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\begin{flushleft} 
{\course}         \hfill {Russell Brown}\\ 

\medskip	
Below are revised versions of the homework problems. Sorry about the
mistakes.

\begin{enumerate}
\item Chapter 6, \#10 continued. 

Use the change of variables $s=t^2$ to express 
\begin{equation}
\int_0 ^ 1 ( 1-t^2)^{(n-1)/2} \, dt
\end{equation}
in terms of the beta function.  The beta function is defined for $
a>0$ and $b>0 $ by
$$
B(a,b) = \int_0 ^ 1(1-s) ^ {a-1} s^{b-1} \, ds.
$$
Use the formula 
$$
B(a,b) = \Gamma (a)\Gamma (b) / \Gamma (a+b) 
$$
to express the integral (1) in terms of the $\Gamma $ function.

Derive and prove a formula for $ v_n$, the volume of the unit ball in
$ \reals ^n$.



\item Assume that $ f\geq 0 $ on $( 0, \infty)$.  Consider the
integral $I$ defined by 
$$
I = \int_{\{x: x_k > 0, k= 1, \dots,n\}} f( x_1 + \dots + x_n )x_1^{
a_1- 1}\dots x_n ^{a_n-1} \,d x.
$$
\begin{enumerate} \item Show that $ f(x_1 +\dots +x_n) $ is measurable
on $\reals ^n$.  (Hint: Imitate the proof that $f(x-t)$ is measurable
on $\reals ^{2n} $ in our study of convolutions.)

\item
Use a linear change of variables $t = x_1+\dots x_n$, and $ t_i =
x_i$, $ i =1,\dots, n-1$ and then a scaling to conclude that
\begin{eqnarray*}
\lefteqn{I = \int_0^\infty f(t) t ^{(\sum_{i=1} ^ n a _i) -1}\, dt} & &  \\
 & & \quad \times \int_{\{ t\in \reals ^{n-1} : t_1 +\dots
t_{n-1} < 1, t_ i > 0 \}}( 1- t_1 -\dots -t_{n-1})^{a_{n}-1} \\
& & \qquad \times  t_1
^{a_1 -1}  \dots t_{n-1} ^{a_{n-1} -1}\, 
dt_1\dots dt_{n-1}.
\end{eqnarray*}


\item Now we want to evaluate the $n-1$ dimensional integral in the
previous part. Our strategy is to choose a particular $f$ for which we
can evaluate both sides of (2) explicitly and then deduce the value of
the integral.  The magic $f$ is $f(t) = e^{-t}$.  Using this conclude
that if $ a_ i > 0$, $i=1\dots n$, then
$$
 \int_{\{ t\in \reals ^{n-1} : t_1 +\dots
t_{n-1} < 1, t_ i > 0 \}}( 1- t_1-\dots -t_{n-1})^{a_n-1} t_1
^{a_1 -1}  \dots t_{n-1} ^{a_{n-1} -1}\, 
dt_1\dots dt_{n-1}
$$

\end{enumerate}

Observe that the last part of this problem establishes the formula for
the beta function used above.

\item Show that if $ g(x) = F(|x|)$, and $ g\geq 0$, then
$$
\int _{\reals ^n } g(x) \,dx = c_n \int_0^\infty F(t) t^{n-1}\,dt.
$$
Hint:  Write $ F(|x|)= F\circ S ( x_1^2+ \dots x_n ^2) $  where $S(t)
=\sqrt t$.  
Consider the integral
$$
\int_ {\{ x_i>0, i =1\dots n\}}  F\circ S(x_1^2+ \dots + x_n^2) \, dx_1\dots dx_n
$$
make the change of variables $x_i^2= t_i$  and apply the previous
problem.
Your proof should give the value of $c_n$.
\end{enumerate}
\end{flushleft}
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