# Calculate Z ∞ 0 Z ∞ 0 sin(x) sin(y) sin(x + y) xy(x + y) dx dy

## Testo completo

(1)

Problem 11953

(American Mathematical Monthly, Vol.124, January 2017) Proposed by C. I. V˘alean (Romania).

Calculate

Z

0

Z

0

sin(x) sin(y) sin(x + y) xy(x + y) dx dy.

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

Solution. We have that I :=

Z

0

Z

0

sin(x) sin(y) sin(x + y)

xy(x + y) dx dy = lim

R→+∞

Z R x=0

Z R y=0

sin(x) sin(y) sin(x + y) xy(x + y) dx dy

= lim

R→+∞2 Z R

x=0

Z x y=0

sin(x) sin(y) sin(x + y) xy(x + y) dy

 dx

= lim

R→+∞

1 2

Z R x=0

Z x y=0

sin(2x) + sin(2y) − sin(2x + 2y)

xy(x + y) dy

 dx.

Then, after letting u = 2x, v = 2y = tu,

I = lim

R→+∞

Z R u=0

Z u v=0

sin(u) + sin(v) − sin(u + v)

uv(u + v) dv

 du

= lim

R→+∞

Z R u=0

Z 1 t=0

sin(u) + sin(tu) − sin(u + tu) t(1 + t)u2 dt



du = lim

R→+∞

Z 1

t=0

fR(t) t(1 + t)dt, where

fR(t) :=

Z R u=0

sin(u) + sin(tu) − sin(u + tu)

u2 du =



−sin(u) + sin(tu) − sin(u + tu) u

R u=0

+ Z R

u=0

cos(u) − cos((1 + t)u)

u du + t

Z R u=0

cos(tu) − cos((1 + t)u)

u du.

By using the Frullani’s integral, Z +∞

0

cos(au) − cos(bu)

u dt = ln(b/a) for a, b > 0, it follows that

R→+∞lim fR(t) = f (t) := ln(1 + t) + t ln((1 + t)/t) = (1 + t) ln(1 + t) − t ln(t).

Note that for any t ∈ [0, 1],

|fR(t) − f (t)| ≤ Z

R

3

u2du = 3 R. Hence, by the Dominated Convergence Theorem, we finally obtain

I = Z 1

0

f (t) t(1 + t)dt =

Z 1

0

ln(1 + t)

t dt − [ln(t) ln(1 + t)]10+ Z 1

0

ln(1 + t) t dt

= 2 Z 1

0

ln(1 + t) t dt = 2

X

n=1

(−1)n+1 n

Z 1

0

tn−1dt = 2

X

n=1

(−1)n+1

n2 = ζ(2) = π2 6 .

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