\documentclass[12pt,reqno]{article} \usepackage[usenames]{color} \usepackage{amssymb} \usepackage{graphicx} \usepackage{amscd} \usepackage[colorlinks=true, linkcolor=webgreen, filecolor=webbrown, citecolor=webgreen]{hyperref} \definecolor{webgreen}{rgb}{0,.5,0} \definecolor{webbrown}{rgb}{.6,0,0} \usepackage{color} \usepackage{fullpage} \usepackage{float} \usepackage{psfig} \usepackage{graphics,amsmath,amssymb} \usepackage{amsfonts} \usepackage{latexsym} \usepackage{epsf} \setlength{\textwidth}{6.5in} \setlength{\oddsidemargin}{.1in} \setlength{\evensidemargin}{.1in} \setlength{\topmargin}{-.5in} \setlength{\textheight}{8.9in} \newcommand{\seqnum}[1]{\href{http://www.research.att.com/cgi-bin/access.cgi/as/~njas/sequences/eisA.cgi?Anum=#1}{\underline{#1}}} \begin{document} \begin{center} \epsfxsize=4in \leavevmode\epsffile{logo129.eps} \end{center} \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} \\ \end{center} \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, $\ln x <0$ 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} &=-\; \frac{q}{p+1} \; I_{p,q-1}\\ \quad &=\left(-\; \frac{q}{p+1}\right) \; \left( - \; \frac {q-1}{p+1} \right) \; I_{p,q-2}\\ \quad &=\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 &= \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 &= \frac 12 \; e^2 - \frac {n}{2} \; \left( \frac {1}{2} e^2 - \frac {n-1}{2} \; I_{n-2} \right)\\ \quad &= \frac 12 \; e^2 \; \left( 1-\frac n2 \right) + \frac {n(n-1)}{4} \; I_{n-2}\\ \quad &= \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 &= \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 & = \dots \\ \quad &= \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 \\ &= \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 & \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 & =\int_0^1 \frac {1}{(x^2+x+1)^n}\; dx, \\ J_n & =\int_0^1 \frac {x}{(x^2+x+1)^n}\; dx, \\ K_n & =\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 &= \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 &=\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}} &= -\frac {x}{(x^2+x+1)^{n+1}} - \frac {2}{(x^2+x+1)^{n+1}} + \frac {2}{(x^2+x+1)^n}\\ \quad &= -\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 &= -\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 &=\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} &= \frac {2(2n-1)}{3n} \; b_n = \frac 23 \; \cdot \; \frac {2n-1}{n} \; b_n\\ \quad &=\left(\frac 23 \right)^2 \; \cdot \; \frac {2n-1}{n} \; \cdot \;\frac {2n-3}{n-1} \; b_{n-1}\\ \quad &=\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 &=\dots\\ \quad &=\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} &=\frac {1}{3n} \left( \frac {1}{3^{n-1}} -1 \right) + \frac {2(2n-1)}{3n} \; a_n\\ \quad &= \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 &=\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 &=\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 & \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 &= \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}& =\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}& \quad \\ b_n &= \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 &=\int_0^1 \frac {1+2x}{(x^2+x+1)^n}\; dx \\ \quad & = \frac{1}{1-n} \left[\frac {1}{(x^2+x+1)^{n+1}} \right]_0^1 \\ \quad &= \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} \bibitem [1]{explicit closed form} Th. Dana-Picard, Explicit closed forms for parametric integrals, \textit{Internat. J. Math. Educ. in Science and Technology} \textbf{35} (2004), 456--467. \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. \bibitem [5]{jm} H. Johnston and J. Mathews, \textit{Calculus}, Addison-Wesley, 2002. \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. \bibitem [7]{sloane} N. J. A. Sloane, \href{http://www.research.att.com /cgi-bin/access.cgi/as/njas}{The On-Line Encyclopedia of Integer Sequences}. \bibitem [8]{numbers} I. M. Vinogradov, \textit{Elements of Number Theory}, 5th rev. ed., Dover, New York, 1954. \bibitem [9]{mathworld double factorial} E. W. Weisstein, \href{http://mathworld.wolfram.com/DoubleFactorial.html}{\textit{Double Factorial}}, MathWorld. \bibitem [10]{wunsch} A. David Wunsch, Complex Variables with Applications, 2nd edition, Addison-Wesley, Reading, Massachussetts, 1994. \end{thebibliography} \bigskip \hrule \bigskip \noindent 2000 {\it Mathematics Subject Classification}: Primary 05A10; Secondary 26A33. \noindent \emph{Keywords:} parametric integrals, double factorials, combinatorics. \bigskip \hrule \bigskip \noindent (Concerned with sequences \seqnum{A000984} and \seqnum{A001147}.) \bigskip \hrule \bigskip \vspace*{+.1in} \noindent Received April 19 2005; revised version received September 29 2005. Published in {\it Journal of Integer Sequences}, September 29 2005. \bigskip \hrule \bigskip \noindent Return to \htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}. \vskip .1in \end{document}