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Problem 11709

(American Mathematical Monthly, Vol.120, May 2013) Proposed by Moubinool Omarjee (France).

Find

I = Z +∞

x=0

1 x

Z x y=0

cos(x − y) − cos(x)

y dy dx.

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

We have that I =

Z +∞

x=0

cos(x) x

Z x y=0

cos(y) − 1 y



dy dx + Z +∞

x=0

sin(x) x

Z x y=0

sin(y) y dy

 dx

= Z +∞

y=0

1 − cos(y)

y Ci(y) dy + Z +∞

x=0

sin(x)

x Si(x) dx where

Ci(y) = − Z +∞

x=y

cos(x)

x dx and Si(x) = Z x

y=0

sin(y) y dy.

Now

Z +∞

x=0

sin(x)

x Si(x) dx = 1

2Si(x)2+∞

x=0=1 2

Z +∞

x=0

sin(x) x dx

2

= π2 8 . Moreover, for t > 0, by letting

F (t) = Z +∞

y=0

1 − cos(y)

y Ci(ty) dy, we obtain

Z +∞

y=0

1 − cos(y)

y Ci(ty) dy = F (1) = Z 1

+∞

F0(t) dt = Z 1

+∞

ln

 1 − 1

t2

 dt 2t = 1

4 Z 1

0

ln(1 − s)

s ds =π2 24 because

F0(t) = Z +∞

y=0

1 − cos(y)

y ·cos(ty) t dy

= 1 2t

Z +∞

y=0

2 cos(ty) − cos((t + 1)y) − cos((t − 1)y)

y dy

= 1 2t

Z +∞

y=0

cos(ty) − cos((t + 1)y)

y dy + 1

2t Z +∞

y=0

cos(ty) − cos((t − 1)y)

y dy

= 1

2tln t + 1 t

 + 1

2tln t − 1 t



= 1 2tln

 1 − 1

t2

 . Hence

I =π2 24+π2

8 =π2 6 .



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