\section{Improper Integrals}
\begin{enumerate}[1.]
\item
{\it Determine whether the integral
$$
\iint_Q \frac{x}{(1+x)(1+y)}dA,
$$
 where $Q$ is the first quadrant of the $xy$-plane, converges or diverges. If
it converges, evaluate it.}


\medskip
We first try to get some intuition for whether the integral should be convergent or divergent, since that will influence our choice of strategy. Notice that for any fixed $y$, the function $\frac{x}{(1+x)(1+y)}$ approaches $\frac{1}{1+y}$ at infinity, so we expect the integral in $x$ to be infinite. We now make this rigorous using inequalities.

We have
$$
\iint_Q \frac{x}{(1+x)(1+y)}dA\geq 
\int_0^{\infty } dy \int_1^\infty \frac{x}{(1+x)(1+y)}dx,
$$
since the function is nonnegative and the second region of integration is smaller. Therefore, if we prove that the second integral is divergent, it follows that so is the first one. We now estimate the second integral from below.
For $x\geq 1$,  we have
$$\frac{x}{1+x}\geq \frac{x}{2x}=\frac{1}{2},
$$
so that 
$$
\int_0^{\infty } dy \int_1^\infty \frac{x}{(1+x)(1+y)}dx
\geq \int_0^{\infty } dy \int_1^\infty \frac{1}{2(1+y)}dx
=\int_0^\infty \infty dy=\infty
$$
and the integral is divergent.


\bigskip

\item {\it Determine whether the integral
$$\displaystyle \iint_{Q} e^{y-x} dA$$ converges or diverges, if
$$
Q : = \{(x,y) : \ x \geq 1, \ 0 \leq y \leq \ln (x^2 + x + 1) \}.
$$
If it converges, evaluate it.}



\medskip
We will prove that the integral converges by direct evaluation.  First, we write the integral over $Q$ as the iterated integral
$$
\int_1^{\infty} \int_0^{\ln (x^2 + x + 1)} e^{y-x} dy dx = \int_{1}^{\infty} e^{-x} \bigg(\int_0^{\ln (x^2 + x + 1)} e^{y} dy \bigg) dx.
$$
Evaluating the inner integral gives
$$
\int_0^{\ln (x^2 + x + 1)} e^{y} dy = e^y \big\vert_{y = 0}^{\ln (x^2 + x + 1)} = (x^2 + x+1)-1=x^2+x.
$$
Therefore
\begin{align*}
\iint_{Q} e^{y-x} d A = \int_{1}^{\infty} (x^2 + x) e^{-x} d x = \int_{1}^{\infty} x^2 e^{-x} dx + \int_{1}^{\infty} x e^{-x} dx
\end{align*}
We will evaluate both of these integrals via integration by parts.  First notice that
$$
\int_{1}^{\infty} x^2 e^{-x} dx = - x^2 e^{-x} \big\vert_{x=1}^{\infty} + 2 \int_{1}^{\infty} x e^{-x} dx = e^{-1} + 2 \int_{1}^{\infty} x e^{-x} dx,
$$
so that
$$
\iint_{Q} e^{y-x} d A = e^{-1} + 3 \int_{1}^{\infty} x e^{-x} dx.
$$
Now, we again use integration by parts.  This gives that
$$
\int_{1}^{\infty} x e^{-x} dx = - x e^{-x} \big\vert_{1}^{\infty} + \int_{1}^{\infty} e^{-x} dx = e^{-1} - e^{-x} \big\vert_{1}^{\infty} = 2 e^{-1}.
$$
Putting together the previous steps, we see that
$$
\iint_{Q} e^{y-x} d A = e^{-1} + 3 \cdot (2 e^{-1}) = 7 e^{-1}.
$$
\bigskip


\item {\it Determine whether the integral
$$
\iint_{D} \frac{e^y }{y^{3/2} }\, dA
$$
 is convergent,
if $D$ is the triangle in the $xy$-plane with vertices $(0,0)$, $(1,0)$, $(1,1)$.
(You do not have to evaluate the integral.)}

\medskip

We estimate:

\begin{align*}
\iint_D \frac{e^y}{y^{3/2} } dA & \geq \iint_D \frac{1}{y^{3/2}}  dA
= \int_0^1 \int_0^x  y^{-3/2} dy dx 
\\
&= \int_0^1  \frac{y^{-1/2}}{-1/2} \Big|_{y=0}^x dx 
= \int_0^1 \infty \, dx = \infty,
\end{align*}
so that the integral is divergent.


\bigskip


\item {\it 
Let $D$ be the region $\{(x,y):\ x\geq 2,\ 0\leq y\leq \pi\}$.
Prove that the integral
$$
\iint_D \frac{1+e^{-x}}{x^2-\sin^2 y}dA
$$
is convergent, by comparing it to a different integral that is easier to evaluate.}

\medskip

We estimate, using that $e^{-x}\leq 1$ for $x\geq 0$, that $\sin^2 y\leq 1$, and that $x^2-1\geq x^2/2$ for $x\geq 2$:
\begin{align*}
\iint_D \frac{1+e^{-x}}{x^2-\sin^2 y}dA
&\leq \iint_D \frac{2}{x^2-\sin^2 y}dA\\
&\leq \int_2^\infty \int_0^\pi  \frac{2}{x^2-1}dydx
\leq \int_2^\infty  \frac{2\pi}{x^2-1}dx 
\\
&\leq \int_2^\infty  \frac{2\pi}{x^2/2}dx
= \int_2^\infty  \frac{4\pi}{x^2}dx = -\frac{4\pi}{x}\Big|_2^\infty = 2\pi.
\end{align*}
Therefore the integral converges.

\end{enumerate}

