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\begin{document}

{\bf Question}

\begin{itemize}
\item[a)]
Prove that, for $|x|&lt;1$

$$\int_0^\infty\frac{\sin
t}{e^t-x}dt=\sum_{n=1}^\infty\frac{x^{n-1}}{n^2+1}.$$

${}$

\item[b)]
The function $J_m(x)$ is defined by

$$J_m(x)=\sum_{n=0}^\infty\frac{(-1)^n}{n!(n+m)!}\left(\frac{x}{2}\right)^{m+2n}.$$

Prove that for all real values of $a$,

$$2\int_0^\infty J_m(2ax)x^{m+1}e^{-x^2}dx=a^me^{-a^2}.$$
\end{itemize}

${}$

[In both cases each step in the calculations should be carefully
explained, and a clear statement given of any theorems applied,
together with a proof that any conditions for the application of
theorems are satisfied.]


\vspace{0.25in}

\newpage

{\bf Answer}

In both examples we use the following theorem, which is a
consequence of Lebesgue's Dominated Convergence Theorem.


${}$

{\bf Theorem A}  Suppose that $\{f_n\}$ is a sequence of
integrable functions.  Then $\int\sum f_n=\sum\int f_n$ provided
either $\sum\int|f_n|$ or $\int\sum|f_n|$ is finite.  (The two
conditions are in fact equivalent.)

${}$

\begin{itemize}
\item[a)]
$\ds \int_0^\infty\frac{\sin t}{e^t-x}dt=\int_0^\infty\frac{\sin
t}{e^t(1-\frac{x}{e^t})}dt$

Now for $\ds t\geq0 \,\,\,\, \left|\frac{x}{e^t}\right|\leq |x|&lt;1$
and so

for all $\ds t\geq0 \,\,\,\,
\frac{1}{1-\frac{x}{e^t}}=1+\frac{x}{e^t}+\frac{x^2}{e^{2t}}+\frac{x^3}{e^{3t}}+\cdots$

Thus for $|x|&lt;1$

$\ds \int_0^\infty\frac{\sin
t}{e^t-x}dt=\int_0^\infty\sum_{n=1}^\infty x^{n-1}\frac{\sin
t}{e^{nt}}dt \hspace{0.5in} (1)$

Now $\ds \sum_{n=1}^\infty\int_0^\infty \left|x^{n-1}\frac{\sin
t}{e^{nt}}\right|dt$

$\ds \leq \sum_{n=1}^\infty\int_0^\infty
|x|^{n-1}\frac{1}{e^{nt}}dt=\sum_{n=1}^\infty\frac{|x|^{n-1}}{n}&lt;\infty\,\,\,$
for $|x|&lt;1$.

We can thus apply theorem A to interchange the order of
integration and summation in equation (1).  We then have

$\ds \int_0^\infty\frac{\sin t}{e^t-x}dt= \sum_{n=1}^\infty
x^{n-1}\int_0^\infty\frac{\sin
t}{e^{nt}}dt=\sum_{n=1}^\infty\frac{x^{n-1}}{n^2+1}$

(The integral $\ds\int_0^\infty\frac{\sin t}{e^t-x}dt$ is easily
evaluated as $\frac{1}{n^2+1}$)

\newpage

\item[b)]
$\ds
J_m(x)=\sum_{n=0}^\infty\frac{(-1)^n}{n!(n+m)!}\left(\frac{x}{2}\right)^{m+2n}$

$\ds
J_m(2ax)=\sum_{n=0}^\infty\frac{(-1)^n}{n!(n+m)!}a^{m+2n}x^{m+2n}$

$\ds 2\int_0^\infty J_m(2ax)x^{m+1}e^{-x^2}dx$

$\ds =2\int_0^\infty\sum_{n=0}^\infty\frac{(-1)^n}{n!(n+m)!}
a^m.a^{2n}.x^{m+2n}.x^{m+1}.e^{-x^2}dx$

$\ds =a^m\sum_{n=0}^\infty\frac{(-1)^na^{2n}}{n!}
\int_0^\infty\frac{2x^{2(m+n)+1}}{(n+m)!}e^{-x^2}dx$

(*)

$\ds =a^m\sum_{n=0}^\infty\frac{(-1)^na^{2n}}{n!}\frac{1}{(n+m)!}
\int_0^\infty t^{m+n}e^{-t}dt$

$\ds =a^m\sum_{n=0}^\infty\frac{(-1)^na^{2n}}{n!}\frac{1}{(n+m)!}
\Gamma(m+n+1)$

$\ds =a^m\sum_{n=0}^\infty\frac{(-1)^na^{2n}}{n!} \hspace{0.75in}
[\Gamma(m+n+1)=(m+n)!]$

$\ds =a^me^{-a^2}$

${}$

The inversion of order of summation and integration at (*) is
justified by the steps after (*) which (with the factor $(-1)^n$
removed, and $a$ replaced by $|a|$) show that
$\sum\int|f_n|&lt;\infty$, thus enabling theorem A to be applied.

\end{itemize}

\end{document}
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