Problem 11903
(American Mathematical Monthly, Vol.123, April 2016) Proposed by P. Perfetti (Italy).
Find a homogeneous polynomialP of degree two in a, b, c, and d such that 0 < a < b < c and d > 0,
Z a 0
px(a − x)(x − b)(x − c)
x + d dx =
Z c b
px(a − x)(x − b)(x − c)
x + d dx
if and only ifP (a, b, c, d) =pd(a + d)(b + d)(c + d).
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
We will show that
P (a, b, c, d) = −a2+ b2+ c2
8 + d2+ab + ac + bc
4 +(a + b + c)d
2 .
Let
f (z) = pz(a − z)(z − b)(z − c) z + d
which is, meromorphic on the entire complex plane C except the intervals [0, a] and [b, c] (here the square root√
· is the branch on C minus the positive real axis with√
−1 = i).
Let γ be the clockwise closed curve γ given by
−d 0 a b c
We have that
Z
γ
f (z)dz = 2 Z a
0
f (x)dx − 2 Z c
b
f (x)dx.
Moreover, by the Residue Theorem Z
γ
f (z)dz = 2πi (Res(f, −d) + Res(f, ∞)) . Now
Res(f, −d) =p
−d(a + d)(−d − b)(−d − c) = ipd(a + d)(b + d)(c + d).
and
Res(f, ∞) = Res
−f (1/z) z2 , 0
= Res
−f (1/z) z2 , 0
= −iRes
p(1 − az)(1 − bz)(1 − cz) z3(1 + dz) , 0
!
= −i[z2]p(1 − az)(1 − bz)(1 − cz) (1 + dz)
= −i[z2]
1 −az
2 −a2z2 8
1 −bz
2 −b2z2 8
1 − cz
2 −c2z2 8
1 − dz + d2z2
= −iP (a, b, c, d).
Hence
Z a 0
f (x) dx − Z c
b
f (x) dx = π
−pd(a + d)(b + d)(c + d) + P (a, b, c, d) which implies that the two integrals on the left are equal if and only if
P (a, b, c, d) =pd(a + d)(b + d)(c + d).
