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\begin{document}

 \centerline{\bf  Improper Integrals \footnote{ This note was written by J. Lewis, and slightly revised by S. Hurder 1/23/2001}
}
\medskip



%%%\prob %%tth:1.
%%\prob
The integrals considered so far $\int_{a}^{b} f(x) \, dx$ assume
implicitly that $a$ and $b$ are finite numbers and that the
function $f(x)$ is nicely behaved on the interval. {\it
Improper\/} integrals arise when
\begin{itemize}
\item The function $f(x)$ blows up (goes to $\pm \infty$) at one of the endpoints, or
\item One of the end points $a$ and/or $b$   is infinite,
\item A combination of both of the above
\end{itemize}

{\bf Examples}.


\begin{itemize}
\item  Find the total area under the curve $y = x e^{-x}$, $0
\leq x < \infty$.
\end{itemize}

The integral to calculate is
$$\int_{0}^{\infty} x
e^{-x} \, dx.$$
 Since $\int x
e^{-x} \, dx = - x e^{-x} - e^{-x}$, the {\it area out to $b$\/}
is $ \int_{0}^{b} t e^{-t} \, dt = \left.(- t e^{-t} -
e^{-t})\right|_{t=0}^{t=b} = - b e^{-b} - e^{-b} + 1$, which tends
to $1$ as $b$ tends to $\infty$. Thus the area is finite and
should be said to be $1$. We say the improper integral
$\int_{0}^{\infty} x e^{-x} \, dx$ {\it converges\/} to the value
$1$.

We are really calculating $\int_{0}^{\infty} x e^{-x} \, dx =
\left.\left( - x e^{-x} - e^{-x}\right)\right|_{0}^{\infty}$ and
interpreting the expression $\left( - x e^{-x} - e^{-x}\right)$ at
$x=\infty$ as $0$ {\it in the sense of limits\/}.


\begin{itemize}
\item  Find the total area under the curve $y =
\dfrac{1}{x\ln(x)}$, $e \leq x < \infty$.
\end{itemize}

 The integral to calculate is
 $$\int_{e}^{\infty} \dfrac{1}{x\ln(x)} \, dx .$$ 
 Since $\int \dfrac{1}{x\ln(x)} \, dx = \ln(\ln(x)) + C$,
 $\int_{e}^{\infty} \dfrac{1}{x\ln(x)} \, dx =
 \left.\ln(\ln(x))\right|_{x=e}^{x=\infty}$, and $\ln(\ln(\infty))$ is to
 be interpreted as $\infty$ {\it in the sense of limits\/}.

 We thus say that the improper integral {\it diverges\/} [to
 $\infty$] and the total area is infinite. Note that
 $$\int_{1}^{\infty} \dfrac{1}{x\ln(x)} \, dx$$
 is improper for an additional reason - at the initial end point
 $x=1$, $\ln(1) = 0$.


\begin{itemize}
\item  The improper integral
$\displaystyle \int_{0}^{1} \dfrac{1}{\sqrt{1- x^2}}\, dx$  
\end{itemize}
converges because
$\left.\arcsin(x)\right|_{x=0}^{x=1}$ can be evaluated because
$\arcsin(1) = \dfrac{\pi}{2}$ {\it in the sense of limits\/}.
The integral is improper since the integrand {\it blows up\/} near
the right hand end point.

\medskip

{\bf The Method}


\begin{itemize}
\item  In each of the examples, we took the $\lim$ as $x \to
{\hbox{\rm singular point}}$ of the anti-derivative function
created {\it after\/} the integration.
\end{itemize}

\eject

{\bf Comparison Tests}

For non-negative functions $f(x)$, the improper integrals
 $\displaystyle \int_{\cdot}^\infty f(x) \, dx$ 
converges if and only if  the approximating integrals
 $\displaystyle \int_{\cdot}^{b} f(t) \, dt$ 
are bounded as $b\to\infty$.
This is because
 $\displaystyle \int_{\cdot}^{b} f(t) \, dt  = F(b) - F(\cdot)$ 
and $F(b)$ is increasing and $F(b)$ has a limit if and only if  $F(b)$ is bounded.

\bigskip

A similar statement can be made for integrals of the form
$\int_{a}^{\cdot}f(x) \, dx$ ($f(x)$ blows up at $a$) or
$\int_{\cdot}^{b} f(x) \, dx$ (f(x) blows up at $b$).



\begin{itemize}
\item 
If $0 \leq f(x) \leq g(x)$, then $\displaystyle 0 \leq \int_{.}^{\cdot} f(x) \, dx
\leq
\int_{.}^{\cdot} g(x)
\, dx$ 
\end{itemize}
\begin{itemize}
\item 
If $0 \leq f(x) \leq g(x)$, and $
\int_{.}^{\cdot} g(x)
\, dx$ converges, then
$\int_{.}^{\cdot} f(x) \, dx$ converges also.
\end{itemize}
\begin{itemize}
\item 
If $0 \leq f(x) \leq g(x)$, and $
\int_{.}^{\cdot} f(x)
\, dx$ diverges, then
$\int_{.}^{\cdot} g(x) \, dx$ diverges also.
\end{itemize}

\bigskip


  {\bf $p$-tests for improper integrals:}

\begin{itemize}
\item $\displaystyle  { \int_{a}^{\infty}  \dfrac{1}{x^p} \, dx  } ~~~ 
\left\{\begin{array} {l}
{\rm converges ~ if \;\;   p >  1,} \\
{\rm diverges ~  if \;\;\;\; p  \leq  1.} 
\end{array} $ 
\end{itemize}

\begin{itemize}
\item $\displaystyle \int_{0}^{b}  \dfrac{1}{x^p} \, dx   ~~~ 
\left\{\begin{array} {l}
{\rm converges ~ if \;\;   p <  1,} \\
{\rm diverges ~  if \;\;\;\; p  \geq  1.} 
\end{array} $ 
\end{itemize}

 
\begin{itemize}
\item $\displaystyle \int_{0}^{\infty}  \dfrac{1}{x^p} \, dx $  ~~~ 
diverges for all p.
\end{itemize}

 
 \bigskip

 
{\bf Examples}

\begin{itemize}
\item 
$\displaystyle \int_{1}^{\infty}\dfrac{\cos^{2}(\phi)}{{\phi}^{2}} \, d\phi$ \;\;\;
 converges by comparison with $\int_{1}^{\infty} \dfrac{1}{\phi^{2}} \,
 d\phi$  or   simply by the  $p$ -- test with $p=2$.\\

\item 
 $\displaystyle \int_{1}^{\infty}\dfrac{\cos^{2}(\phi)}{{\phi}} \, d\phi$ 
 diverges, but cannot be handled directly by the comparison
 with $\int_{1}^{\infty} \dfrac{1}{\phi} \,
 d\phi$. \\

\item 
$\displaystyle \int_{0}^{\infty} e^{-{x^2}\over{2}} \, dx$ 
 The integral converges since $e^{-{x^2}\over{2}} \ll e^{-x}$ for large
 $x$, and $\int_{\cdot}^{\infty} e^{-x} \, dx$ converges.\\

\item 
$\displaystyle \int_{-4}^{3} \dfrac{1}{x^2}\, dx$ DIVERGES.\\

\end{itemize}



\end{document}
