%%ezz.tex: Two Definite Integrals That Are Surpisingly Equal
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{\bf  
Two Definite Integrals That Are Definitely (and Surprisingly!) Equal
}
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{\it Shalosh B. EKHAD, Doron ZEILBERGER and Wadim ZUDILIN}
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\qquad \qquad \qquad \qquad {\it To our good friend Gert Almkvist (1934-2018), In Memoriam}

{\bf Proposition}: For  real $a>b>0$  and non-negative integer $n$, the following beautiful and surprising identity holds.
$$
\int_0^1 \, \frac{x^n\,(1-x)^n}{ ((x+a)(x+b))^{n+1}} \, dx \, = \,
\int_0^1 \, \frac{x^n\,(1-x)^n}{ ((a-b)x +(a+1)b)^{n+1}} \, dx \quad .
$$

{\bf Proof}: Fix $a$ and $b$, and let $L(n)$ and $R(n)$ be the {\it integrals} on the left and right sides respectively,
and let $F_1(n,x)$, and $F_2(n,x)$ be the corresponding {\it integrands}, so that
$L(n)=\int_{0}^{1} F_1(n,x)\, dx$ and  $R(n)=\int_{0}^{1} F_2(n,x)\, dx$.
We cleverly construct the rational functions
$$
R_1(x) \, = \,
{\frac {x \left( -1+x \right)  \left( 2\,abx+a{x}^{2}+b{x}^{2}-ba+{x}^{2} \right) }{ \left( x+b \right)  \left( x+a \right) }} \, , \,
R_2(x) \, = \,
{\frac {x \left( -1+x \right)  \left( 2\,abx+a{x}^{2}-b{x}^{2}-ba+2\,xb-b \right) }{ba+xa-xb+b}} \, ,
$$
with the motive that (check!)
$$
 \left( n+1 \right) F_1(n,x) - \left( 2\,n+3 \right)  \left( 2\,ba+a+b \right) F_1 \left( n+1,x \right) 
+ \left( a-b \right) ^{2} \left( n+2 \right) F_1 \left( n+2 ,x\right) \, = \,
\frac{d}{dx} (R_1(x)F_1(n,x)) \quad ,
$$
and
$$
 \left( n+1 \right) F_2(n,x) - \left( 2\,n+3 \right)  \left( 2\,ba+a+b \right) F_2 \left( n+1,x \right) 
+ \left( a-b \right) ^{2} \left( n+2 \right) F_2 \left( n+2 ,x\right) \, = \,
\frac{d}{dx} (R_2(x)F_2(n,x)) \quad .
$$
Integrating both identities from $x=0$ to $x=1$, and noting that the right sides vanish, we have
$$
 \left( n+1 \right) L \left( n \right) - \left( 2\,n+3 \right)  \left( 2\,ba+a+b \right) L \left( n+1 \right) 
+ \left( a-b \right) ^{2} \left( n+2 \right) L \left( n+2\right) \, = \, 0 \quad ,
$$
$$
 \left( n+1 \right) R \left( n \right) - \left( 2\,n+3 \right)  \left( 2\,ba+a+b \right) R \left( n+1 \right) 
+ \left( a-b \right) ^{2} \left( n+2 \right) R \left( n+2\right) \, = \, 0 \quad .
$$
Since $L(0)=R(0)$ and $L(1)=R(1)$ (check!), the proposition follows by mathematical induction. \halmos

{\bf Comments};

{\bf 1}. This beatiful identity is equivalent to an identity buried in Bailey's classic book [B], section 9.5, formula (2), 
but you need an expert (like the third-named author) to realize that!

{\bf 2.} Our proof was obtained by the first named-author, running a Maple program, \hfill\break
{\tt http://sites.math.rutgers.edu/\~{}zeilberg/tokhniot/EKHAD.txt} , written by the second-named author, that
implements the Almkvist-Zeilberger algorithm [AZ] designed by Zeilberger and our good mutual friend Gert Almkvist,
to whose memory this note is dedicated.

{\bf References}

[AZ] Gert Almkvist and Doron Zeilberger,
{\it The method of differentiating under the
integral sign}, J. Symbolic Computation {\bf 10}(1990), 571-591; \hfill\break
{\tt http://www.math.rutgers.edu/\~{}zeilberg/mamarimY/duis.pdf} 


[B] W. N. Bailey, {\it ``Generalized hypergeometric series''}, Cambridge University Press, 1935.


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Shalosh B. Ekhad, Department of Mathematics, Rutgers University (New Brunswick), Hill Center-Busch Campus, 110 Frelinghuysen
Rd., Piscataway, NJ 08854-8019, USA. \hfill\break
Email: {\tt Shalosh B. Ekhad [at] gmail  [dot] com}   \quad .

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Doron Zeilberger, Department of Mathematics, Rutgers University (New Brunswick), Hill Center-Busch Campus, 110 Frelinghuysen
Rd., Piscataway, NJ 08854-8019, USA. \hfill\break
Email: {\tt DoronZeil [at] gmail  [dot] com}   \quad .


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Wadim Zudilin, Institute for Mathematics, Astrophysics and Particle Physics
Radboud Universiteit, PO Box 9010
6500 GL Nijmegen, The Netherlands
Email:  {\tt  wzudilin [at] gmail [dot] com} \quad .

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Written: Oct. 31, 2019.
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