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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. 1 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). 2 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! 3