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\noindent {\bf Question}

\noindent Determine whether the following improper integrals
converge or diverge, and evaluate those which converge.
\begin{enumerate}
\item $\int_0^4 {\rm d}x/x^{3/2}$;
\item $\int_1^\infty {\rm d}x/(x+1)$;
\item $\int_5^\infty {\rm d}x/(x-1)^{3/2}$;
\item $\int_0^9 {\rm d}x/(9-x)^{3/2}$;
\item $\int_{-\infty}^{-2} {\rm d}x/(x+1)^3$;
\item $\int_{-1}^8 {\rm d}x/x^{1/3}$;
\item $\int_2^\infty {\rm d}x/(x-1)^{1/3}$;
\item $\int_{-\infty}^\infty x {\rm d}x/(x^2+4)$;
\item $\int_0^1 e^{\sqrt{x}}{\rm d}x/\sqrt{x}$;
\item $\int_1^\infty {\rm d}x/x\ln(x)$;
\end{enumerate}

\medskip

\noindent {\bf Answer}

\noindent \begin{enumerate}
\item this is an improper integral because $1/x^{3/2}$ is continuous
on $(0,4]$ and $\lim_{x\rightarrow 0+} 1/x^{3/2} =\infty$.  So, we
evaluate:
\begin{eqnarray*}
\int_0^4 \frac{1}{x^{3/2}} \: {\rm d}x & = & \lim_{c\rightarrow
0+} \int_c^4 \frac{1}{x^{3/2}} \: {\rm d}x \\
 & = & \lim_{c\rightarrow 0+} \int_c^4 x^{-3/2} \: {\rm d}x \\
 & = & \lim_{c\rightarrow 0+} \left( -\frac{2}{\sqrt{4}} +
\frac{2}{\sqrt{c}}\right) \\
 & = & -1 + 2\lim_{c\rightarrow 0+} \frac{1}{\sqrt{c}} =\infty,
\end{eqnarray*}
and so this improper integral {\bf diverges}.
\item this is an improper integral because the interval of integration
is $[1,\infty)$, which is not a closed interval.  So, we evaluate:
\begin{eqnarray*}
\int_1^\infty \frac{1}{x+1} \: {\rm d}x & = &
\lim_{M\rightarrow\infty} \int_1^M \frac{1}{x+1}\: {\rm d}x \\
 & = & \lim_{M\rightarrow\infty} \left[ \ln(M+1) -\ln\left(
\frac{1}{2} \right) \right] =\infty,
\end{eqnarray*}
and so this improper integral {\bf diverges}.
\item this is an improper integral, as the interval of integration is
$[5,\infty)$, which is not a closed interval.  So, we evaluate:
\begin{eqnarray*}
\int_5^\infty \frac{1}{(x-1)^{3/2}}\: {\rm d}x & = &
\lim_{M\rightarrow\infty} \int_5^M \frac{1}{(x-1)^{3/2}}\: {\rm
d}x \\
 & = & \lim_{M\rightarrow\infty} \int_5^M (x-1)^{-3/2}\: {\rm d}x \\
 & = & \lim_{M\rightarrow\infty} \left[ -\frac{2}{\sqrt{M-1}} +
1\right] = 1,
\end{eqnarray*}
and so this improper integral {\bf converges to $1$}.
\item this is an improper integral because $1/(9-x)^{3/2}$ is
continuous on $[0,9)$ and $\lim_{x\rightarrow 9-} 1/(9-x)^{3/2}
=\infty$.  So, we evaluate:
\begin{eqnarray*}
\int_0^9 \frac{1}{(9-x)^{3/2}}\: {\rm d}x & = & \lim_{c\rightarrow
9-} \int_0^c \frac{1}{(9-x)^{3/2}}\: {\rm d}x \\
 & = & \lim_{c\rightarrow 9-} \int_0^c (9-x)^{-3/2}\: {\rm d}x \\
 & = & \lim_{c\rightarrow 9-} \left[ -\frac{2}{3}
+\frac{2}{\sqrt{9-c}} \right] =\infty,
\end{eqnarray*}
and so this improper integral {\bf diverges}.
\item this is an improper integral, since the interval of integration
is $(-\infty, -2]$ and so is not a closed interval.  So, we
evaluate:
\begin{eqnarray*}
\int_{-\infty}^{-2} \frac{1}{(x+1)^3}\: {\rm d}x & = &
\lim_{M\rightarrow -\infty} \int_{M}^{-2} \frac{1}{(x+1)^3}\: {\rm
d}x
\\
 & = & \lim_{M\rightarrow -\infty} \left[ -\frac{1}{2} \frac{1}{(-2
+1)^2} +\frac{1}{2} \frac{1}{(M+1)^2}\right] = -\frac{1}{2},
\end{eqnarray*}
and so this improper integral {\bf converges to $-\frac{1}{2}$}.
\item this is an improper integral, since the integrand is not
continuous on $[-1,8]$ as it has a discontinuity at $0$.  Hence,
we can break it up as the sum of two improper integrals:
\[ \int_{-1}^8 {\rm d}x/x^{1/3} = \int_{-1}^0 {\rm d}x/x^{1/3} +
\int_0^8 {\rm d}x/x^{1/3}, \] and we have that $\int_{-1}^8 {\rm
d}x/x^{1/3}$ converges if both $\int_{-1}^0 {\rm d}x/x^{1/3}$ and
$\int_0^8 {\rm d}x/x^{1/3}$ converge.  So, we evaluate:
\begin{eqnarray*}
\int_{-1}^0 \frac{1}{x^{1/3}} {\rm d}x & = & \lim_{c\rightarrow
0-} \int_{-1}^c \frac{1}{x^{1/3}} {\rm d}x \\
 & = & \lim_{c\rightarrow 0-} \int_{-1}^c x^{-1/3} {\rm d}x \\
 & = & \lim_{c\rightarrow 0-} \left[ \frac{3}{2} c^{2/3} - \frac{3}{2}
\right] = -\frac{3}{2},
\end{eqnarray*}
and
\begin{eqnarray*}
\int_0^8 \frac{1}{x^{1/3}} {\rm d}x & = & \lim_{c\rightarrow 0+}
\int_c^8 \frac{1}{x^{1/3}} {\rm d}x \\
 & = & \lim_{c\rightarrow 0+} \int_c^8 x^{-1/3} {\rm d}x \\
 & = & \lim_{c\rightarrow 0+} \left[ \frac{3}{2} 8^{2/3} - \frac{3}{2}
c^{2/3} \right] = 6.
\end{eqnarray*}
Since both these improper integrals converge, we see that the
original improper integral $\int_{-1}^8 {\rm d}x/x^{1/3}$ {\bf
converges to $\frac{9}{2}$}.
\item this is an improper integral, since the interval of integration
is $[2,\infty)$ and hence is not a closed interval.  So, we
evaluate:
\begin{eqnarray*}
\int_2^\infty \frac{1}{(x-1)^{1/3}}\: {\rm d}x & = &
\lim_{M\rightarrow\infty} \int_2^M \frac{1}{(x-1)^{1/3}}\: {\rm
d}x \\
 & = & \lim_{M\rightarrow\infty} \int_2^M (x-1)^{-1/3}\: {\rm d}x \\
 & = & \lim_{M\rightarrow\infty} \left[ \frac{3}{2} (M-1)^{2/3} -
\frac{3}{2} \right] =\infty,
\end{eqnarray*}
and so this improper integral {\bf diverges}.
\item this is an improper integral since the interval of integration
is $(-\infty, \infty)$ and hence is not a closed interval.  We
evaluate this improper integral by breaking it up as the sum of
two improper integrals $\int_{-\infty}^\infty x {\rm d}x/(x^2+4) =
\int_{-\infty}^0 x {\rm d}x/(x^2+4) + \int_0^\infty x {\rm
d}x/(x^2+4)$, and evaluating the two resulting improper integrals
separately.  So,
\begin{eqnarray*}
\int_{-\infty}^0 \frac{x}{x^2+4}\: {\rm d}x & =&
\lim_{M\rightarrow -\infty} \int_M^0 \frac{x}{x^2+4}\: {\rm d}x \\
 & = & \lim_{M\rightarrow -\infty} \left[ \frac{1}{2}\ln(M^2 +4) -
\frac{1}{2}\ln(4)\right] =\infty.
\end{eqnarray*}
Since one of these two improper integrals diverges, we don't need
to evaluate the other one, as the original improper integral
$\int_{-\infty}^0 x {\rm d}x/(x^2 +4)$ necessarily {\bf diverges}.
\item this is an improper integral, as the integrand is continuous on
$(0,1]$ and $\lim_{x\rightarrow 0+} e^{\sqrt{x}}/\sqrt{x}=
\infty$. So, we evaluate:
\begin{eqnarray*}
\int_0^1 \frac{e^{\sqrt{x}}}{\sqrt{x}}\: {\rm d}x & = &
\lim_{c\rightarrow 0+} \int_c^1 \frac{e^{\sqrt{x}}}{\sqrt{x}}\:
{\rm d}x \\
 & = & \lim_{c\rightarrow 0+} (2 -2\sqrt{c}) = 2,
\end{eqnarray*}
and so this improper integral {\bf converges to $2$}.
\item this is an improper integral, as the interval of integration is
$[1,\infty)$ and so is not a closed interval.  Moreover, the
integrand is not continuous at $0$ but $\lim_{x\rightarrow 1+}
1/x\ln(x) =\infty$, and so we need to break this improper integral
into the sum of two improper integrals $\int_1^\infty {\rm
d}x/x\ln(x) = \int_1^2 {\rm d}x/x\ln(x) + \int_2^\infty {\rm
d}x/x\ln(x)$, and evaluate the two resulting improper integrals
separately.  So,
\begin{eqnarray*}
\int_1^2 \frac{1}{x\ln(x)} \: {\rm d}x & = & \lim_{c\rightarrow
1+} \int_c^2 \frac{1}{x\ln(x)} \: {\rm d}x \\
 & = & \lim_{c\rightarrow 1+} (\ln(\ln(2)) -\ln(\ln(c))) = \infty,
\end{eqnarray*}
and so this improper integral diverges, and so the original
improper integral $\int_1^\infty {\rm d}x/x\ln(x)$ necessarily
{\bf diverges}.
\end{enumerate}


\end{document}