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\begin{center}
\vskip 1cm{\LARGE\bf 
Sequences of Definite Integrals, Factorials \\
\vskip .1in
and Double Factorials}

\vskip 1cm
\large
Thierry Dana-Picard \\
Department of Applied Mathematics \\
Jerusalem College of Technology\\
Havaad Haleumi Str.\ 21 \\
POB 16031\\ 
Jerusalem 91160 \\
Israel \\
\href{mailto:dana@jct.ac.il}{\tt dana@jct.ac.il} \\
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\vskip .2 in
\begin{abstract}
We study sequences of definite integrals. Some of them provide closed
forms involving factorials and/or double factorials. Other examples are
associated with either sequences or pairs of sequences of rational
numbers, for which summations are found.
\end{abstract}



\newtheorem{theorem}{Theorem}[section]
\newtheorem{proposition}{Proposition}[section]
\newtheorem{corollary}{Corollary}[section]
\newtheorem{lemma}{Lemma}[section] 



\newtheorem{example}[theorem]{Example}
\newtheorem{remark}[theorem]{Remark}




\section{Introduction.} 
The study  of sequences of  either definite or improper  integrals has
connections with  various fields, such  as combinatorics (see \cite{sury},
\cite{catalan}),  infinite series  (see \cite{glaister}), and  others.  In
this paper, we  study sequences of integrals, depending  mostly on one
parameter,  sometimes  on  two  parameters.
The method generally  used is a telescopic  method (see
\cite[p.\ 579]{jm});  when   a   recurrence   relation  based   on
multiplication by a homographic function exists, the implementation of
the  method  is  easy.  Otherwise  it  can  be  very  hard,  sometimes
impossible.

In Section  \ref{section log}, improper integrals  depending either on
one or two parameters (that are non-negative integers) are considered, where the integrand involves a logarithm. In these examples, various situations are described (different roles and influences of the parameters, existence or non-existence of a closed expression for the general term of the sequence, etc.).

In Section  \ref{section rational},  three sequences of  integrals are
studied; the main one is the sum of a sequence of rational numbers and
a sequence of  rational multiples of $\sqrt{3}$. Then formulas for related
integrals are derived.


\section{Logarithmic integrals with a parameter.}
\label{section log}


For every natural number $n$, we define the improper integral
\begin{equation}I_n=\int_0^1 x \; (\ln x )^n \; dx.
\end{equation}
As 
\begin{equation*}
\underset{x \rightarrow 0^+}{\lim} \; x^p \; (\ln x)^q =0 
\end{equation*}
holds
 for any two positive integers $p$ and $q$, we can work with $I_n$ as if it
 is a definite integral, i.e., by writing ``ordinary'' expressions for the integrals and not writing limits for $\lambda$ arbitrarily close to 0 of $\int_{\lambda}^1 x \; (\ln x )^n \; dx$.

We take $u(x)=(\ln x )^n$ and $v(x)=x^2/2$ in order to perform an integration by parts; we have
\begin{equation*}
I_n = \left[ \frac 12 x^2 \; (\ln x)^n \right]_0^1 -\frac n2 \; 
   \int_0^1 x \; (\ln x )^{n-1} \; dx,
\end{equation*}
thus,
\begin{equation}
I_n=-\; \frac n2 I_{n-1}.
\end{equation}
By telescoping, we have
\begin{equation*}
I_n=-\frac n2 I_{n-1}=\left(-\frac n2 \right)\left(-\frac {n-1}{2} \right) I_{n-2}=\dots
=(-1)^n \; \frac {n!}{2^n}\; I_0.
\end{equation*}
A straightforward  computation provides $I_0= 1/2$,
and we have proved the following proposition:
\begin{proposition}
For any natural number $n$,
\begin{equation*}
\int_0^1 x \; (\ln x )^n \; dx=(-1)^n \; \frac {n!}{2^{n+1}}.
\end{equation*}
\end{proposition}

Note that the sequence whose general term is $(-1)^n\; 2^{n+1}I_n$ provides an integral representation of factorials.


Now consider the sequence of integrals defined by
\begin{equation}
I_n=\int_0^1 x \; (\ln x )^{n+1/2} \; dx.
\end{equation}

The natural logarithm is a negative function over the interval $(0,1)$, thus the value of this integral, if it exists, is a pure imaginary complex number. Actually, the square root function is a multi-valued function with two branches. Each branch is analytic, thus an integral of the form
\begin{equation*}
\int_{\varepsilon}^1 x \; (\ln x )^{n+1/2} \; dx \qquad \text{where} \quad \varepsilon \neq 0, 
\end{equation*}
 is path-independent (in fact, as $x$ is  a positive real number, $\ln x &lt;0$ and  the involved path can be a segment on the $y-$axis), when computed along a path which does not intersect the standard branch cut (see \cite{wunsch}).
 Moreover, for any non-negative integer $n$, we have
\begin{equation*}
\underset{x \rightarrow 0^+}{\lim} \; x\; (\ln x)^{n+1/2} = 0. 
\end{equation*}
Therefore the given integral 
$I_n$ is well-defined (we use a method as in \cite[p.\ 362]{wunsch}).

In a manner similar to the method used above,
we obtain the following recurrence relation:
\begin{equation}
I_n=-\frac 12 \left( n+ \frac 12 \right) \; I_{n-1}=- \frac {2n+1}{4} \; I_{n-1}.
\end{equation}
We need now to compute the first integral of the sequence:
\begin{equation*}
I_0=\int_0^1 x \; (\ln x )^{1/2} \; dx = \frac {i \; \sqrt{2 \pi}}{8}.
\end{equation*}
 
We recall the definition of the double factorial of an odd number (see  
Sloane's sequence \seqnum{A001147} and \cite{mathworld double factorial}):
\begin{equation*}
\forall n \in \mathbb{N}, \; (2n-1)!!=1 \cdot 3 \cdot 5 \cdot \dots \cdot (2n-1),
\end{equation*}
and for an even number
\begin{equation*}
\forall n \in \mathbb{N}, \; (2n)!!=2 \cdot 4 \cdot 6 \cdot \dots \cdot (2n).
\end{equation*}

Therefore the following formula holds:
\begin{proposition}
For any natural number $n$:
\begin{equation*}
\int_0^1 x \; (\ln x )^{n+1/2} \; dx =\frac {(-1)^n \; n!!}{2^{2n+3}}\; i \sqrt{2 \pi }.
\end{equation*}
\end{proposition}


Now let $p$ and $q$ be non-negative integers. We define
\begin{equation}
I_{p,q}=\int_0^1 x^p \; (\ln x)^q \; dx.
\end{equation}
We take $u(x)=(\ln x )^q$ and $v(x)=x^{p+1}/(p+1)$ in order to perform an integration by parts and get
\begin{equation*}
I_{p,q}=\underbrace{\left[ \frac {1}{p+1}x^{p+1}\; (\ln x)^q\right]_0^1}_{=0}
-\frac {q}{p+1} \; \int_0^1 \frac 1x \cdot x^{p+1} \; (\ln x)^{q-1} \; dx,
\end{equation*}
i.e.,
\begin{equation}
I_{p,q}=-\; \frac{q}{p+1} \; I_{p,q-1}.
\end{equation}

Now we have
\begin{align*}
I_{p,q}
&amp;=-\; \frac{q}{p+1} \; I_{p,q-1}\\
\quad
&amp;=\left(-\; \frac{q}{p+1}\right) \; \left( - \; \frac {q-1}{p+1} \right) \;  I_{p,q-2}\\
\quad
&amp;=\left(-\; \frac{q}{p+1}\right) \; \left( - \; \frac {q-1}{p+1} \right) \; \left( - \; \frac {q-2}{p+1} \right) \; I_{p,q-3}\\ 
\quad &amp;= \dots
\end{align*}
As $I_{p,0}=1/(p+1)$, we have finally
\begin{proposition}
For any pair $(p,q)$ of natural numbers,
\begin{equation*}
\label{eq int two parameters}
\int_0^1 x^p \; (\ln x)^q \; dx=\frac {(-1)^q \; q!}{(p+1)^{q+1}}.
\end{equation*}
\end{proposition}

This example of a parametric integral depending on two parameters that
are non-negative integers, together with another integral described in \cite{sequences of definite integrals}, shows the great difference between the influences of the parameters: the whole computation is concentrated on one parameter only, and the other one is ``passive''.  Nevertheless, the final result is depends on both parameters. 

\begin{remark}
{\rm Equation (\ref{eq int two parameters}) is equivalent to
\begin{equation*}
q!=(-1)^q\; (p+1)^{q+1} \; \int_0^1 x^p \; (\ln x)^q \; dx.
\end{equation*}
This integral form for a factorial is surprising, as it contains a parameter without influence.}
\end{remark}

In the examples studied above, the reason for the computation of a closed form to be so easy lies in the fact that, when performing the integration by parts, the integrated part of the result is equal to 0. This provides a recurrence relation for the sequence $(I_n)$ of the form
\begin{equation}
\label{eq homographic relation}
I_n=f(n)\; I_{n-1},
\end{equation}
where $f$ is a homographic function of $n$ with integer coefficients.
Other examples of this kind have been studied in
\cite{glaister,explicit closed form,sequences of definite integrals}.
When this situation does not occur,
computations can be more complicated, as the next example shows.

For every natural number $n$, we define the integral
\begin{equation}
I_n=\int_1^e x \; (\ln x )^n \; dx.
\end{equation}
We have
\begin{equation*}
I_0=\int_1^e x  \; dx = \left[ \frac 12 x^2 \right]_1^e=\frac 12 (e^2-1).
\end{equation*}

Choosing as above $u(x)=(\ln x )^n$ and $v(x)=x^2/2$, an integration by parts yields 
\begin{equation*}
I_n =\left[ \frac 12 x^2 \; (\ln x )^n \right]_1^e
      - \frac n2 \int_1^e x \; (\ln x )^{n-1} \; dx 
\end{equation*}
and leads to the following recurrence relation:
\begin{equation}
I_n = \frac 12 e^2 - \frac n2 \; I_{n-1}.
\end{equation}

The presence of a non-zero integrated term makes the work harder than in previous examples. We have
\begin{align*}
I_n &amp;= \frac 12 \; e^2 - \frac {n}{2} \; \left( \frac {1}{2} e^2 
       - \frac {n-1}{2} \; I_{n-2} \right)\\
\quad 
&amp;=  \frac 12 \; e^2 \; \left( 1-\frac n2 \right)
     + \frac {n(n-1)}{4} \; I_{n-2}\\
\quad
&amp;= \frac 12 \; e^2 \; \left( 1-\frac n2 \right)
     + \frac {n(n-1)}{4} \left(  \frac 12 e^2 - \frac {n-2}{2}\; 
     I_{n-3} \right)\\
\quad 
&amp;= \frac 12 \; e^2 \;  \left( 1-\frac {n}{2} + \frac {n(n-1)}{4} \right)
     - \frac {n(n-1)(n-2)}{8} \; I_{n-3} \\
\quad &amp; = \dots \\
\quad
&amp;= \frac 12 \; e^2 \; \left(
     1-\frac {n}{2} +  \frac {n(n-1)}{4} + \dots +(-1)^{n-1} \frac {n!}{2^{n-1}}
    \right) +\frac {(-1)^n \; n!}{2^n}\; I_0 \\
&amp;= \frac 12 \; e^2 \; \left(1-\frac n2 +\frac {n(n-1)}{2^2}-\frac{n(n-1)(n-2)}{2^3}
   + \dots + (-1)^{n-1} \frac {n!}{2^{n-1}} +(-1)^n \; \frac {n!}{2^n} \right)\\
\quad &amp; \qquad 
   + (-1)^{n+1} \; \frac {n!}{2^{n+1}}.
\end{align*}
Recall that for any two non-negative integers $n$ and $k$ such that $0 \leq k \leq n$,
\begin{equation*}
A_n^k = \frac {n!}{(n-k)!}
\end{equation*}
($A_n^k$ is the number of arrangements without repetition of $n$ elements by $k$).
Hence, the following holds:
\begin{proposition}
\begin{equation*}
\int_1^e \; x \; (\ln x )^n \; dx =
\frac 12 \; e^2 \; \underset{k=0}{\overset{n}{\sum}} (-1)^k \; \frac {A_n^k}{2^k} 
+ (-1)^{n+1} \; \frac {n!}{2^{n+1}}.
\end{equation*}
\end{proposition}


\section{Three related parametric rational integrals}
\label{section rational}
For $n$ a positive integer, we define the integrals
\begin{align}
I_n &amp; =\int_0^1 \frac {1}{(x^2+x+1)^n}\; dx, \\
J_n &amp; =\int_0^1 \frac {x}{(x^2+x+1)^n}\; dx, \\
K_n &amp; =\int_0^1 \frac {x^2}{(x^2+x+1)^n}\; dx.
\end{align}

\subsection{First integral: complete computations}

As in the previous examples, we wish to find  a recurrence relation for the sequence $(I_n)$, then a closed form for the general term, if possible. We perform an integration by parts; let
\begin{equation*}
u(x)=\frac {1}{(x^2+x+1)^n} \qquad \text{and} \qquad v(x)=x,
\end{equation*}
whence
\begin{equation*}
u'(x)=\frac {-n(2x+1)}{(x^2+x+1)^{n+1}}\qquad \text{and} \qquad v'(x)=1.
\end{equation*}
It follows that
\begin{align*}
I_n &amp;= \left[ \frac {x}{(x^2+x+1)^n} \right]_0^1
+n \; \int_0^1 \frac{x(2x+1)}{(x^2+x+1)^{n+1}} \; dx\\
\quad &amp;=\frac{1}{3^n} + n \; \underbrace{\int_0^1 \frac{x(2x+1)}{(x^2+x+1)^{n+1}} \; dx}_{=T_n}.
\end{align*}

In order to compute $K_n$, we decompose the integrand into partial fractions:
\begin{align*}
\forall x \in \mathbb{R}, 
\frac{x(2x+1)}{(x^2+x+1)^{n+1}}
&amp;= -\frac {x}{(x^2+x+1)^{n+1}} - \frac {2}{(x^2+x+1)^{n+1}} + \frac {2}{(x^2+x+1)^n}\\
\quad
&amp;= -\frac 12 \cdot \frac {2x+1}{(x^2+x+1)^{n+1}} - \frac 32 \cdot \frac {1}{(x^2+x+1)^{n+1}} + \frac {2}{(x^2+x+1)^n}.
\end{align*}

Thus,
\begin{align*}
T_n
&amp;= -\frac 12 \; \left[ -\frac 1n \cdot \frac {1}{(x^2+x+1)^n} \right]_0^1 -\frac 32 \; I_{n+1} + 2\; I_n\\
\quad
&amp;=\frac{1}{2n} \; \left( \frac{1}{3^n}-1 \right) -\frac 32 I_{n+1}+2I_n. 
\end{align*}
By re-arranging the terms, we obtain the following relation of recurrence:
\begin{equation}
\label{eq recurrence relation}
I_{n+1}=\frac{1}{3n} \; \left( \frac{1}{3^{n-1}}-1\right) + \frac {2(2n-1)}{3n}\; I_n.
\end{equation}

We first compute the integral $I_1$:
\begin{equation*}
I_1=
\int_0^1 \frac{1}{x^2+x+1}\; dx= \int_0^1 \frac{1}{\left(x+\frac 12 \right)^2+\frac 34}\; dx= \int_0^1  \frac {1}{\frac 34 \; \left[ \frac 43 \; \left(x+\frac 12 \right)^2+1 \right]} \; dx.
\end{equation*}

Using the substitution $t=(2/\sqrt{3})(x+1/2)$, we obtain:
\begin{equation}
\label{eq first integral}
I_1=\frac {2}{\sqrt{3}} \; \int_{1/\sqrt{3}}^{\sqrt{3}} \frac {1}{t^2+1} \; dt = \frac {2}{\sqrt{3}} \left( \arctan \sqrt{3}-\arctan \frac {1}{\sqrt{3}} \right) 
= \frac {\pi \; \sqrt{3}}{9}.
\end{equation}

From Equations (\ref{eq recurrence relation}) and (\ref{eq first integral}) follows that $I_n$ is given by a relation of the form
\begin{equation}
I_n=a_n+b_n \pi \sqrt{3}
\end{equation}
where $a_n$ and $b_n$ are rational numbers.
We study separately the sequences $(a_n)$ and $(b_n)$.

Consider $b_n$ first:
\begin{align*}
b_{n+1}
&amp;= \frac {2(2n-1)}{3n} \; b_n = \frac 23 \; \cdot \; \frac {2n-1}{n} \; b_n\\
\quad
&amp;=\left(\frac 23 \right)^2 \; \cdot \; \frac {2n-1}{n} \; \cdot \;\frac {2n-3}{n-1} \; b_{n-1}\\
 \quad
&amp;=\left(\frac 23 \right)^3 \; \cdot \; \frac {2n-1}{n} \; \cdot \;\frac {2n-3}{n-1} \; 
\cdot \;\frac {2n-5}{n-2} \; b_{n-2}\\
 \quad
&amp;=\dots\\
\quad
&amp;=\left( \frac 23 \right)^n \; \cdot \; 
\frac{(2n-1)(2n-3)(2n-5) \dots 3 \cdot 1}{n(n-1)(n-2)\dots 2 \cdot 1} \; b_1.
\end{align*}
Inserting suitable factors into the numerator, and dividing out by the same factors, we obtain a closed factorial form for $b_{n+1}$:
\begin{equation}
b_{n+1}=\left( \frac 23 \right)^n \; \cdot \; \frac {(2n)!}{2^n \cdot (n!)^2} \; b_1
= \frac {1}{3^n} \cdot \frac {(2n)!}{(n!)^2}\; b_1=\frac{1}{3^{n+2}} \cdot \frac {(2n)!}{(n!)^2}.
\end{equation}
Note that 
\begin{equation*}
\frac {(2n)!}{(n!)^2}= \begin{pmatrix} 2n \\ n \end{pmatrix}.
\end{equation*}
The interested reader will find concrete occurences of these numbers 
(special paths in graphs, etc.) in Sloane's encylopedia,
sequence \seqnum{A000984}. 

Another representation can be given for the sequence $(b_n)$,
using the \textit{double factorial} (see Sloane's encyclopedia,
\seqnum{A001147} and \cite{mathworld double factorial}). 
We have
\begin{equation}
\forall n \in \mathbb{N}, \; b_n=\frac {2^{n-1}}{3^{n+1}} \cdot \frac {(2n-3)!!}{(n-1)!}.
\end{equation}


A closed form for $a_n$ is harder to derive. From Equation (\ref{eq recurrence relation}), we derive the following recurrence relation for $a_n$:
\begin{equation}
\label{eq recurrence bn}
a_{n+1}=\frac{1}{3n} \left( \frac {1}{3^{n-1}} -1 \right) + \frac {2(2n-1)}{3n} \; a_n.
\end{equation}
By Equation (\ref{eq first integral}), $a_1=0$, whence the sequence $(a_n)$ is well defined by the above relation.

Let's now use a telescopic process:
\begin{align*}
a_{n+1}
&amp;=\frac {1}{3n} \left( \frac {1}{3^{n-1}} -1 \right) + \frac {2(2n-1)}{3n} \; a_n\\
\quad
&amp;= \frac {1}{3n} \left( \frac {1}{3^{n-1}} -1 \right) + \frac {2(2n-1)}{3n} \;
\left[ \frac {1}{3(n-1)} \left( \frac {1}{3^{n-2}} -1 \right) + \frac {2(2n-3)}{3(n-1)} a_{n-1} \right]\\
\quad 
&amp;=\frac {1}{3n} \left( \frac {1}{3^{n-1}} -1 \right) 
   +\frac {2}{3^2} \; \frac{(2n-1)}{n(n-1)} \; \left( \frac {1}{3^{n-2}} -1 \right)
   + \left( \frac 23 \right)^2 \; \frac{(2n-1)(2n-3)}{n(n-1)} \; a_{n-1} \\ 
\quad
&amp;=\frac {1}{3n} \left( \frac {1}{3^{n-1}} -1 \right) 
   +\frac {2}{3^2} \; \frac{(2n-1)}{n(n-1)} \; \left( \frac {1}{3^{n-2}} -1 \right)\\
\quad
&amp; \qquad 
   + \left( \frac 23 \right)^2 \; \frac{(2n-1)(2n-3)}{n(n-1)} \;
   \left[  \frac {1}{3(n-2)} \left( \frac {1}{3^{n-3}} -1 \right) + \frac {2(2n-5)}{3(n-2)} a_{n-2} \right] \\
\quad
&amp;= \dots \quad .
\end{align*}
Iterations are needed until $a_2$ is reached, because $a_2=a_1=0$.
Finally, the following formula is derived:
\begin{equation*}
a_{n+1}=\frac {1}{3n} \left( \frac {1}{3^{n-1}} -1 \right) 
+ \underset{k=2}{\overset{n-2}{\sum}} 2^{k-1}\; (3^{-n}-3^{-k}) \; 
\frac {(2n-1)(2n-3) \dots (2n-2k+3)}{n(n-1)(n-2) \dots (n-k+1)}. 
\end{equation*}
The rational fraction on the right can be turned into a 
closed factorial formula. Shifting the index $n+1$ to $n$, we obtain
  \begin{equation}
a_{n}=\frac {1}{3(n-1)} \left( \frac {1}{3^{n-2}} -1 \right) 
+ \frac 12 \; \underset{k=2}{\overset{n-3}{\sum}} (3^{-n+1}-3^{-k}) \; 
\frac {(2(n-1))!}{(2n-2k-1)!} \; \left( \frac {(n-k-1)!}{(n-1)!} \right)^2.
\end{equation}

A formula involving double factorials looks a little more compact (see 
Sloane's sequence \seqnum{A001147} and \cite{mathworld double factorial}):
\begin{equation}
a_{n}=\frac {1}{3(n-1)} \left( \frac {1}{3^{n-2}} -1 \right) 
+ \frac 12 \; \underset{k=2}{\overset{n-3}{\sum}} 2^{k}\; (3^{-n+1}-3^{-k}) \;
\frac {(2n-3)!! \; (n-k-1)!}{(2n-2k-1)!! \; (n-1)!} .
\end{equation}

In conclusion, we have
\begin{proposition}
For any non-negative integer $n$,
\begin{equation*}
\int_0^1 \frac {dx}{(x^2+x+1)^n} = a_n+ b_n \; \pi \; \sqrt{3},
\end{equation*}
with 
\begin{align*}
a_{n}&amp; =\frac {1}{3(n-1)} \left( \frac {1}{3^{n-2}} -1 \right) 
+ \frac 12 \; \underset{k=2}{\overset{n-3}{\sum}} 2^{k}\; (3^{-n+1}-3^{-k}) \;
\frac {(2n-3)!! \; (n-k-1)!}{(2n-2k-1)!! \; (n-1)!}\\
\text{and}&amp; \quad \\
b_n &amp;= \frac {2^{n-1}}{3{n+1}} \cdot \frac {(2n-3)!!}{(n-1)!}.
\end{align*}
\end{proposition}

\subsection{Extensions} 
From the results above, the two related parametric integrals $J_n$ and $K_n$ can be computed:

We have
\begin{align*}
I_n+2J_n &amp;=\int_0^1 \frac {1+2x}{(x^2+x+1)^n}\; dx \\
\quad &amp; =  \frac{1}{1-n} \left[\frac {1}{(x^2+x+1)^{n+1}} \right]_0^1 \\
\quad &amp;= \frac {1}{1-n} \; \left( \frac {1}{3^{n-1}}-1 \right),
\end{align*}
i.e.,
\begin{equation}
J_n=\frac {1}{2(1-n)} \; \left( \frac {1}{3^{n-1}}-1 \right) - \frac 12 \; I_n.
\end{equation}
An expression of $J_n$ as a function of $n$ is obtained by substitution.

A closed form for $K_n$ is obtained by substitution, according to the following remark:
\begin{equation}
I_n+J_n+K_n =\int_0^1 \frac {1+x+x^2}{(x^2+x+1)^n}\; dx
= \int_0^1 \frac {1}{(x^2+x+1)^{n-1}}\; dx =I_{n-1}.
\end{equation}


\section{Acknowledgements}

The author wishes to thank the referee for his remarks and
appreciations, and the editor for his care, including long-distance
phone calls.

\bibliographystyle{amsalpha}

\begin{thebibliography}{AA}

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\bibitem [2]{catalan} Th. Dana-Picard,
Parametric   integrals  and   Catalan  numbers, 
\textit{Internat. J. Math. Educ. in Science and Technology}
\textbf{36} (2005), 410--414.

\bibitem [3] {sequences of definite integrals} Th. Dana-Picard,
Sequences of definite integrals, preprint, 2004.

\bibitem [4]{glaister} P. Glaister, Factorial sums,
\textit{Internat. J. Math. Educ. in Science and Technology} \textbf{34}
(2003), 250--257.

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\bibitem [6]{sury} B. Sury, T. Wang and F.-Z. Zhao, 
\href{http://www.cs.uwaterloo.ca/journals/JIS/Sury/sury99.pdf}{Identities
involving reciprocal of binomial coefficients} \textit{J. Integer Sequences}
\textbf{7} (2004), Article 04.2.8. 

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\href{http://www.research.att.com
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\end{thebibliography}



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\noindent 2000 {\it Mathematics Subject Classification}:
Primary 05A10; Secondary 26A33.

\noindent \emph{Keywords:}
parametric integrals, double factorials, combinatorics.

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\noindent (Concerned with sequences
\seqnum{A000984} and
\seqnum{A001147}.)

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\vspace*{+.1in}
\noindent
Received April 19 2005;
revised version received  September 29 2005.
Published in {\it Journal of Integer Sequences}, September 29 2005.

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