\documentclass[12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\p}{\partial}
\parindent=0pt
\begin{document}

{\bf Question}

State Lebesgue's theorem on dominated convergence for a sequence
of functions $\{f_n\}$.

The functions $f(x,t)$ has the following properties:

\begin{itemize}
\item[i)]
$f(x,t)$ is an integrable function of $x$, for each t;

\item[ii)]
there is a function $h(x)$ such that for each $x$, $\ds\lim_{t\to
a}f(x,t)=h(x)$, where $a$ is a fixed real number;

\item[iii)]
there is an integrable function $g(x)$ with the property that, for
each $x$, there exists $\delta>0$ such that for $|t-a|<\delta,$

$$|f(x,t)|\leq g(x).$$

\end{itemize}

Use Lebesgue's theorem to show that

$$ \lim_{t\to a}\int f(x,t)dx=\int h(x)dx.$$

Suppose that $\phi(x,t)$ is an integrable function of $x$, for
each $t$, and that the partial derivative $\frac{\p\phi}{\p t}$
exists for all $x,t$.

Suppose also that there is an integrable function $\psi(x)>0$ such
that

$$\left|\frac{\p\phi}{\p t}\right|\leq\psi(x)$$

for all $x$.  Prove that

$$\frac{d}{dt}\int\phi(x,t)dx=\int\frac{\p\phi}{\p t}dx.$$

If $\ds I(\alpha)=\int_0^\infty x^\alpha e^{-x}dx, \hspace{0.75in}
\alpha>0,$

show that

$$ I(\alpha)=I'(\alpha+1)-(\alpha+1)I'(\alpha).$$


\vspace{0.25in}
\newpage

{\bf Answer}

Lebesgue's theorem on dominated convergence states that if
$\{f_n\}$ is a sequence of measurable functions with the property
that $|f_n(x)|\leq g(x)$ for all $n,x$ where $g(x)$ is integrable,
and if $f_n(x)\rightarrow f(x)$ as $n\rightarrow\infty$ for all
$x$, then $f$ is integrable, and $\ds \lim_{n\to\infty}\int
f_n(x)dx=\int f(x)dx$.

${}$

Let $\{t_n(x)\}$ be an arbitrary sequence of real numbers
satisfying $|t_n-a|<\delta(x)$ and converging to $a$.

Let $f_n(x)=f(x,t_n(x))$

Then $|f_n(x)|\leq g(x)$ for all $x$.

Thus $\ds\lim_{n\to\infty}\int f(x,t_n)dx=\int h(x)dx$ by
Lebesgue's theorem.

Hence $\ds\lim_{t\to\infty}\int f(x,t)dx=\int h(x)dx$

Now let $\Phi(t)=\int\phi(x,t)dx$

Consider $\ds \frac{\Phi(t+h)-\Phi(t)}{h}$

$\ds =\int\frac{\phi(x,t+h)-\phi(x,t)}{h}dx=\int X(x,h)dx$

Now $\ds X(x,h)\rightarrow \left.\frac{\p\phi}{\p t}\right|_x
\hspace{0.5in}$ as $h\rightarrow 0$

Thus there exists $\delta$ such that for all $h$ satisfying
$0<|h|<\delta$,

$\ds \left|X(x,h)-\left.\frac{\p\phi}{\p
t}\right|_x\right|<\psi(x)$

Thus for all $h$ satisfying $0<|h|<\delta$, we have

$\ds \left|X(x,h)\right|\leq\left.\left|\frac{\p\phi}{\p
t}\right|_x\right|+\psi(x)\leq 2\psi(x)$

Since $\psi$ is integrable $2\psi$ is integrable and so the above
result shows that

$\ds \lim_{h\to0}\int X(x,h)dx=\int\frac{\p\phi}{\p t}dx$

i.e. $\ds\frac{d}{dt}\int\phi(x,t)dx=\int\frac{\p\phi}{\p t}dx$

$\ds I(\alpha)=\int_0^\infty x^\alpha e^{-x}dx$

$\ds \frac{\p}{\p\alpha}(x^\alpha e^{-x})=x^\alpha(\log x)e^{-x}$

$=x^\alpha(\log
x)e^{-\frac{1}{2}x}e^{-\frac{1}{2}x}<e^{-\frac{1}{2}x}
\hspace{0.5in}$ for $x\geq x_0$

since $x^\alpha(\log x)e^{-\frac{1}{2}x} \rightarrow 0$ as
$x\rightarrow\infty$

also $x^\alpha(\log x)e^{-\frac{1}{2}x}\rightarrow 0$ as
$x\rightarrow 0$ and so

$|x^\alpha(\log x)e^{-\frac{1}{2}x}|<1$ if $0<x\leq x_1$.

Thus if

$\psi(x)=\left\{ \begin{array}{cr} 1& 0<x\leq x_1\\ |x^\alpha(\log
x)e^{-\frac{1}{2}x}|& x_1<x<x_2\\ e^{-\frac{1}{2}x}& x\geq
x_2\end{array}\right.$

$\ds \left|\frac{\p}{\p\alpha}(x^\alpha e^{-x})\right|\leq\psi(x)$
and $\psi$ is integrable,

thus $\ds I'(\alpha)=\int_0^\infty\frac{\p}{\p\alpha}(x^\alpha
e^{-x})$

$\ds (\alpha+1)I'(\alpha)=\int_0^\infty(\alpha+1)x^\alpha(\log
x)e^{-x}dx$

$\ds =\left[x^{\alpha+1}(\log
x)e^{-x}\right]_0^\infty-\int_0^\infty
x^{\alpha+1}\left[e^{-x}\frac{1}{x}-(\log x)e^{-x}\right]dx$

$\ds =\int_0^\infty x^{\alpha+1}(\log x)e^{-x}dx-\int_0^\infty
x^\alpha e^{-x}dx$

$=I'(\alpha+1)-I(\alpha)$

Hence $I(\alpha)=I'(\alpha+1)-(\alpha+1)I'(\alpha)$

\end{document}