Solutions: Practice Problems for Friday, October 24, 2014.

The William Lowell Putnam Mathematical Competition
October 31, 2014
1. Evaluate
Z
0
1
ln(x + 1)
dx.
x2 + 1
Solution: First solution: We make the substitution x = tan θ, rewriting the desired
integral as
Z π/4
log(tan(θ) + 1) dθ.
0
Write
log(tan(θ) + 1)
= log(sin(θ) + cos(θ)) − log(cos(θ))
√
and then note that sin(θ) + cos(θ) = 2 cos(π/4 − θ). We may thus rewrite the integrand
as
1
log(2) + log(cos(π/4 − θ)) − log(cos(θ)).
2
But over the interval [0, π/4], the integrals of log(cos(θ)) and log(cos(π/4−θ)) are equal,
so their contributions cancel out. The desired integral is then just the integral of 21 log(2)
over the interval [0, π/4], which is π log(2)/8.
Second solution: (by Roger Nelsen) Let I denote the desired integral. We make the
substitution x = (1 − u)/(1 + u) to obtain
1
(1 + u)2 log(2/(1 + u)) 2 du
2(1 + u2 )
(1 + u)2
0
Z 1
log(2) − log(1 + u)
=
du
1 + u2
0
Z 1
du
= log(2)
− I,
2
0 1+u
Z
I=
1
yielding
1
I = log(2)
2
1
Z
0
du
π log(2)
=
.
2
1+u
8
Third solution: (attributed to Steven Sivek) Define the function
Z 1
log(xt + 1)
dx
f (t) =
x2 + 1
0
so that f (0) = 0 and the desired integral is f (1). Then by differentiation under the integral,
Z 1
x
0
f (t) =
dx.
2
0 (xt + 1)(x + 1)
By partial fractions, we obtain
x=1
2t arctan(x) − 2 log(tx + 1) + log(x2 + 1) f (t) =
2(t2 + 1)
x=0
πt + 2 log(2) − 4 log(t + 1)
,
=
4(t2 + 1)
0
whence
log(2) arctan(t) π log(t2 + 1)
+
−
2
8
f (t) =
and hence
π log(2)
f (1) =
−
4
1
Z
0
Z
0
t
log(t + 1)
dt
t2 + 1
log(t + 1)
dt.
t2 + 1
But the integral on the right is again the desired integral f (1), so we may move it to the
left to obtain
π log(2)
2f (1) =
4
and hence f (1) = π log(2)/8 as desired.
Fourth solution: (by David Rusin) We have
Z
0
1
log(x + 1)
dx =
x2 + 1
Z
1
0
∞
X
(−1)n−1 xn
n=1
n(x2 + 1)
!
dx.
We next justify moving the sum through the integral sign. Note that
∞ Z
X
1
n=1 0
(−1)n−1 xn dx
n(x2 + 1)
Page 2
is an alternating series whose terms strictly decrease to zero, so it converges. Moreover,
its partial sums alternately bound the previous integral above and below, so the sum of the
series coincides with the integral.
Put
Z
1
xn dx
;
x2 + 1
Jn =
0
then J0 = arctan(1) =
π
4
and J1 =
1
2
log(2). Moreover,
Z 1
1
Jn + Jn+2 =
xn dx =
.
n+1
0
Write
Am =
Bm =
m
X
(−1)i−1
i=1
m
X
i=1
2i − 1
(−1)i−1
;
2i
then
J2n = (−1)n (J0 − An )
J2n+1 = (−1)n (J1 − Bn ).
Now the 2N -th partial sum of our series equals
N
X
J2n
J2n−1
−
2n − 1
2n
n=1
=
N
X
(−1)n−1
n=1
2n − 1
(J1 − Bn−1 ) −
(−1)n
(J0 − An )
2n
= AN (J1 − BN −1 ) + BN (J0 − AN ) + AN BN .
As N → ∞, AN → J0 and BN → J1 , so the sum tends to J0 J1 = π log(2)/8.
Fifth solution: (suggested by Alin Bostan) Note that
Z 1
x dy
log(1 + x) =
,
1
+ xy
0
so the desired integral I may be written as
Z 1Z 1
x dy dx
I=
.
2
0
0 (1 + xy)(1 + x )
Page 3
We may interchange x and y in this expression, then use Fubini’s theorem to interchange
the order of summation, to obtain
Z 1Z 1
y dy dx
I=
.
2
0
0 (1 + xy)(1 + y )
We then add these expressions to obtain
Z 1Z 1
y
dy dx
x
+
2I =
2
2
1+x
1+y
1 + xy
0
0
Z 1Z 1
2
2
x + y + xy + x y dy dx
=
(1 + x2 )(1 + y 2 ) 1 + xy
0
0
Z 1Z 1
(x + y) dy dx
=
.
(1
+ x2 )(1 + y 2 )
0
0
By another symmetry argument, we have
Z 1Z 1
2I = 2
0
so
Z
I=
0
1
x dx
1 + x2
x dy dx
,
(1 + x2 )(1 + y 2 )
0
Z
1
0
1
1 + y2
= log(2) ·
π
.
8
Remarks: The first two solutions are related by the fact that if x = tan(θ), then 1 −
x/(1 + x) = tan(π/4 − θ). The strategy of the third solution (introducing a parameter
then differentiating it) was a favorite of physics Nobelist (andR Putnam Fellow) Richard
∞
Feynman. The fifth solution resembles Gauss’s evaluation of −∞ exp(−x2 ) dx. Noam
Elkies notes that this integral is number 2.491#8 in Gradshteyn and Ryzhik, Table of integrals, series, and products. The Mathematica computer algebra system (version 5.2)
successfully computes this integral, but we do not know how.
2. Find a nonzero polynomial P (x, y) such that P (bac, b2ac) = 0 for all real numbers a. (Note:
bνc is the greatest integer less than or equal to ν.)
Solution: Take P (x, y) = (y − 2x)(y − 2x − 1). To see that this works, first note that
if m = bac, then 2m is an integer less than or equal to 2a, so 2m ≤ b2ac. On the other
hand, m + 1 is an integer strictly greater than a, so 2m + 2 is an integer strictly greater
than 2a, so b2ac ≤ 2m + 1.
Page 4

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