\input homework_macros.tex

%\distiller
\printer
\def\dd #1#2{\frac{d#1}{d#2}}

\toks0={
7.6  p. 603:  1,  5,  7, 13, 17, 25, 37, 39, 47, 53, 55.
}

\SN{\jobname}

\np$\int^\infty_0\Frac{dx}{x^2+1}$.
This is improper with the only singularity at $\infty$.
Hence
$\int^\infty_0\Frac{dx}{x^2+1}=\lim_{t\to\infty}
\int^t_0\Frac{dx}{x^2+1}$.
Since $\int^t_0\Frac{dx}{x^2+1}=\arctan t-\arctan 0=\arctan t$
and since $\displaystyle\lim_{t\to\infty}\arctan t=\frac\pi2$,
$\int^\infty_0\Frac{dx}{x^2+1}$ converges and its value is $\frac\pi2$.

\np$\int^1_{-1}\Frac{dx}{x^{2/3}}$.
This is improper because the function we are integrating is undefined
at $0$, a point in the interval of integration.
Hence
$\int^1_{-1}\Frac{dx}{x^{2/3}}=
\int^0_{-1}\Frac{dx}{x^{2/3}}+\int^1_{0}\Frac{dx}{x^{2/3}}=
\lim_{t\to 0^-}\int^t_{-1}\Frac{dx}{x^{2/3}}+
\lim_{t\to0^+}\int^1_{t}\Frac{dx}{x^{2/3}}$.
Next do the indefinite integral:
$\int\Frac{dx}{x^{2/3}}=\int x^{-2/3}dx=\Frac{x^{-2/3+1}}{-2/3+1}+C=
3x^{1/3}+C$.
Since $\displaystyle\lim_{t\to 0} 3x^{1/3}=0$,
$\displaystyle\lim_{t\to 0^-}\int^t_{-1}\Frac{dx}{x^{2/3}}=
\lim_{t\to 0^-}3x^{1/3}-3{-1}^{1/3}=0-3(-1)=3$;
$\displaystyle\lim_{t\to0^+}\int^1_{t}\Frac{dx}{x^{2/3}}=
3(1)^{1/3}-\lim_{t\to0^+}3x^{1/3}=3$ so
$\int^1_{-1}\Frac{dx}{x^{2/3}}$ converges and its value
is $3+3=6$.

\np$\int^1_0\Frac{dx}{\sqrt{1-x^2}}$.
This integral is improper because the function we are integrating 
is undefined at $1$.
Hence $\int^1_0\Frac{dx}{\sqrt{1-x^2}}=
\lim_{t\to1^-}\int^t_0\Frac{dx}{\sqrt{1-x^2}}=
\lim_{t\to1^-}\arcsin t-\arcsin 0=\arcsin 1=\frac\pi2$.
Hence the integral converges and its value is $\frac\pi2$.

\np$\int^{\infty}_{-\infty}\Frac{2x \> dx}{(x^2+1)^2}$.
This integral is improper because the end points of the interval of
integration are $\pm\infty$.
Hence $\int^{\infty}_{-\infty}\Frac{2x \> dx}{(x^2+1)^2}=
\int^0_{-\infty}\Frac{2x \> dx}{(x^2+1)^2}+
\int^{\infty}_{0}\Frac{2x \> dx}{(x^2+1)^2}=
\lim_{t\to-\infty}\int^{0}_{t}\Frac{2x \> dx}{(x^2+1)^2}+
\lim_{t\to\infty}\int^{t}_{0}\Frac{2x \> dx}{(x^2+1)^2}$.
Do the indefinite integral by substitution: $w=x^2+1$, $dw=2x\> dx$,
so
$\int\Frac{2x \> dx}{(x^2+1)^2}=\int\Frac{dw}{w^2}=\Frac{w^{-1}}{-1}+C=
\Frac{-1}{w}+C=\Frac{-1}{x^2+1}+C$.
Since $\displaystyle\lim_{x\to\pm\infty}\Frac{-1}{x^2+1}=0$,
$\int^{\infty}_{-\infty}\Frac{2x \> dx}{(x^2+1)^2}=
\br2{\Frac{-1}{0^2+1}-\lim_{t\to-\infty}\Frac{-1}{t^2+1}}+
\br2{\lim_{t\to\infty}\Frac{-1}{t^2+1}-\Frac{-1}{0^2+1}}=
(-1-0)-(0-(-1))=0$.
Hence the integral converges and its value is $0$.

\np$\int^\infty_0\Frac{dx}{(1+x)\sqrt x}$.
This integral is improper because of the $\infty$ {\sl and\/}
because the function we are integrating is not defined at $0$.
Hence
$\int^\infty_0\Frac{dx}{(1+x)\sqrt x}=
\lim_{t\to\infty}\int^t_1\Frac{dx}{(1+x)\sqrt x}+
\lim_{t\to0^+}\int^1_t\Frac{dx}{(1+x)\sqrt x}$.
Next we do the indefinite integral: substitute $u=\sqrt x$,
$du=\frac 12\Frac{dx}{\sqrt x}$ so
$\int\Frac{dx}{(1+x)\sqrt x}=\int\Frac{2du}{1+u^2}=2\arctan u+C=
\arctan\br1{\sqrt x}+C$.
Hence $\lim_{t\to\infty}\int^t_1\Frac{dx}{(1+x)\sqrt x}=
\lim_{t\to\infty}\arctan\br1{\sqrt t}-\arctan(\sqrt1)=
\frac\pi2-\arctan1$ and
$\lim_{t\to0^+}\int^1_t\Frac{dx}{(1+x)\sqrt x}=
\arctan(\sqrt 1)-\lim_{t\to0^+}\arctan\br1{\sqrt t}=
\arctan(1)-\arctan(0)=\arctan(1)-0$ so
$\int^\infty_0\Frac{dx}{(1+x)\sqrt x}$ converges and its value is
$\frac\pi4$.

\np$\int^1_0x\ln x\> dx$.
This integral is improper because the function we are integrating
is not defined at $0$.
Hence $\int^1_0x\ln x\> dx=\lim_{t\to0^+}\int^1_tx\ln x\> dx$.
Next we do the improper integral $\int x\ln x\> dx$ by parts:
$u=\ln x$, $dw=x\>dx$.
Then $du=\frac{dx}x$ and $w=\frac{x^2}2$ so
$\int x\ln x\> dx=\Frac{x^2\ln x}2-\int \frac{x^2}2\frac{dx}{x}=
\Frac{x^2\ln x}2-\frac12\int x\>dx=\Frac{x^2\ln x}2-\Frac{x^2}4+C$.
We will have to evaluate $\displaystyle\lim_{t\to0^+} t^2\ln t$.
This limit has the form $0\cdot-\infty$ so we may use l'H\^ opital's
Rule.
$\displaystyle\lim_{t\to0^+} t^2\ln t=
\displaystyle\lim_{t\to0^+} \Frac{\ln t}{1/t^2}=
\displaystyle\lim_{t\to0^+} \Frac{1/t}{-2/t^3}=
-\displaystyle\lim_{t\to0^+} \Frac{t^2}{2}=0$.
Hence
$\int^1_0x\ln x\> dx=
\lim_{t\to0^+}\br1{\Frac{x^2\ln x}2-\Frac{x^2}4}\Big\vert^1_t=
\br1{\Frac{1^2\ln(1)}2-\Frac{1^2}4}-
\br1{\Frac{\lim_{t\to0^+}t^2\ln t}2-\Frac{(\lim_{t\to0^+}t)^2}4}=
\frac{-1}4$.

\np$\int^\pi_0\Frac{\sin\theta}{\sqrt{\pi-\theta}}\>d\theta$
is improper because the function we are integrating is undefined at
$\pi$.
This time we see no way to do the integral, so one of the comparison 
theorems is required.
Between $0$ and $\pi$, $0\leq\sin\theta\leq1$
and $\sqrt{\pi-\theta}\geq 0$ so
$0\leq\Frac{\sin\theta}{\sqrt{\pi-\theta}}\leq
\Frac{1}{\sqrt{\pi-\theta}}$.
We have checked that the first comparison theorem applies.
Hence, if $\int^\pi_0\Frac{d\theta}{\sqrt{\pi-\theta}}$ converges, so does
$\int^\pi_0\Frac{\sin\theta}{\sqrt{\pi-\theta}}\>d\theta$.
The integral $\int^\pi_0\Frac{d\theta}{\sqrt{\pi-\theta}}$ is improper
because the function we are integrating is undefined at $\pi$.
Hence $\int^\pi_0\Frac{d\theta}{\sqrt{\pi-\theta}}=
\lim_{t\to\pi^-}\int^t_0\Frac{d\theta}{\sqrt{\pi-\theta}}=
\lim_{t\to\pi^-}\int^t_0(\pi-\theta)^{-1/2}d\theta=
\lim_{t\to\pi^-}\Frac{-\sqrt{\pi-\theta}}2\Big\vert^t_0=
\Frac{-\sqrt{0}}2-\Frac{-\sqrt{\pi}}2=\frac{\sqrt{\pi}}2$.
Hence $\int^\pi_0\Frac{d\theta}{\sqrt{\pi-\theta}}$ converges
and therefore so does 
$\int^\pi_0\Frac{\sin\theta}{\sqrt{\pi-\theta}}\>d\theta$.

\def\sfrac#1#2{{{\scriptscriptstyle#1}\over{\scriptscriptstyle#2}}}
\np$\int^{\ln2}_0x^{-2}e^{^{\sfrac{-1}x}}dx$ is improper because the
function we are integrating is undefined at $0$.
This integral can actually be done: substitute $u=-x^{-1}$,
$du=\frac{dx}{x^2}$ so
\vfill\eject
$\int^{\ln2}_0x^{-2}e^{^{\sfrac{-1}x}}dx=
\int^{-1/\ln2}_{-\infty}e^u du$.
This can be verified as follows:
$\int^{\ln2}_0x^{-2}e^{^{\sfrac{-1}x}}dx=
\lim_{t\to0^+}\int^{\ln2}_{t}x^{-2}e^{^{\sfrac{-1}x}}dx=
\lim_{t\to0^+}\int^{-1/\ln2}_{-1/t}e^u du=
\lim_{t\to-\infty}\int^{-1/\ln2}_{t}e^u du=
\int^{-1/\ln2}_{-\infty}e^u du$.
Moreover
$\displaystyle\lim_{t\to-\infty}\int^{-1/\ln2}_{t}e^u du=
e^{^{-1/\ln2}}-\lim_{t\to-\infty}e^t=e^{^{-1/\ln2}}$.
Hence $\int^{\ln2}_0x^{-2}e^{^{\sfrac{-1}x}}dx$ converges
(which is all we are asked - but the value is $e^{^{-1/\ln2}}$).

\np$\int^\infty_1\Frac{dx}{x^3+1}$ is improper because we are 
integrating to $\infty$.
We do not see how to do the integral, so we use a comparison theorem.
First note $x^3<x^3+1$ and $0\leq x^3$ for all $x$ in the interval
of integration, $[1,\infty)$.
Hence $0\leq\Frac 1{x^3}\leq\Frac1{x^3+1}$ so the first comparison
theorem applies and
$\int^\infty_1\Frac{dx}{x^3+1}$ converges provided
$\int^\infty_1\Frac{dx}{x^3}$ converges.
But $\int^\infty_1\Frac{dx}{x^3}=
\lim_{t\to\infty}\Frac{t^{-2}}{-2}-\Frac{1}{-2}=
0+\frac12$.
Hence $\int^\infty_1\Frac{dx}{x^3}$ converges and therefore
so does $\int^\infty_1\Frac{dx}{x^3+1}$.

\np$\int^\infty_1\Frac{\sqrt{x+1}}{x^2}\>dx$ is improper
because we are integrating to $\infty$.
This time we use the limit comparison theorem.
First note that $\Frac{\sqrt{x+1}}{x^2}\geq0$:
we wish to compare this function to
$\Frac{\sqrt{x}}{x^2}$.
Check that
$\displaystyle\lim_{x\to\infty}
\Frac{\frac{\sqrt{x+1}}{x^2}}{\frac{\sqrt{x}}{x^2}}=
\displaystyle\lim_{x\to\infty}
\Frac{\sqrt{x+1}}{\sqrt{x}}=
\sqrt{\displaystyle\lim_{x\to\infty}
\Frac{x+1}{x}}$.
So we need to evaluate
$\displaystyle\lim_{x\to\infty}\Frac{x+1}{x}$
This has the form $\frac\infty\infty$ so we may apply
l'H\^opital's Rule:
$\displaystyle\lim_{x\to\infty}\Frac{x+1}{x}=
\displaystyle\lim_{x\to\infty}\Frac{1}{1}=1$.
Hence the limit comparison theorem may be applied and
it says that $\int^\infty_1\Frac{\sqrt{x+1}}{x^2}\>dx$
converges if and only if
$\int^\infty_1\Frac{\sqrt{x}}{x^2}\>dx$ does.
But
$\int^\infty_1\Frac{\sqrt{x}}{x^2}\>dx=
\int^\infty_1x^{-3/2}\>dx=\lim_{t\to\infty}\Frac{t^{-1/2}}{1/2}-
\Frac{1}{1/2}=0-2$.
This means that $\int^\infty_1\Frac{\sqrt{x}}{x^2}\>dx$
converges and hence so does
$\int^\infty_1\Frac{\sqrt{x+1}}{x^2}\>dx$.

\np$\int^\infty_\pi\Frac{2+\cos x}x\>dx$ is improper since we are
integrating to $\infty$.
We do not know how to do the integral so we use a comparison
theorem.
Since $\cos x\geq-1$ and since $\frac 1x>0$ on 
the interval of integration $[\pi,\infty)$,
$0\leq\Frac{2-1}{x}\leq\Frac{2+\cos x}{x}$.
Hence the comparison theorem applies so if
$\int^\infty_\pi\Frac{2-1}{x}\>dx$ diverges, so does
$\int^\infty_\pi\Frac{2+\cos x}x\>dx$.
Now $\int^\infty_\pi\Frac{2-1}{x}\>dx=
\lim_{t\to\infty}\int^t_\pi\Frac{1}{x}\>dx=
\lim_{t\to\infty}\ln t-\ln1$.
But $\displaystyle\lim_{t\to\infty}\ln t=\infty$ so
$\int^\infty_\pi\Frac{2-1}{x}\>dx$ diverges and hence so does
$\int^\infty_\pi\Frac{2+\cos x}x\>dx$.

\end