Skip to main content

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

Notes

  1. Availble online at http://sites.math.rutgers.edu/~zeilberg/tokhniot/EKHAD.txt.

  2. Available online at https://twitter.com/gregeganSF/status/1192309179119104000.

  3. See https://specfun.inria.fr/bostan/publications/EZZ.pdf and https://specfun.inria.fr/bostan/publications/EZZ2.pdf.

References

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

    MathSciNet  MATH  Google Scholar 

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

    MathSciNet  Article  Google Scholar 

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

Download references

Author information

Authors and Affiliations

Authors

Additional information

Publisher's Note

Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.

This column is a place for those bits of contagious mathematics that travel from person to person in the community, because they are so elegant, surprising, or appealing that one has an urge to pass them on.

Contributions are most welcome.

Submissions should be uploaded to http://tmin.edmgr.com or sent directly to Sergei Tabachnikov, e-mail: tabachni@math.psu.edu

Rights and permissions

Open Access This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. The images or other third party material in this article are included in the article's Creative Commons licence, unless indicated otherwise in a credit line to the material. If material is not included in the article's Creative Commons licence and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder. To view a copy of this licence, visit http://creativecommons.org/licenses/by/4.0/.

Reprints and Permissions

About this article

Verify currency and authenticity via CrossMark

Cite this article

Ekhad, S.B., Zeilberger, D. & Zudilin, W. Two Definite Integrals That Are Definitely (and Surprisingly!) Equal. Math Intelligencer 42, 10–11 (2020). https://doi.org/10.1007/s00283-020-09972-2

Download citation

  • Published:

  • Issue Date:

  • DOI: https://doi.org/10.1007/s00283-020-09972-2