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Prove Z 1 0 Z 1 0  x 1 − xy k dx dy = Z 1 0  1 x k dx

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(1)

Problem 12031

(American Mathematical Monthly, Vol.125, March 2018) Proposed by O. Furdui (Romania).

(a) Prove

Z 1 0

Z 1 0

 x

1 − xy



dx dy = 1 − γ, where {a} denotes the fractional part of a, and γ is Euler’s constant.

(b) Let k be a nonnegative integer. Prove Z 1

0

Z 1 0

 x

1 − xy

k

dx dy = Z 1

0

 1 x

k

dx.

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

Solution. (a) follows from (b). For k = 1, by letting t = 1/x we have, Z 1

0

Z 1 0

 x

1 − xy



dx dy = Z 1

0

 1 x

 dx =

Z

1

{t}

t2 dt =

X

n=1

Z n+1 n

t − n t2 dt

= lim

N →∞

N −1

X

n=1



ln(n + 1) − ln(n) − 1 n + 1



= lim

N →∞(ln(N ) − HN + 1) = 1 − γ where HN :=PN

n=11/n is the N -th harmonic number.

As regards (b), let t = 1x− y then x = t+y1 ,

∂(x,y)

∂(t,y)

=

1 (t+y)2, Z 1

0

Z 1 0

 x

1 − xy

k

dx dy = Z 1

t=0

Z 1 y=1−t

 1 t

k

dt dy (y + t)2 +

Z

t=1

Z 1 y=0

 1 t

k

dt dy (y + t)2

= Z 1

0

 1 t

k 1 − 1

t + 1

 dt +

Z

1

1 tk

 1 t − 1

t + 1

 dt

= Z 1

0

 1 t

k

dt − Z 1

0

 1 t

k

dt t + 1+

Z

1

dt tk+1(t + 1)

= Z 1

0

 1 t

k

dt where the last equality holds because by letting s = 1/t,

Z 1 0

 1 t

k

dt t + 1 =

Z

1

{s}k

(1/s + 1)s2ds =

X

n=1

Z n+1 n

(s − n)k s(s + 1)ds

= lim

N →∞

N −1

X

n=1

Z 1 0

sk

 1

s + n− 1 s + 1 + n

 ds

= lim

N →∞

Z 1 0

sk

N −1

X

n=1

 1

s + n− 1 s + 1 + n

 ds

= lim

N →∞

Z 1 0

sk

 1

s + 1− 1 s + N



= Z 1

0

sk s + 1ds

= Z

1

(1/t)k (1/t + 1)t2dt =

Z

1

dt tk+1(t + 1).



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