
\magnification 1200
\documentstyle{amsppt}      
\document

\centerline{\bf Improper Integrals}
\bigskip
\noindent
Assume that $f:[a, \infty) \longrightarrow \Bbb R$ is a function such that for any $b
\in (a, \infty), f$ is integrable on $[a, b].$  If
$$\lim_{b\to \infty} \int_a^b f(x) dx.$$
\noindent
exists and is finite, then we say that the \underbar{improper integral} of $f$ on
$(a, \infty)$ converges and we write
$$\int_a^\infty f(x) dx = \lim_{b\to \infty} \int_a^b f(x) dx.$$

\noindent
{\bf \underbar{Example.}}  Let $p > 0.$  If $p \neq 1$ then 

$$I(b) = \int_1^b \frac {1}{x^p} dx = \frac {x^{-p+1}}{-p+1} |_1^b = \frac {1}{1 - p}
[b^{1-p} - 1]$$

\noindent
If $p>1$ then $I(b) \longrightarrow \frac {1}{p-1}$ as $b \longrightarrow \infty$ and
therefore
$$\int_1^\infty \frac{1}{x^p} dx = \frac {1}{p-1}, \;\; p > 1.$$

\noindent
If $0 < p < 1,$ then $I(b) \longrightarrow \infty$ and the improper integral
$\int_1^ \infty \frac {1}{x^p} dx$ diverges.
\newline
\noindent
If $p=1$ then $I(b) = lnx |_1^b = lnb \longrightarrow \infty$ as $b \longrightarrow
\infty.$

\noindent
Therefore we have the 
\bigskip
\noindent
{\bf \underbar{Theorem 1}}
$$\int_1^\infty \frac {1}{x^p} dx = \cases \frac {1}{p-1} &\text{if $p > 1$}\\
\infty &\text{if $p \leq 1$}. \endcases$$
\bigskip
\noindent
{\bf \underbar{Example.}}  Let $a > 0.$  Then

$$I(b) = \int_0^b e^{-ax} dx = -\frac {1}{a} e ^{-ax} |_0^b = \frac {1}{a} (1 -
e^{-ab}) \longrightarrow \frac {1}{a} \;\; {\text{as}} \;  b \longrightarrow \infty.$$

\noindent
Therefore, $\int_0^ \infty e^{-ax} dx$ converges to $\frac 1a$ if $a>0.$  It diverges
if $a < 0.$ \newline

\noindent
Similarly, we define $$\int_{- \infty}^b f(x) dx = \lim_{a \to -\infty} \int_a^b f(x)
dx$$
\noindent
if $f$ is integrable on $[a, b]$ for any $a \leq b$ and the above limits exist and
is finite.
\newline

\noindent
Also, we define
$$\int_{- \infty}^{\infty} f(x) dx = \lim_{b \to \infty} \int_c^b f(x) dx + \lim_{a
\to - \infty} \int_a^c f(x) dx$$

\noindent
if $f$ is integrable on any interval $[a,b]$ and both limits above exist and are
finite.

\noindent
If a function $f$ is unbounded and on (a,b] it is integrable on $[a + \epsilon, b]$ for
any $\epsilon > 0$ and $\lim_{\epsilon \to 0} \int_{a + \epsilon}^b f(x) dx$ exists
and is finite, then we say that the \underbar{improper integral} of $f$ over $(a, b]$
converges and we write $$\int_a^b f(x) dx = \lim_{\epsilon \to 0} \int_{a +
\epsilon}^b f(x) dx.$$

\noindent
{\bf \underbar{Example.}}  Let $p > 0.$  We have $$I(\epsilon) = \int_{\epsilon}^{1}
\frac{1}{x^p} dx = \cases \frac {1}{1-p} [a - \epsilon^{1-p}] & {\text {if}} \; p \neq
1 \\ 
-ln \; \epsilon \;  & {\text {if}} \; p = 1 \endcases$$

\noindent
If $0 < \epsilon < 1$ then $I(\epsilon) \longrightarrow \frac {1}{1-p}$ as $\epsilon
\longrightarrow 0$ and if $1 \leq p$ then $I(\epsilon) \longrightarrow \infty$ as
$\epsilon \longrightarrow 0.$ 
\bigskip
\noindent
Therefore, we have 
\bigskip
\noindent
{\bf {\underbar{Theorem 2.}}}
$$\int_0^1 \frac {1}{x^p} dx = \cases \frac {1}{1-p} &\text{if $0 < p < 1$}\\
\infty &\text {if $p \geq 1.$} \endcases $$

\noindent
If a function $f$ is integrable on $[a, b - \epsilon]$ for any $\epsilon > 0$ and
unbounded on $[a, b),$ then we define $$\int_b^a f(x) dx = \lim_{\epsilon
\longrightarrow 0} \int_a^{b - \epsilon} f(x) dx$$

\noindent
if the above limit exists and is finite.

\bigskip
\noindent
Finally, we define 

$$\int_a^b f(x) dx = \lim_{\epsilon \longrightarrow 0}
\int _{a + \epsilon}^c f(x) dx + \lim_{\epsilon \longrightarrow
0} \int_c^{b - \epsilon} f(x) dx$$
\noindent
if $f$ is integrable on any interval $[a + \epsilon, \; b - \epsilon]$ but unbounded
on $(a, c]$ and $[c, b),$ $c \epsilon (a, b),$ and if both limits above exist and
are finite.
\bigskip
\noindent
{\bf {\underbar{Example.}}}  The integral
$$\int_{-1}^{1} \frac {1}{(1 - x^2)^\frac 14}$$
\noindent
is improper since $$f(x) = \frac {1}{(1 - x^2)^\frac 14}$$ is unbounded on both
intervals $[0,1)$ and $(-1,0].$  We will show later that it converges.

\newpage

{\bf {\centerline{Convergence Tests for Improper Integrals}}}
\bigskip
\noindent
Here we will discuss only infinite integrals of the form $$\int_a^\infty f(x) dx$$
\noindent
where the function $f$ is integrable over any interval $[a, b], b \geq a.$  The tests
for the integrals of the form $\int_{- \infty}^a f(x) dx$ are reduced to the above
case by the substitution $t = -x.$  Also, the substitution $t = \frac {1}{x-a}$ or
$t = \frac {1}{x - b}$ reduces the improper integrals $\int_a^b f(x) dx$ to the
above case.
\bigskip
{\bf {\underbar{Theorem 1}}} (Cauchy criterion).  Suppose that the function $f$ is
integrable over any interval $[0, b), b \leq a.$  Then the improper integral
$\int_a^ \infty f(x) dx$ converges if and only if for every $\epsilon > 0$ there
exists $N = N(\epsilon)$ such that
$$ d \geq c \geq N \Longrightarrow | \int_c^d f(x) dx | <
\epsilon.\tag{$*$}$$  

{\bf \underbar{Remark.}}  In other words, the Cauchy criterion says that
$\int_a^{\infty} f(x) dx$ converges if and only if $$\int_c^d f(x) dx \longrightarrow
0 \; {\text {as}} \; c \longrightarrow \infty.$$
\noindent
{\bf {\underbar{Example.}}}  The infinite integral $\int_1^ {\infty} \frac {dx}{x^2}
dx$ converges since $$| \int_c^d \frac {dx}{x^2} | = \frac {1}{d} - \frac {1}{c} \leq
\frac {2}{c} \longrightarrow 0 \; {\text{as}} \;\;c \longrightarrow \infty.$$ 

\bigskip
\noindent
{\bf {\underbar{Example.}}} The integral $\int_1^{\infty} \frac{dx}{x}$ does not
converge since for $c \geq 1$ we have 
$$| \int_c^{c^2} \frac{dx}{x} | = ln \; c^2 - ln \; c = ln \; c \longrightarrow \infty
\; {\text {as}} \; c \longrightarrow \infty.$$ 

\noindent
{\bf {\underbar{Proof.}}} \;\; Suppose first that the integral converges, i.e.
$$I(b) = \int_a^b f(x) dx \longrightarrow I \;\;{\text {as}} \; b \longrightarrow
\infty.$$

\noindent
Then by the definition of the limit for any $\epsilon > 0$ there exists
$N=N(\epsilon) \geq a$ such that 
$$b \geq N \Rightarrow |I(b) - I | < \frac {\epsilon}{2}$$
\noindent
Therefore, if $d \geq c \geq N$ then
$$| \int_c^d f(x) dx | = |I(d) - I(c)| \leq |I(d) - I| + |I(c) - I| < \frac
{\epsilon}{2} + \frac {\epsilon}{2}= \epsilon $$

\noindent
Conversely, we assume that (*) holds.  The sequence of numbers {$I(k)$} is a \newline 
\underbar{Cauchy sequence}.  Such sequences always converge.  This is a fact
following from the construction of the real numbers.  Therefore, there exists a real
number $I$ such that $I(k) \longrightarrow I$ as $k \rightarrow \infty.$
Thus for any $b \geq a$ we have
$$I(b) = [I(b) - I(k)] + I(k) \longrightarrow I \; {\text {if}} \;\; k \geq b
\longrightarrow \infty$$
\noindent
since $I(k) \longrightarrow I$ and $I(k) - I(b) \longrightarrow 0$ by (*).

\bigskip
\noindent
Next we have 
\bigskip
\noindent
{\bf {\underbar{Theorem 2.}}}  Assume that $f(x) \geq 0$ for $x \geq a$ and that it
is integrable on $[a, b]$ for any $b \geq a.$  Then $\int_a^{\infty} f(x) dx$
converges if and only if there exists M such that 
$$\int_a^b f(x) dx \leq M \; {\text{for any}} \;\; b \geq a.$$

\noindent
{\bf {\underbar{Proof.}}}  Since $f(x) \geq 0$ the function $$I(b) = \int_a^b f(x)
dx, \;\; a \geq b$$

\noindent
is increasing.  Therefore, $I(b)$ converges as $b \longrightarrow \infty$ if and only
if $I(b)$ is bounded. 

\bigskip
\noindent
{\bf \underbar{Example.}}  The integral $\int_1^{\infty} \frac{1}{x^2 + x + 1} dx$
converges since
$$\int_1^b \frac {1}{x^2 + x + 1} dx \leq \int_1^b \frac {1}{x^2} dx = 1 - \frac 1b
\leq 1 \;\; {\text {for any}} \;\; b \geq 1.$$

\bigskip
\noindent
{\bf \underbar{Theorem 3.}}  (Comparison Test).  Suppose that $f(x), g(x) {\text
{and}} \; h(x)$ are integrable on $[a, b]$ for any $b \geq a$ and that $0 \leq g(x)
\leq 1 |f(x)| \leq h(x)$ for any $x \geq A,$ where $A$ is some number with $A \geq
a.$  We have the following:  
\bigskip
\noindent
a)  If $\int_a^{\infty} h(x) dx$ converges, then $\int_a^{\infty} f(x) dx$ converges.
\bigskip
\noindent
b)  If $\int_a^{\infty} g(x) dx = \infty$ and $f(x) \geq 0$ then $\int_a^{\infty}
f(x) dx = \infty.$
\bigskip
\noindent
{\bf \underbar{Example.}}  The integral $\int_1^{\infty} \frac {\sin x}{x} dx$
converges since for $x \geq 1$ we have $| \frac {\sin x}{x^2} | \leq \frac {1}{x^2}$


and $\int_1^{\infty} \frac {1}{x^2} dx < \infty.$
 
\bigskip
\noindent
{\bf \underbar{Example.}}  The integral $\int_1^{\infty} \frac {\sqrt x}{2x + 1} dx$
diverges since for $x \geq 1$
$$\frac{\sqrt x}{2x + 1} \geq \frac {\sqrt x}{2x + x} = \frac 13 \frac {1}{\sqrt x}$$
and $\int_1^{\infty} \frac 13 \frac {1}{\sqrt{x}}$ diverges.

\bigskip
\noindent
{\bf \underbar{Proof.}}  We will use the Cauchy criterion for $g(x).$  Let $\epsilon >
0.$  Since $\int_a^{\infty} g(x) dx$ converge, there exists $N = N(\epsilon) \geq a$
such that $$d \geq c \geq N \Longrightarrow \int_c^d g(x) dx < \epsilon.$$
\noindent
Then for $d \geq c \geq N$ we have
$$| \int_c^d f(x) dx | \leq \int_c^d |f(x) |dx \leq \int_c^d g(x) dx < \epsilon.$$

\noindent
Therefore $\int_a^{\infty} f(x) dx$ converges.  
\bigskip
\noindent
\underbar{Example.}  (Gamma Function)  Let $s>0$ and
$$f(x) = x^{s-1} e^{-x}, \;\;x \in (0, \infty).$$
\noindent
For $s \geq 1$ the function $f(x)$ is bounded on any interval $[0,a], \;\; a >
0.$  Since
$$\frac {x^{s-1}e^{-x}}{e^{-\frac 12}} = x^{s-1}e^{- \frac 12 x}
\longrightarrow 0 \;\; {\text {as}} \;\; x \rightarrow \infty$$ 
we have that there is $A > 0$ such that
$$\frac {x^{s-1}e^{-x}}{e^{-\frac 1{2x}}} \leq 1, \;\; {\text {for}} \;\; x
\geq A.$$

\noindent
Therefore 
$$0 \leq x^{s-1}e^{-x} \leq e^{-\frac {1}{2}x}, \;\; {\text {for}} \;\; x 
\geq A.$$

\noindent
By the comparison test, the integral $\int_0^{\infty} x^{s-1}e^{-x}dx$ converges to a
finite number which we denote by $\Gamma(s),$ i.e. $$\Gamma(s) = \int_0^{\infty} x^{s-1}
e^{-x} dx, s \geq 1.$$



\noindent
If $0 < s < 1$ then the function $f(x) = x^{s-1}e^{-x}$ is unbounded near $0.$ 
Since $$0 \leq x^{s-1}e^{-x} \leq \frac{1}{x^{1-s}}, \; 0 < x \leq 1 $$

\noindent
and $\int_0^1 \frac {dx}{x^{1-s}}$ converges for $0 < s < 1$ we have, by the
comparison test, that $\int_0^1 x^{s-1}e^{-x} dx$ converges.


\bigskip
\noindent
If $0 < s < 1$ and $x \geq 1$ then $0 \leq x^{s-1}e^{-x} \leq e^{-x}$ and
since $\int_1^\infty e^{-x} dx$ converges, by the comparison test we conclude
that $\int_1^{\infty} x^{s-1}e^{-x} dx$ converges.  Therefore, we have that
$\int_0^{\infty} x^{s-1}e^{-x} dx$ converges for any $s>0$ to a finite number
which is denoted by $\Gamma(s)$ and it is called the {\bf {\underbar{Gamma
function}}}, i.e.

$$\Gamma (s) = \int_0^{\infty} x^{s-1}e^{-x} dx, \;\; s > 0.$$

\noindent
If $s > 1,$ then by integrating by parts we obtain

$$
\aligned 
\Gamma (s) &= \int_0^{\infty} x^{s-1}d(-e^{-x}) = -x^{s-1}e^{-x} |_0^{\infty} + (s-1)
\int_0^{\infty} x^{s-2}e^{-x} dx\\  
&= (s-1) \int_0^{\infty} x^{(s-1)-1} e^{-x} dx = (s-1)
\Gamma (s-1). \endaligned$$

\noindent
Thus, if $s > n$ by integrating by parts repeatedly we obtain

$$\Gamma (s) = (s-1)(s-2) \cdots (s-n) \Gamma (s-n)$$
 
\noindent
In particular we obtain $\Gamma (n+1) = n(n-1) \cdot 1 \cdot \Gamma (1) = n!$ since $\Gamma (1) = 1.$

\bigskip
\noindent
{\bf {\underbar{Example.}}}  Since $e^{-x^2} \leq e^{-x}$ for $x > 1$ the integral
$\int_0^{\infty} e^{-x^2}dx$ converges.  It can be shown that $$\int_0^{\infty} e^{-x^2}
dx = \frac 12 \sqrt {\pi}.$$ \noindent
(See p. 371, no. 54.)

\bigskip

{\bf \underbar{Theorem 4.}}  (Limit Comparison Test)  Suppose $f(x) \geq 0$ and $g(x) > 0$ and that
they are integrable on any interval $[a, b], \; b \geq a.$  If

$$\lim_{x \to\infty} \frac {f(x)}{g(x)} = l\neq 0$$

\noindent
then either both integrals
$$\int_a^{\infty} f(x) dx \;\; {\text {and}} \;\; \int_a^{\infty} g(x) dx$$

\noindent
converge or they both diverge.

\bigskip
\noindent
{\bf {\underbar{Example.}}}  The integral $\int_1^{\infty} \frac {2x - 1}{x^5 + x^3 + x}
dx$ converges since 

$$\frac {\frac {2x-1}{x^5 + x^3 + 1}}{\frac {1}{x^4}} = \frac {2x^5 - x^4}{x^5 + x^3 + 1}  \;\;
{@>>{x \to \infty}>} \;\;     2 \neq 0$$

\noindent
and since $\int_1^ {\infty} \frac {1}{x^4} dx < \infty.$

\bigskip
\noindent
{\bf {\underbar{Example.}}}  The integral $$\int_1^{\infty} \frac {1}{\sqrt {x^2 + 5x +
8}} dx$$ diverges since

$$\frac {\frac{1}{\sqrt {x^2 +5x + 8}}}{\frac 1x} = \frac {x}{\sqrt {x^2 + 5x + 8}} = \frac
{1}{\sqrt {1 + \frac 5x + \frac 8{x^2}}} {@>>{x \to \infty}>} 1 \neq 0$$

and $\int_1^{\infty} \frac 1x dx = \infty.$

\bigskip
\noindent
{\bf \underbar{Proof of Theorem 4}.}  Since $\displaystyle {\lim_{x \to \infty}} \frac {f(x)}{g(x)} =
l > 0,$ by the definition of the limit for $\epsilon = \frac l2$ there exists $A = A(\epsilon)$ such
that $$x \geq A \Longrightarrow \; | \frac {f(x)}{g(x)} - l | \; < \frac l2.$$

\noindent
The last relation is written as


$$-\frac l2 < \frac {f(x)}{g(x)} - l < \frac l2, \;\; x \geq A$$

\noindent
or

$$\frac l2 < \frac {f(x)}{g(x)} < \frac {3l}{2}, \;\; x \geq A$$

\noindent
or 

$$\frac l2 g(x) < f(x) < \frac {3l}{2} g(x), \;\; x \geq A $$

\bigskip
\noindent
By Thoeorem 4 the last relation implies that
$$\int_a^{\infty} f(x) dx \and \int_a^{\infty} g(x) dx$$

\noindent
are either both converging or both diverging.

\bigskip
\noindent
{\bf {\underbar{Example.}}}  None of the convergence tests we discussed above can be used
to show that $$\int_1^{\infty} \frac {\sin x}{x} dx$$ converges.  This integral converges
because of two reasons: \bigskip
\noindent
i)  For any $b \geq 1, \;\; | \int_1^b \sin x dx | \;\; \leq 2.$

\bigskip
\noindent
ii)  The function $g(x) = \frac {1}{x}$ is decreasing on any interval $[1, b], \; b \geq 1.$

\bigskip
\noindent
The convergence of the above integral follows from:

\bigskip
\noindent
{{\bf {\underbar{Theorem 5}}} (Dirichlet's Test)  If \newline 

\noindent
i) $f$ is continuous on $[a, \infty)$ and there is $M$ such that
$$| \int_a^b f(x) dx | \leq M, \;\; {\text {for any}} b \geq a$$

\noindent
and  \newline
ii)  $g$ is a decreasing-continuously differentiable function on $[a, \infty)$ with $g(x)
\rightarrow 0 \;\; {\text {as}} \;\; x \rightarrow \infty,$ then

$$ \int_a^{\infty} f(x) g(x) dx$$ converges.

\bigskip
\noindent
{\underbar{Remark.}}  The condition that $g$ is continuously differentiable can be dropped.\newline
To prove Theorem 5, we need the version of the \newline
{\bf \underbar{Second Mean Value Theorem for
Integrals.}}  Let $f(x)$ be continuous on $[a, b]$ and $g(x)$ be decreasing and continuously
differentiable on $[a, b].$  Then there exists $\xi \in [a, b]$ such that



$$\int_a^b f(x) g(x) dx = g(a) \int_a^{\xi} f(x) dx + g(b) \int_{\xi}^b f(x) dx.$$

\noindent
{\bf \underbar{Proof.}} Let $$F(x) = \int_a^x f(t) dt.$$
\noindent
Then by the Fundamental Theorem of Calculus, we have $F'(x) = f(x).$  Integrating by parts, we
obtain 
$$\int_a^b f(x) g(x) dx = \int_a^b g(x) d(F(x)) = g(x) F(x)|_a^b - \int_a^b F(x)g'(x) dx.$$

\noindent
Since $-g'(x) \geq 0$ on $[a, b]$ by the Mean Value Theorem for integrals we obtain that there is
$\xi \in [a, b]$ such that
$$\int_a^b F(x)(-g'(x)] dx = F(\xi) \int_a^b (-g'(x)) dx.$$

\noindent
Therefore
$$\aligned
\int_a^b f(x) g(x) dx &= g(b)F(b) - g(a)F(a) - F(\xi) \int_a^b g'(x) dx\\
&=g(b)F(b) - F(\xi) [g(b) - g(a)]\\
&=g(a)F(\xi) + g(b) [F(b) - F(\xi)]\\
&=g(a) \int_a^{\xi} f(x) dx + g(b) \int_{\xi}^b f(x) dx \endaligned $$

\noindent
{\bf \underbar{Proof of Theorem 5.}}  Let $d \geq c \geq a.$  By the last result there exists $\xi
\in [c, d]$ such that 

$$\int_c^d f(x) g(x) dx = g(c) \int_c^{\xi} f(x) dx + g(d) \int_{\xi}^d f(x) dx$$

\noindent
since

$$| \int_c^{\xi} f(x) dx | = | \int_a^{\xi} f(x) dx - \int_a^c f(x) dx | \leq | \int_a^{\xi} f(x)
dx | + | \int_a^{\xi} f(x) | dx \leq 2M.$$

\noindent
and similarly

$$| \int_{\xi}^d f(x) dx | \leq 2M $$

\noindent
we have that

$$| \int_c^d f(x) g(x) dx | \leq g(c) 2M + g(d) 2M = 4M g(c) {@>>{c \to \infty}>} \;0$$

\noindent
Therefore, by the Cauchy criterion $\int_a^{\infty} f(x) g(x) dx$ converges. 

\bigskip  
\noindent
{\bf \underbar{Example.}}  If $p > 0$ then the integral
$$\int_1^{\infty} \frac {\sin x}{x^p} dx$$

\noindent
converges since, if we let $f(x) = \sin x$ and $g(x) = \frac {1}{x^p}$, then $f$ and $g$ satisfy the
assumptions of Theorem 5.  In fact, for $p > 1$ it follows by the Comparison Test since 

$$| \frac {\sin x}{x^p} | \leq \frac 1{x^p} \;\; {\text {and}} \;\; \int_1^{\infty} \frac 1{x^p}
dx < \infty.$$

\noindent
Also, the integral $\int_0^{\infty} \frac {\sin x}{x} dx$ converges since for $x \geq 1$ we can use
the Dirichlet's test and since $\frac {\sin x}{x}$ is continuous on $[0, 1].$  It can be shown (see
p. 372, no. 55) that

$$\int_0^{\infty} \frac {\sin x}{x} dx = \frac {\pi}{2}.$$

\noindent
{\bf {\underbar{Remark}}}  We remark here that $$\int_0^{\infty} \frac {|\sin x|}{x} dx =
\infty.$$

\noindent
In fact, for any $k = 2, 3, 4, \cdots $

$$ \aligned
\int_0^{\infty} \frac {| \sin x |}{x} dx & \geq \int_{\pi}{2 \pi} \frac {| \sin x|}{x} dx + \int_{2
\pi}^{3 \pi} \frac {| \sin x|}{x} dx + \cdots + \int_{(k-1) \pi}^{k \pi} \frac {|sin x|}{x} dx\\
& \geq \frac {1}{2 \pi} \int_{\pi}^{2 \pi} |\sin x| dx + \frac {1}{3 \pi} \int_{2 \pi}^{3 \pi} |\sin
x| dx + \cdots + \frac {1}{k \pi} \int_{(k - 1) \pi}^{k \pi} |\sin x| dx\\
& = \frac {1}{\pi} \int_0^{\pi} \sin x dx [\frac 12 + \frac 13 + \cdots + \frac 1k]\\
&=\frac 2{\pi} [ \frac 12 + \frac 13 + \cdots + \frac 1k] \endaligned$$

\noindent
Since $$\frac {1}{j} > \int_j^{j + 1} \frac 1x dx$$

\bigskip
\noindent
We have $$\int_0^{\infty} \frac {|\sin x|}{x} \geq \frac 2{\pi} [ \int_2^3 \frac 1x dx + \int_3^4
\frac 1x dx + \cdots + \int_k^{k + 1} \frac 1x dx] \geq \frac 2{\pi} \int_2^{k + 1} \frac 1x dx.$$

\noindent
Thus  $$\int_0^{\infty} \frac {| \sin x|}{x} \geq \frac 2{\pi} \int_2^{\infty} \frac 1x dx =
\infty.$$

\noindent
By the comparison test we have seen that if $\int_a^\infty |f(x) | dx$ converges, then
so does the integral $\int f(x) dx.$ 
\bigskip
\noindent
The last remark shows that the converse is not true.  If $\int_a^{\infty} | f(x)| dx$ converges,
then we say that the improper integral of $f$ is {\bf {\underbar{absolutely
convergent.}}}

\bigskip
\noindent
{\bf {\underbar{Exercises.}}}

\bigskip
\noindent
1.  Show that $\int_0^{\infty} \sin (x^2) dx$ is convergent.  (Hint:  Let the substitution $t =
x^2.)$
\bigskip
\noindent
2.  Show that the integral $\int_{-1}^{\infty} (1 - x^2) ^{-\frac 14} dx$ is convergent.
\bigskip
\noindent
3.  Compute the integral $\int_1^{\infty} \frac {x^3 + x^2 + 1}{x^6 + x^3} dx.$



 \bye



 
                                                                                                                                                                                

                                                                                                                                                                                                                                                                            
 

