FormalPara Proposition 1.

For real \(a>b>0\)and nonnegative integer n, the following beautiful and surprising identity holds:

$$ \int _0^1 \frac{x^n (1-x)^n}{ ((x+a)(x+b))^{n+1}} \, dx = \int _0^1 \frac{x^n (1-x)^n}{ ((a-b)x +(a+1)b)^{n+1}} \, dx.$$
FormalPara Proof.

Fix a and b, let L(n) and R(n) be the integrals on the left and right sides respectively, and let \(F_1(n,x)\) and \(F_2(n,x)\) be the corresponding 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

$$ R_1(x) = {\frac{x \left( x-1 \right)((a+b+1) x^2+ 2abx -ab) }{ \left( x+b \right) \left( x+a \right) }}$$

and

$$R_2(x) = {\frac{x \left( x-1 \right) \bigl ( (a-b)x^2+2b(a+1)x-(a+1)b \bigr ) }{(a-b)x - (a+1)b}}$$

motivated by the fact that (check!)

$$\left( n+1 \right) F_1(n,x) - \left( 2 n+3 \right) \left( 2 ba+a+b \right) F_1 \left( n+1,x \right) + \left( a-b \right) ^{2} \left( n+2 \right) F_1 \left( n+2 ,x\right) = \frac{d}{dx} (R_1(x)F_1(n,x))$$

and

$$\left( n+1 \right) F_2(n,x) - \left( 2 n+3 \right) \left( 2 ba+a+b \right) F_2 \left( n+1,x \right) + \left( a-b \right) ^{2} \left( n+2 \right) F_2 \left( n+2 ,x\right) = \frac{d}{dx} (R_2(x)F_2(n,x)).$$

Integrating both identities from \(x=0\) to \(x=1\) and noting that the right-hand sides vanish, we have

$$\begin{aligned}&\left(n+1\right) L \left(n\right) - \left(2n+3\right) \left(2ba+a+b\right) L \left(n+1\right) \\ &\quad +\left(a-b\right)^{2} \left(n+2\right) L \left(n+2\right) = 0\end{aligned}$$

and

$$ \begin{aligned}&\left( n+1 \right) R \left( n \right) - \left( 2 n+3 \right) \left( 2 ba+a+b \right) R \left( n+1 \right) \\ &\quad + \left( a-b \right) ^{2} \left( n+2 \right) R \left( n+2\right) = 0.\end{aligned}$$

Since \(L(0)=R(0)\) and \(L(1)=R(1)\) (check!), the proposition follows by mathematical induction. \(\square \)

FormalPara Remark 1.

This beautiful identity is equivalent to an identity buried in Bailey’s classic book [3, Section 9.5, formula (2)], but you need an expert (like the third-named author) to realize that!

FormalPara Remark 2.

Our proof was obtained by the first-named author by running a Maple programFootnote 1 written by the second-named author that implements the Almkvist–Zeilberger algorithm [2] designed by Zeilberger and our good mutual friend Gert Almkvist, to whose memory this note is dedicated.

FormalPara Remark 3.

Our integrals are not taken from a pool of no-one-cares-about-them analytic creatures: the right-hand side covers a famous sequence of rational approximations to \(\log (1+(a-b)/((a+1)b))\) [1], and hence the left-hand side does, too.

FormalPara Remark 4.

We thank Greg Egan for spotting a sign error in an earlier version.

FormalPara Remark 5.

Watch Greg Egan’s beautiful animation.Footnote 2

FormalPara Remark 6.

To our surprise, the identity turned out to be not as surprising as we had believed. Mikael Sundquist noticed that the change of variable \(x=b(1-u)/(b+u)\) gives a “Calculus 1 proof.” Indeed,

$$dx=-\frac{b(1+b)}{(b+u)^2}\, du,$$

and we have

$$\int _0^1 \frac{x^n (1-x)^n}{ ((x+a)(x+b))^{n+1} } \, dx = \int _0^1 \frac{(b(1-u)/(b+u))^n (1- (b(1-u)/(b+u)))^n}{ (b(1-u)/(b+u)+a)(b(1-u)/(b+u)+b))^{n+1} } \frac{b(1+b)}{(b+u)^2}\, du = \int _0^1 \frac{(1-u)^n u^n b^{n+1} (1+b)^{n+1}}{( b(1-u)+a(b+u))(b(1-u)+b(b+u)) )^{n+1}} \, du = \int _0^1 \frac{u^n(1-u)^n}{((a-b)u+(a+1)b)^{n+1}} \, du.$$

Note that n does not have to be an integer.

FormalPara Remark 7.

Alin Bostan has two further insightful proofs.Footnote 3