Answer
(a) The function $f\left( x \right) = \frac{{{{\rm{e}}^{ - x}} - {{\rm{e}}^{ - ax}}}}{x}$ is not defined at $x=0$.
Here, we show that it can be made continuous by assigning the value $f\left( 0 \right) = a - 1$. That is,
$f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}
{\frac{{{{\rm{e}}^{ - x}} - {{\rm{e}}^{ - ax}}}}{x}}&{{\rm{for{\ } }}x \ne 0}\\
{a - 1}&{{\rm{for{\ } }}x = 0}
\end{array}} \right.$
(b) We prove that $\left| {f\left( x \right)} \right| \le {{\rm{e}}^{ - x}} + {{\rm{e}}^{ - ax}}$ for $x>1$.
Applying the Comparison Theorem, we show that $I\left( a \right) = \mathop \smallint \limits_0^\infty \frac{{{{\rm{e}}^{ - x}} - {{\rm{e}}^{ - ax}}}}{x}{\rm{d}}x$ converges.
(c) We show that $I\left( a \right) = \mathop \smallint \limits_0^\infty \mathop \smallint \limits_1^a {{\rm{e}}^{ - xy}}{\rm{d}}y{\rm{d}}x$
(d) Using Fubini's Theorem, we prove that
$I\left( a \right) = \ln a - \mathop {\lim }\limits_{T \to \infty } \mathop \smallint \limits_1^a \frac{{{{\rm{e}}^{ - Ty}}}}{y}{\rm{d}}y$
(e) Using the Comparison Theorem, we show that
$\mathop {\lim }\limits_{T \to \infty } \mathop \smallint \limits_1^a \frac{{{{\rm{e}}^{ - Ty}}}}{y}{\rm{d}}y = 0$
Therefore, $I\left( a \right) = \ln a$
Work Step by Step
(a) Write $f\left( x \right) = \frac{{{{\rm{e}}^{ - x}} - {{\rm{e}}^{ - ax}}}}{x}$.
We notice that $f\left( x \right)$ is not defined at $x=0$, hence it is not continuous at $x=0$.
Now, we evaluate the limit
$\mathop {\lim }\limits_{x \to 0} f\left( x \right) = \mathop {\lim }\limits_{x \to 0} \frac{{{{\rm{e}}^{ - x}} - {{\rm{e}}^{ - ax}}}}{x}$
Using L'Hôpital's Rule (Theorem 1 of Section 7.7), we obtain
$\mathop {\lim }\limits_{x \to 0} f\left( x \right) = \mathop {\lim }\limits_{x \to 0} \frac{{{{\rm{e}}^{ - x}} - {{\rm{e}}^{ - ax}}}}{x} = \mathop {\lim }\limits_{x \to 0} \frac{{ - {{\rm{e}}^{ - x}} + a{{\rm{e}}^{ - ax}}}}{1} = - 1 + a$
If we define the function $f\left( x \right)$ such that
$f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}
{\frac{{{{\rm{e}}^{ - x}} - {{\rm{e}}^{ - ax}}}}{x}}&{{\rm{for{\ } }}x \ne 0}\\
{a - 1}&{{\rm{for{\ } }}x = 0}
\end{array}} \right.$
then $f\left( x \right)$ is continuous by definition of continuity (Section 2.4).
Hence, $f\left( x \right) = \frac{{{{\rm{e}}^{ - x}} - {{\rm{e}}^{ - ax}}}}{x}$, though not defined at $x=0$, can be made continuous by assigning the value $f\left( 0 \right) = a - 1$.
(b) From the result in part (a), we have the function:
$f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}
{\frac{{{{\rm{e}}^{ - x}} - {{\rm{e}}^{ - ax}}}}{x}}&{{\rm{for{\ } }}x \ne 0}\\
{a - 1}&{{\rm{for{\ } }}x = 0}
\end{array}} \right.$
such that $f\left( x \right)$ is continuous.
For $x>1$, we can simply write $f\left( x \right) = \frac{{{{\rm{e}}^{ - x}} - {{\rm{e}}^{ - ax}}}}{x}$.
$f\left( x \right) = \frac{{{{\rm{e}}^{ - x}}}}{x} + \left( { - \frac{{{{\rm{e}}^{ - ax}}}}{x}} \right)$
Using the triangle inequality (Eq. (1) of Section 1.1), we obtain
$\left| {f\left( x \right)} \right| = \left| {\frac{{{{\rm{e}}^{ - x}}}}{x} + \left( { - \frac{{{{\rm{e}}^{ - ax}}}}{x}} \right)} \right| \le \left| {\frac{{{{\rm{e}}^{ - x}}}}{x}} \right| + \left| { - \frac{{{{\rm{e}}^{ - ax}}}}{x}} \right|$
$\left| {f\left( x \right)} \right| \le \frac{1}{{\left| x \right|}}\left( {\left| {{{\rm{e}}^{ - x}}} \right| + \left| {{{\rm{e}}^{ - ax}}} \right|} \right)$
Since ${{\rm{e}}^{ - x}} > 0$ and ${{\rm{e}}^{ - ax}} > 0$; for $x>1$ we can remove the absolute signs. So,
$\left| {f\left( x \right)} \right| \le \frac{1}{x}\left( {{{\rm{e}}^{ - x}} + {{\rm{e}}^{ - ax}}} \right)$
Since $\frac{1}{x}\left( {{{\rm{e}}^{ - x}} + {{\rm{e}}^{ - ax}}} \right) \le {{\rm{e}}^{ - x}} + {{\rm{e}}^{ - ax}}$, so
(1) ${\ \ \ \ }$ $\left| {f\left( x \right)} \right| \le {{\rm{e}}^{ - x}} + {{\rm{e}}^{ - ax}}$
Recall the Comparison Theorem (Theorem 5 of Section 5.2) which states that if $f$ and $g$ are integrable and $f\left( x \right) \le g\left( x \right)$, then
$\mathop \smallint \limits_a^b f\left( x \right){\rm{d}}x \le \mathop \smallint \limits_a^b g\left( x \right){\rm{d}}x$
Applying the Comparison Theorem to equation (1) we obtain
$\mathop \smallint \limits_0^\infty \left| {f\left( x \right)} \right|{\rm{d}}x \le \mathop \smallint \limits_0^\infty \left( {{{\rm{e}}^{ - x}} + {{\rm{e}}^{ - ax}}} \right){\rm{d}}x$
Since $f\left( x \right) = \frac{{{{\rm{e}}^{ - x}} - {{\rm{e}}^{ - ax}}}}{x}$, so
$\mathop \smallint \limits_0^\infty \left| {\frac{{{{\rm{e}}^{ - x}} - {{\rm{e}}^{ - ax}}}}{x}} \right|{\rm{d}}x \le \mathop \smallint \limits_0^\infty \left( {{{\rm{e}}^{ - x}} + {{\rm{e}}^{ - ax}}} \right){\rm{d}}x$
For $a>0$ and $x>1$, we have ${{\rm{e}}^{ - x}} \ge {{\rm{e}}^{ - ax}}$. So, we can remove the absolute sign and write
$\mathop \smallint \limits_0^\infty \frac{{{{\rm{e}}^{ - x}} - {{\rm{e}}^{ - ax}}}}{x}{\rm{d}}x \le \mathop \smallint \limits_0^\infty \left( {{{\rm{e}}^{ - x}} + {{\rm{e}}^{ - ax}}} \right){\rm{d}}x$
We know that the right-hand side yields
$\mathop \smallint \limits_0^\infty \left( {{{\rm{e}}^{ - x}} + {{\rm{e}}^{ - ax}}} \right){\rm{d}}x = \mathop \smallint \limits_0^\infty {{\rm{e}}^{ - x}}{\rm{d}}x + \mathop \smallint \limits_0^\infty {{\rm{e}}^{ - ax}}{\rm{d}}x$
$ = \mathop {\lim }\limits_{P \to \infty } \mathop \smallint \limits_0^P {{\rm{e}}^{ - x}}{\rm{d}}x + \mathop {\lim }\limits_{P \to \infty } \mathop \smallint \limits_0^P {{\rm{e}}^{ - ax}}{\rm{d}}x$
$ = - \mathop {\lim }\limits_{P \to \infty } \left( {{{\rm{e}}^{ - x}}|_0^P} \right) - \frac{1}{a}\mathop {\lim }\limits_{P \to \infty } \left( {{{\rm{e}}^{ - ax}}|_0^P} \right)$
$ = - \mathop {\lim }\limits_{P \to \infty } \left( {{{\rm{e}}^{ - P}} - 1} \right) - \frac{1}{a}\mathop {\lim }\limits_{P \to \infty } \left( {{{\rm{e}}^{ - aP}} - 1} \right)$
$ = 1 + \frac{1}{a}$
Therefore, $\mathop \smallint \limits_0^\infty \frac{{{{\rm{e}}^{ - x}} - {{\rm{e}}^{ - ax}}}}{x}{\rm{d}}x \le 1 + \frac{1}{a}$.
This implies that $I\left( a \right) = \mathop \smallint \limits_0^\infty \frac{{{{\rm{e}}^{ - x}} - {{\rm{e}}^{ - ax}}}}{x}{\rm{d}}x$ converges.
(c) Evaluate $\mathop \smallint \limits_1^a {{\rm{e}}^{ - xy}}{\rm{d}}y$.
$\mathop \smallint \limits_1^a {{\rm{e}}^{ - xy}}{\rm{d}}y = - \frac{1}{x}{{\rm{e}}^{ - xy}}|_1^a = - \frac{1}{x}{{\rm{e}}^{ - ax}} + \frac{1}{x}{{\rm{e}}^{ - x}} = \frac{{{{\rm{e}}^{ - x}} - {{\rm{e}}^{ - ax}}}}{x}$
So, we write the double integral:
$\mathop \smallint \limits_0^\infty \mathop \smallint \limits_1^a {{\rm{e}}^{ - xy}}{\rm{d}}y{\rm{d}}x = \mathop \smallint \limits_0^\infty \frac{{{{\rm{e}}^{ - x}} - {{\rm{e}}^{ - ax}}}}{x}{\rm{d}}x$
Since $I\left( a \right) = \mathop \smallint \limits_0^\infty \frac{{{{\rm{e}}^{ - x}} - {{\rm{e}}^{ - ax}}}}{x}{\rm{d}}x$, we conclude that
$I\left( a \right) = \mathop \smallint \limits_0^\infty \mathop \smallint \limits_1^a {{\rm{e}}^{ - xy}}{\rm{d}}y{\rm{d}}x$
(d) From part (c) we obtain
$I\left( a \right) = \mathop \smallint \limits_0^\infty \mathop \smallint \limits_1^a {{\rm{e}}^{ - xy}}{\rm{d}}y{\rm{d}}x$
Using Fubini's Theorem to change the order of integration and re-write the integral:
$I\left( a \right) = \mathop \smallint \limits_0^\infty \mathop \smallint \limits_1^a {{\rm{e}}^{ - xy}}{\rm{d}}y{\rm{d}}x = \mathop \smallint \limits_{y = 1}^a \left( {\mathop \smallint \limits_{x = 0}^\infty {{\rm{e}}^{ - xy}}{\rm{d}}x} \right){\rm{d}}y$
$ = \mathop \smallint \limits_{y = 1}^a \left( {\mathop {\lim }\limits_{T \to \infty } \mathop \smallint \limits_{x = 0}^T {{\rm{e}}^{ - xy}}{\rm{d}}x} \right){\rm{d}}y$
$ = \mathop \smallint \limits_{y = 1}^a \left( {\mathop { - \lim }\limits_{T \to \infty } \frac{1}{y}\left( {{{\rm{e}}^{ - Ty}} - 1} \right)} \right){\rm{d}}y$
$ = \mathop \smallint \limits_{y = 1}^a \left( {\mathop { - \lim }\limits_{T \to \infty } \frac{1}{y}{{\rm{e}}^{ - Ty}} + \frac{1}{y}} \right){\rm{d}}y$
$ = - \mathop \smallint \limits_{y = 1}^a \left( {\mathop {\lim }\limits_{T \to \infty } \frac{1}{y}{{\rm{e}}^{ - Ty}}} \right){\rm{d}}y + \mathop \smallint \limits_{y = 1}^a \frac{1}{y}{\rm{d}}y$
$ = - \mathop {\lim }\limits_{T \to \infty } \mathop \smallint \limits_1^a \frac{{{{\rm{e}}^{ - Ty}}}}{y}{\rm{d}}y + \ln y|_1^a$
$ = - \mathop {\lim }\limits_{T \to \infty } \mathop \smallint \limits_1^a \frac{{{{\rm{e}}^{ - Ty}}}}{y}{\rm{d}}y + \ln a$
Hence, $I\left( a \right) = \ln a - \mathop {\lim }\limits_{T \to \infty } \mathop \smallint \limits_1^a \frac{{{{\rm{e}}^{ - Ty}}}}{y}{\rm{d}}y$.
(e) From the result in part (d), we obtain $I\left( a \right) = \ln a - \mathop {\lim }\limits_{T \to \infty } \mathop \smallint \limits_1^a \frac{{{{\rm{e}}^{ - Ty}}}}{y}{\rm{d}}y$.
Here we show that: $\mathop {\lim }\limits_{T \to \infty } \mathop \smallint \limits_1^a \frac{{{{\rm{e}}^{ - Ty}}}}{y}{\rm{d}}y$ is equal to zero.
Case 1. $a \ge 1$
Since ${{\rm{e}}^{ - Ty}} \ge 0$, for $y \ge 1$ we have $0 \le \frac{{{{\rm{e}}^{ - Ty}}}}{y} \le {{\rm{e}}^{ - T}}$.
Applying the Comparison Theorem as in part (b) gives
$0 \le \mathop \smallint \limits_1^a \frac{{{{\rm{e}}^{ - Ty}}}}{y}{\rm{d}}y \le \mathop \smallint \limits_1^a {{\rm{e}}^{ - T}}{\rm{d}}y$
Since $\mathop \smallint \limits_1^a {{\rm{e}}^{ - T}}{\rm{d}}y = {{\rm{e}}^{ - T}}\left( {a - 1} \right)$, so
$0 \le \mathop \smallint \limits_1^a \frac{{{{\rm{e}}^{ - Ty}}}}{y}{\rm{d}}y \le {{\rm{e}}^{ - T}}\left( {a - 1} \right)$
Applying the limit gives
$0 \le \mathop {\lim }\limits_{T \to \infty } \mathop \smallint \limits_1^a \frac{{{{\rm{e}}^{ - Ty}}}}{y}{\rm{d}}y \le \mathop {\lim }\limits_{T \to \infty } {{\rm{e}}^{ - T}}\left( {a - 1} \right)$
But $\mathop {\lim }\limits_{T \to \infty } {{\rm{e}}^{ - T}}\left( {a - 1} \right) = 0$. Thus, by Squeeze Theorem (Theorem 1 of Section 2.6):
$\mathop {\lim }\limits_{T \to \infty } \mathop \smallint \limits_1^a \frac{{{{\rm{e}}^{ - Ty}}}}{y}{\rm{d}}y = 0$
Case 2. $0 < a < 1$
Since ${{\rm{e}}^{ - Ty}} \ge 0$, for $a \le y \le 1$ we have $0 \le \frac{{{{\rm{e}}^{ - Ty}}}}{y} \le \frac{{{{\rm{e}}^{ - aT}}}}{a}$.
Applying the Comparison Theorem as in part (b) gives
$0 \le \mathop \smallint \limits_1^a \frac{{{{\rm{e}}^{ - Ty}}}}{y}{\rm{d}}y \le \mathop \smallint \limits_1^a \frac{{{{\rm{e}}^{ - aT}}}}{a}{\rm{d}}y$
Since $\mathop \smallint \limits_1^a \frac{{{{\rm{e}}^{ - aT}}}}{a}{\rm{d}}y = \frac{{{{\rm{e}}^{ - aT}}}}{a}\left( {a - 1} \right)$, so
$0 \le \mathop \smallint \limits_1^a \frac{{{{\rm{e}}^{ - Ty}}}}{y}{\rm{d}}y \le \frac{{{{\rm{e}}^{ - aT}}}}{a}\left( {a - 1} \right)$
Applying the limit gives
$0 \le \mathop {\lim }\limits_{T \to \infty } \mathop \smallint \limits_1^a \frac{{{{\rm{e}}^{ - Ty}}}}{y}{\rm{d}}y \le \mathop {\lim }\limits_{T \to \infty } \frac{{{{\rm{e}}^{ - aT}}}}{a}\left( {a - 1} \right)$
But $\mathop {\lim }\limits_{T \to \infty } \frac{{{{\rm{e}}^{ - aT}}}}{a}\left( {a - 1} \right) = 0$. Thus, by Squeeze Theorem:
$\mathop {\lim }\limits_{T \to \infty } \mathop \smallint \limits_1^a \frac{{{{\rm{e}}^{ - Ty}}}}{y}{\rm{d}}y = 0$
Hence, from both cases we conclude that
$\mathop {\lim }\limits_{T \to \infty } \mathop \smallint \limits_1^a \frac{{{{\rm{e}}^{ - Ty}}}}{y}{\rm{d}}y = 0$
Therefore, $I\left( a \right) = \ln a - \mathop {\lim }\limits_{T \to \infty } \mathop \smallint \limits_1^a \frac{{{{\rm{e}}^{ - Ty}}}}{y}{\rm{d}}y = \ln a$.
