Problem 11966
(American Mathematical Monthly, Vol.124, March 2017) Proposed by C. I. V˘alean (Romania).
Prove that
Z 1
0
xln(1 + x)
1 + x2 dx= π2
96 +(ln(2))2 8 .
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Solution. Let
F(a) = Z 1
0
xln(1 + ax) 1 + x2 dx, then the derivative is
F′(a) = Z 1
0
x2
(1 + x2)(1 + ax)dx= 1 a2+ 1
Z 1
0
ax− 1 1 + x2 + 1
1 + ax
dx
= 1
a2+ 1
a
2ln(1 + x2) − arctan(x) +ln(1 + ax) a
1
0
= 1
a2+ 1
a
2ln(2) −π
4 +ln(1 + a) a
=
1 2ln(2)a
a2+ 1 −
π 4
a2+ 1 −
aln(1 + a)
a2+ 1 +ln(1 + a)
a .
Let
I:=
Z 1
0
xln(1 + x) 1 + x2 dx, then it follows that
I= F (1) = F (0) + Z 1
0
F′(a) da
= 0 + 1 2ln(2)
Z 1 0
a
a2+ 1da −π 4
Z 1 0
1
a2+ 1da − I + Z 1
0
ln(1 + a)
a da
= 1
2ln(2) ln(a2+ 1) 2
1
0
− π
4 [arctan(a)]10− I + π2 12
= −I + π2
48+(ln(2))2 4 , where we used the known fact
Z 1
0
ln(1 + a) a da=
∞
X
k=1
(−1)k+1 Z 1
0
ak−1 k da=
∞
X
k=1
(−1)k+1 k2 = π2
12. Therefore, we may conclude
I= π2
96 +(ln(2))2 8 .
