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UI Putnam Training Sessions, Advanced Level
Problem Set 12: Integrals, II Solutions
http://www.math.illinois.edu/contests.html
The Problems
Z
1.
π/2
ln(sin x) dx
0
Solution:
− 12 π ln 2 Let I be the given integral. Using the identities sin(π − x) = sin x and
sin x = 2 sin(x/2) cos(x/2), we have
Z
π/2
π/2
Z
ln sin x dx =
0
Z π
Z
2I =
ln sin x dx =
0
Z
π/2
Z
0
π/2
ln sin y dy + 2
= π ln 2 + 2
π
ln sin(π − y) dy =
ln sin x dx,
0
π/2
Z π
Z π
π
ln 2 dx +
ln sin(x/2) dx +
ln cos(x/2) dx
I=
0
Z
0
ln cos y dy
0
0
= π ln 2 + 2I1 + 2I2 ,
say. Setting y = π/2 − u and using the relation cos(π/2 − u) = sin u, we see that I2 = I1 = I. Hence
the above relation implies 2I = π ln 2 + 4I, and solving for I gives I = −(1/2)π ln 2, as claimed.
2. (B1, Putnam 1965) Evaluate
Z
1Z 1
lim
n→∞ 0
1
Z
cos2
...
0
0
nπ
o
(x1 + · · · + xn ) dx1 . . . dxn .
2n
(Hint: Integrate “backwards”.)
Solution:
1/2 Change variables yk = 1 − xk and note that this changes cos2 to sin2 . Adding the
two integrals gives an integral over cos2 (. . . ) + sin2 (. . . ) = 1, which is equal to 1. Hence the given
n-dimensional integrals are all 1/2, so the limit is 1/2 as well.
3. (A3, Putnam 1982) Evaluate
Z
0
∞
arctan(πx) − arctan(x)
dx.
x
(Hint: Write as limit of finite integrals
RT
0
and split up the finite integral into two.)
Solution: (π/2) ln π The given improper integral is the limit, as T → ∞, of the integrals over
the range 0 ≤ x ≤ T . Let I(T ) denotes the latter finite integral over [0, T ]. Splitting this integral
into a difference of two integrals
and making the change of variables y = πx Rin the first of these
R πT arctan(y)
πT
dy. Since arctan y → π/2 as y → ∞, and T (1/y) = ln π, we
integrals, we get I(T ) = T
y
get limT →∞ I(T ) = (π/2) ln π.
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UI Putnam Training
Problem Set 12: Integrals, II Solutions
4. Find the limit
n
X
lim
n→∞
i=1
Fall 2016
1
.
n+i
(Hint: Riemann sums!)
Solution: ln 2 Writing 1/(n + i) as (1/n)1/(1 + i/n), we see that the given sum is a Riemann
R1
sum for the integral 0 1/(1 + x)dx, corresponding to a partition of [0, 1] into n equal subintervals
of length 1/n each, and sample points 1/n, 2/n, . . . , 1. Its limit, as n → ∞, is therefore the value
of the integral, ln 2.
5. (B1, Putnam 1970) Find the limit
2n
1/n
1 Y 2
lim 4
n + i2
.
n→∞ n
i=1
(Hint: Riemann sums!)
Solution: exp(2 ln 5 − 4 + 2 arctan 2) Factoring out n2 from each term, the product becomes
2n
2n
Y
1/n
1 Y 2
2 1/n
P (n) = 4
=
.
n +i
1 + (i/n)2
n
i=1
i=1
Taking logarithms, we get
2n
1 X
log P (n) = 2 ·
log 1 +
2n
i=1
2 !
i
, = 2R(n),
n
R2
where R(n) is a Riemann sum for the integral 0 ln(1 + x2 )dx. The latter integral can be evaluated
by integrating by parts:
2 Z
Z 2
2
2x
ln(1 + x2 )dx = x ln(1 + x2 ) −
x
dx = ln 5 − 2 + arctan 5.
1 + x2
0
0
0
Hence
lim log P (n) = 2 lim R(n) = 2(ln 5 − 4 + 2 arctan 5).
n→∞
n→∞
Exponentiating this expression we get the limit of the product P (n).
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UI Putnam Training
Problem Set 12: Integrals, II Solutions
Fall 2016
Challenge Problem of the Week
The Broken Stick Problem: Pick two random points on the interval [0, 1], and
break up this interval at these two points to get three pieces.
(a) What is the probability that these pieces form a triangle?
(b) (Hard!) What is the probability that these pieces form an acute triangle?
Solution: (a) Answer: 1/4
This part can be solved by the following geometric argument: Let x and y denote the two breaking points.
Then choosing x and y randomly from [0, 1] is equivalent to choosing the point (x, y) randomly from the
unit square [0, 1] × [0, 1]. Let R denotes the set of such points (x, y) for which a triangle can be formed.
Then R is a region inside the unit square, and the probability that a triangle can be formed is equal to
the area of this region.
Thus it remains to determine the region R and its area. To this end note that the three pieces have
lengths x, y − x, 1 − y (if x ≤ y) or y, x − y, 1 − x (if x > y). They form a triangle if and only if all
three pieces are < 1/2. It is easy to check that these conditions hold if and only if (x, y) falls into one
of the triangles with vertices (0, 0.5), (0.5, 0.5), (0.5, 1) and (0.5, 0), (0.5, 0.5), (1, 0.5), respectively. The
combined area of these triangles is 2(1/8) = 1/4; this is the desired probability.
(b) Answer: −2 + 3 ln 2 .
We start out as before, considering the pair (x, y) as a random point in the unit square. By symmetry,
it suffices to consider the case x ≤ y. In this case, the lengths of the pieces, from left to right, are
x, y − x, 1 − y. A further symmetry argument shows that we may assume that the three pieces are in
increasing order, i.e., that x ≤ y − x ≤ 1 − y, or equivalently, (∗) 2x ≤ y ≤ (1 + x)/2.
Now note that a triangle with sides a, b, c is acute if and only if it satisfies a2 < b2 + c2 , b2 <
2
a + c2 , c2 < a2 + b2 . If a, b, c are in increasing order, then these three conditions simplify to the single
condition c2 < a2 + b2 . We apply this with the numbers x, y − x, 1 − y as a, b, c to conclude that under
our assumption (∗∗), an acute triangle can be formed if and only if (1 − y)2 ≤ x2 + (y − x)2 . The latter
condition simplifies to (∗∗) y ≥ (1 − 2x2 )/(2 − 2x).
Thus, the problem reduces to computing the area of the region inside the unit square determined by
the two conditions (∗) and (∗∗). This can be done by a routine, but somewhat tedious integration; the
result is −1/6 + (1/4) ln 2. Multiplying by 2 · 3! to compensate for the two symmetry reductions gives the
final answer, −2 + 3 ln 2 ≈ 0.0794 . . . .
Happy Problemsolving!
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