# Two Definite Integrals That Are Definitely (and Surprisingly!) Equal

To our good friend Gert Almkvist (1934–2018) In Memoriam.

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

## References

1. K. Alladi and M. L. Robinson. Legendre polynomials and irrationality. J. Reine Angew. Math. 318 (1980), 137–155.

2. Gert Almkvist and Doron Zeilberger. The method of differentiating under the integral sign. J. Symbolic Computation 10 (1990), 571–591.

3. W. N. Bailey. Generalized Hypergeometric Series. Cambridge University Press, 1935.

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