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(1)

Problem 12228

(American Mathematical Monthly, Vol.128, January 2021) Proposed by H. Grandmontagne (France).

Prove

Z 1 0

(ln(x))2ln 2√

x/(x2+ 1)

x2− 1 dx = 2G2, whereG is Catalan’s constantP

n=0(−1)n/(2n + 1)2.

Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.

Solution.

We consider, for x > 0, the digamma function ψ(x) and its derivatives of order m ≥ 1:

ψ(x) = Γ(x)

Γ(x) = −γ +

X

n=0

 1

n + 1− 1 n + x



, ψ(m)(x) = (−1)m+1m!

X

n=0

1 (n + x)m+1. We recall some known values of ψ(x) and ψ(m)(x) up to order 3 at x = 1/2, 1/4, 3/4:

ψ 1 2



= −γ − 2 ln(2), ψ 1 2



= π2

2 , ψ′′ 1 2



= −14ζ(3), ψ′′′ 1 2



= π4, ψ 1

4



= −γ − 3 ln(2) −π

2, ψ 1 4



= 8G + π2, ψ′′ 1 4



= −56ζ(3) − 2π3, ψ 3

4



= −γ − 3 ln(2) +π

2, ψ 3 4



= −8G + π2, ψ′′ 3 4



= −56ζ(3) + 2π3. Moreover

ψ′′′ 1 4



+ ψ′′′ 1 4



= 16π4. We need also the following two identities. Let Hk=Pk

n=1 1

n then, for x > 0, S2(x) :=

X

k=1

Hk

(k + x)2 = (γ + ψ(x)) ψ(x) −ψ′′(x)

2 , (1)

and

S3(x) :=

X

k=1

Hk

(k + x)3 = −(ψ(x))2

2 −(γ + ψ(x)) ψ′′(x)

2 +ψ′′′(x)

4 . (2)

Indeed, we have that S2(x) =

X

k=1

Hk

(k + x)2 =

X

k=1

1 (k + x)2

k

X

n=1

1 n =

X

n=1

1 n

X

k=n

1 (k + x)2 =

X

n=1

ψ(x + n) n

= −

X

n=1

1 n

Z 1 0

tx+n−1ln(t) 1 − t dt = −

Z 1 0

tx−1ln(t) 1 − t

X

n=1

tn n dt =

Z 1 0

tx−1ln(t) ln(1 − t) 1 − t dt

= lim

y→0+

2

∂x∂y Z 1

0

tx−1(1 − t)y−1dt = lim

y→0+

2

∂x∂y

Γ(x)Γ(y) Γ(x + y)

= lim

y→0+

∂y

 Γ(x)Γ(y)

Γ(x + y)(ψ(x) − ψ(x + y))



= lim

y→0+

Γ(x)Γ(y)

Γ(x + y)((ψ(y) − ψ(x + y))(ψ(x) − ψ(x + y)) − ψ(x + y))

= lim

y→0+

 1

y − γ + O(y)

 

−1

y − γ + O(y) − ψ(x)

 

−ψ(x)y −ψ′′(x)

2 y2+ O(y3)



− (ψ(x) + ψ′′(x)y + O(y2))

= (γ + ψ(x)) ψ(x) −ψ′′(x) 2 ,

(2)

and the proof of (1) is complete. We obtain (2) by differentiating (1) with respect to x.

The given integral can be written as Z 1

0

(ln(x))2ln 2√

x/(x2+ 1)

x2− 1 dx = ln(2)I1+I2

2 − I3= 2G2 where at the last step we applied

I1:=

Z 1 0

ln2(x)

x2− 1dx = −

X

k=0

Z 1 0

ln2(x)x2kdx = −2

X

k=0

1

(2k + 1)3 = −7ζ(3) 4 , I2:=

Z 1 0

ln3(x) x2− 1dx = 6

X

k=0

1

(2k + 1)4 = π4 16,

and, by using (2) with the given evaluations listed at the beginning,

I3:=

Z 1 0

ln2(x) ln(x2+ 1) x2− 1 dx = 2

X

k=0

Z 1 0

ln2(x)x2k

k

X

n=1

(−1)n n dx = 2

X

k=0

1 (2k + 1)3

k

X

n=1

(−1)n n

= −2

X

k=1

Hk

(2k + 1)3+ 2

X

k=1

Hk

(4k + 1)3+ 2

X

k=1

Hk

(4k + 3)3

= −1 4S3

 1 2

 + 1

32

 S3

 1 4

 + S3

 3 4



= −1 4



−π4

8 − 14 ln(2)ζ(3) +π4 4



+ 1 32



−(8G + π2)2

2 +(3 ln(2) + π2)(−56ζ(3) − 2π3)

2 −(−8G + π2)2

2 +(3 ln(2) −π2)(−56ζ(3) + 2π3)

2 + 4π4



= −2G24

32−7 ln(2)ζ(3)

4 .



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