SOLUTIONS

No problem is ever permanently closed. Any comments, new solutions, or new insights on old problems are always welcomed by the problem editor.

152. [2004, 130; 2005, 201–202] Proposed by Joe Flowers and Doug Martin (student), Texas Lutheran University, Seguin, Texas.

Let

F (s) = L[f(t)] = ∫_{0}^{∞} e^{−st}f(t) dt

denote the Laplace transform of f(t). Find L[sin^{n} bt], where b is any real constant and n is any non-negative integer.

Also solved by Kenneth B. Davenport, Dallas, Pennsylvania.

153. [2005, 52] Proposed by Joe Howard, Portales, New Mexico.

Let n ≥ 2 be an integer. Prove that

n^{n} > (n + 1)^{n−1} + \frac{n}{n + 1}.

Solution by Thomas P. Dence, Ashland University, Ashland, OH. The inequality is clearly true for n = 2. Since

\frac{n}{n + 1} < 1

and n^{n}, (n + 1)^{n−1} ∈ ℕ, then it suffices to show that

n^{n} ≥ (n + 1)^{n−1} + 1

or equivalently,

n^{n} > (n + 1)^{n−1}

for all n ≥ 3. But this is true if and only if

\frac{n^{n}}{(n + 1)^{n−1}} > 1

if and only if

n + 1 > (\frac{n + 1}{n})^{n} = (1 + \frac{1}{n})^{n}.

Expanding the latter binomial gives

(1 + \frac{1}{n})^{n} = 1 + 1 + \binom{n}{2}\frac{1}{n^{2}} + \binom{n}{3}\frac{1}{n^{3}} + ⋯ + \binom{n}{n}\frac{1}{n^{n}}

< 1 + 1 + \frac{1}{2!} \frac{n − 1}{n} + \frac{1}{3!} \frac{n − 1}{n} \frac{n − 2}{n} + ⋯ + \frac{1}{n!} \frac{n − 1}{n} \frac{n − 2}{n} ⋯ \frac{1}{n}

< ∑_{k=0}^{n} \frac{1}{k!} < 1 + ∑_{k=1}^{∞} \frac{1}{k!} < 1 + ∑_{k=0}^{∞} \frac{1}{2^{k}} = 3 < n + 1.

for all n ≥ 3.

Also solved by Kenneth B. Davenport, Dallas, Pennsylvania; Ovidiu Furdui, Western Michigan University, Kalamazoo, Michigan; Mohamed Akkouchi, Université Cadi Ayyad, Marrakech, Morocco (2 solutions); C. Wesley Nevans, Truman State University, Kirksville, Missouri (student); Huizeng Qin, Shandong University of Technology, Shandong, People’s Republic of China; Said Amghibech, Sainte Foy (qc), Canada; and the proposer.

154. [2005, 52] Proposed by Ovidiu Furdui, Western Michigan University, Kalamazoo, Michigan.

Let n ≥ 1 be an integer and m > 0, m < 2n. Prove that

∫_{0}^{∞}∫_{0}^{∞} ⋯ ∫_{0}^{∞} \frac{sin x_{1} sin x_{2} ⋯ sin x_{n}}{(x_{1} + x_{2} + ⋯ + x_{n})^{m}} dx_{1}dx_{2} ⋯ dx_{n} = \frac{1}{2Γ(m)}·B (\frac{m}{2}; n − \frac{m}{2}) ,

where Γ(·) is the Gamma function and B(·, ·) is the Beta function.

Solution by Said Amghibech, Sainte Foy (qc), Canada . We have

Γ(m) = ∫_{0}^{∞} t^{m−1} exp(−t)dt = a^{m} ∫_{0}^{∞} t^{m−1} exp(−at)dt

for all a > 0. By choosing a = x_{1} + x_{2} + ⋯ + x_{n} we get

Γ(m) ∫_{0}^{∞}∫_{0}^{∞} ⋯ ∫_{0}^{∞} \frac{sin x_{1} sin x_{2} ⋯ sin x_{n}}{(x_{1} + x_{2} + ⋯ + x_{n})^{m}} dx_{1}dx_{2} ⋯ dx_{n} =: I

and

I = ∫_{0}^{∞}(∫_{0}^{∞} sin x exp(−tx)dx)^{n}t^{m−1}dt = ∫_{0}^{∞} \frac{t^{m−1}}{(1 + t^{2})^{n}} dt.

By putting tan s = t we obtain

I = ∫_{0}^{π/2} cos^{2n−m−1} s sin^{m−1} sds.

By putting sin^{2} s = y we get

2I = ∫_{0}^{1} y^{\frac{m}{2} −1}(1 − y)^{n− \frac{m}{2} −1}dy = B (\frac{m}{2}, n − \frac{m}{2})

which gives the result.

Also solved by the proposer.

155. [2005, 53] Proposed by José Luis Díaz-Barrero, Universidad Politècnica de Cataluña, Barcelona, Spain.

Let P be any point inside △ABC and let A′ = AP ∩ BC, B′ = BP ∩ AC, and C′ = CP ∩ AB. Prove that

\frac{1}{AA′^{2}} + \frac{1}{BB′^{2}} + \frac{1}{CC′^{2}} ≥ \frac{4\sqrt{3}}{3}\sqrt{\frac{1}{AP^{4} + BP^{4} + CP^{4}}} .

Solution by Ovidiu Furdui, Western Michigan University, Kalamazoo, Michigan. First we show that

\frac{AP}{AA′} + \frac{BP}{BB′} + \frac{CP}{CC′} = 2.

To see this we notice that

\frac{AP}{AA′} = 1 − \frac{P A′}{AA′} = 1 − \frac{σ[BP C]}{σ[ABC]},

where σ[T] is the area of triangle T. Let E ∈ BC such that P E ⊥ BC and D ∈ BC such that AD ⊥ BC. Then △P EA′ ∼ △ADA′ implies that

\frac{P A′}{AA′} = \frac{P E}{AD} = \frac{P E · BC · \frac{1}{2}}{AD · BC · \frac{1}{2}} = \frac{σ[BP C]}{σ[ABC]}.

Therefore,

\begin{matrix}\frac{AP}{AA′} + \frac{BP}{BB′} + \frac{CP}{CC′} = 3 − \frac{P A′}{AA′} − \frac{P B′}{BB′} − \frac{P C′}{CC′} \\ = 3 − \frac{σ[BP C] + σ[P CA] + σ[AP B]}{σ[ABC]} = 3 − 1 = 2. \end{matrix}

Next we notice that

AP^{4} + BP^{4} + CP^{4} ≥ \frac{(AP^{2} + BP^{2} + CP^{2})^{2}}{3}

since

a^{2} + b^{2} + c^{2} ≥ \frac{(a + b + c)^{2}}{3} .

Therefore,

\begin{matrix}\frac{4\sqrt{3}}{3} · \sqrt{\frac{1}{AP^{4} + BP^{4} + CP^{4}}} ≤ \frac{4\sqrt{3}}{3} · \frac{\sqrt{3}}{AP^{2} + BP^{2} + CP^{2}} \\ = \frac{4}{AP^{2} + BP^{2} + CP^{2}} . \end{matrix}

Let us denote

AA′ = x; BB′ = y; CC′ = z; \frac{AP}{AA′} = α; \frac{BP}{BB′} = β; \frac{CP}{CC′} = γ.

We observe then that it suffices to show

\frac{4}{AP^{2} + BP^{2} + CP^{2}} ≤ \frac{1}{AA′^{2}} + \frac{1}{BB′^{2}} + \frac{1}{CC′^{2}} .

In view of the above notations this inequality is equivalent to

\frac{4}{α^{2}x^{2} + β^{2}y^{2} + γ^{2}z^{2}} ≤ \frac{1}{x^{2}} + \frac{1}{y^{2}} + \frac{1}{z^{2}}

We know that

\frac{1}{x^{2}} + \frac{1}{y^{2}} + \frac{1}{z^{2}} = \frac{α^{2}}{α^{2}x^{2}} + \frac{β^{2}}{β^{2}y^{2}} + \frac{γ^{2}}{γ^{2}z^{2}} .

In view of the Cauchy-Buniakowsky-Schwartz inequality we get that

(α^{2}x^{2} + β^{2}y^{2} + γ^{2}z^{2}) · (\frac{α^{2}}{α^{2}x^{2}} + \frac{β^{2}}{β^{2}y^{2}} + \frac{γ^{2}}{γ^{2}z^{2}} )

≥ (αx · \frac{1}{x} + βy · \frac{1}{y} + γz · \frac{1}{z})^{2}

= (α + β + γ)^{2} = 4.

Therefore,

\frac{1}{x^{2}} + \frac{1}{y^{2}} + \frac{1}{z^{2}} ≥ \frac{4}{α^{2}x^{2} + β^{2}y^{2} + γ^{2}z^{2}} .

Also solved by Huizeng Qin, Shandong University of Technology, Shandong, People’s Republic of China and the proposer.

156. [2005, 53] Proposed by Ovidiu Furdui, Western Michigan University, Kalamazoo, Michigan.

Prove that

I = ∫0^{\frac{π}{2}} cos^{2} x (ln^{2}(2 cos x) − ln(2 cos x)) dx = \frac{π^{3}}{48} − \frac{π}{4} .

Solution by Joe Howard, Portales, New Mexico. From [1],

∫0^{\frac{π}{2}} ln^{2}(2 sin θ)dθ = \frac{π^{3}}{24} .

By letting θ = \frac{π}{2} − x,

∫0^{\frac{π}{2}} ln^{2}(2 cos x)dx = \frac{π^{3}}{24} .  (1)

By L’Hopital’s Rule

lim_{x→ \frac{π}{2}} sin 2x ln^{k}(2 cos x) = 0

for k = 1, 2. Hence for k = 1, 2,

sin 2x ln^{k}(2 cos x)|0^{\frac{π}{2}} = 0.  (2)

Integrating by parts where u = x

∫0^{\frac{π}{2}} x · tan x · ln(2 cos x)dx = −\frac{x}{2} ln^{2}(2 cos x)|0^{\frac{π}{2}} + \frac{1}{2} ∫0^{\frac{π}{2}} ln^{2}(2 cos x)dx. (3)

Integrating by parts (u = ln^{2}(2 cos x)) and using (1), (2), and (3),

I_{1} = ∫0^{\frac{π}{2}} cos^{2} x ln^{2}(2 cos x)dx

= \frac{1}{4} sin 2x ln^{2}(2 cos x)|0^{\frac{π}{2}}

+ (\frac{1}{2}x ln^{2}(2 cos x)|0^{\frac{π}{2}} + ∫0^{\frac{π}{2}} x tan x ln(2 cos x)dx)

+ ∫0^{\frac{π}{2}} sin^{2} x ln(2 cos x)dx

= 0 + \frac{π^{3}}{48} + ∫0^{\frac{π}{2}} sin^{2} x ln(2 cos x)dx.

Integrating by parts (u = ln(2 cos x)) and using (2),

I_{2} = ∫0^{\frac{π}{2}} cos 2x ln(2 cos x)dx = \frac{1}{2} sin 2x ln(2 cos x)|0^{\frac{π}{2}}

+ ∫0^{\frac{π}{2}} sin^{2} xdx = 0 + \frac{π}{4} .

Finally,

I = I_{1} − ∫0^{\frac{π}{2}} cos^{2} x ln(2 cos x)dx

= \frac{π^{3}}{48} + ∫0^{\frac{π}{2}} (sin^{2} x − cos^{2} x) ln(2 cos x)dx

= \frac{π^{3}}{48} − I_{2} = \frac{π^{3}}{48} − \frac{π}{4} .

\underline{Reference}

1. F. Bowman, “Note on the Integral ∫0^{\frac{π}{2}} (ln sin θ)^{n}dθ,” Journal of the London Mathematical Society, 22 (1947), 172.

Also solved by Kenneth B. Davenport, Dallas, Pennsylvania; Huizeng Qin and Yousim Lu (jointly), Shandong University of Technology, Shandong, People’s Republic of China; and the proposer.