the interior of the interval of integration (an infinite
discontinuity).  We need to see whether 
$$\lim_{b\to 2^-} \int_1^b \frac{1}{(x-2)^2}\ dx$$
and
$$\lim_{a\to 2^+} \int_a^3 \frac{1}{(x-2)^2}\ dx$$
exist.  But actually neither of them does because
$$\int \frac{1}{(x-2)^2}\ dx = -\frac{1}{x-2}$$
still has an infinite discontinuity at $2$.  This says that \emph{neither limit above}
exists.  This integral \emph{diverges}.  (Note: The conclusion would be the same
if either one of the limits failed to exist; it is not necessary that both fail to exist
for the integral to diverge).  
\end{enumerate}
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