Question

Given,
$\displaystyle\Omega(n)=\sum_{k=2}^{n-1} \int_{0}^{\frac{\pi}{4}} \frac{\sin ^{2 k+1} x \cos ^{k} x+\sin ^{k} x \cos ^{2 k+1} x}{\sin ^{3 k+3} x+\cos ^{3 k+3} x} d x$
Then find,
$\displaystyle\lim _{n \rightarrow \infty} \frac{\Omega(n)}{n}$

Answer 1

Let $\rm s_a$ = sin(ax) and $c_a$= cos(ax)

so the integrand is

$\rm \dfrac{{s_1}^{2k+1} {c_1}^k + {s_1}^k {c_1}^{2k+1}}{{s_1}^{3k+3} + {c_1}^{3k+3}}$

Factorize everything and recall the identities

2 s₁ c₁ = s₂

c₁² = (1 + c₂)/2

s₁² = (1 - c₂)/2

$\rm= \dfrac{{s_1}^k {c_1}^k \left({s_1}^{k+1} + {c_1}^{k+1}\right)}{\left({s_1}^{k+1} + {c_1}^{k+1}\right) \left({s_1}^{2k+2} - {s_1}^{k+1} {c_1}^{k+1} + {c_1}^{2k+2}\right)}$

$\rm= \dfrac{\left(s_1c_1\right)^k}{\left({s_1}^2\right)^{k+1} - \left(s_1 c_1\right)^{k+1} + \left({c_1}^2\right)^{k+1}}$

$\rm= \dfrac{\left(\frac{s_2}2\right)^k}{\left(\frac{1-c_2}2\right)^{k+1} - \left(\frac{s_2}2\right)^{k+1} + \left(\frac{1+c_2}2\right)^{k+1}}$

$\rm= 2 \times \dfrac{{s_2}^k}{(1-c_2)^{k+1} - {s_2}^{k+1} + (1+c_2)^{k+1}}$

After simplifying, substitute y = 2x. Then

$\begin{gathered}\displaystyle \rm \int_0^{\frac\pi4} \frac{\sin^{2k+1}(x) \cos^k(x) + \sin^k(x) \cos^{2k+1}(x)}{\sin^{3k+3}(x) + \cos^{3k+3}(x)} \, dx \\\\\\ \rm= 2 \int_0^{\frac\pi4} \frac{\sin^k(2x)}{(1-\cos(2x))^{k+1} - \sin^{k+1}(2x) + (1 + \cos(2x))^{k+1}} \, dx\\\\\\ \rm = \rm\int_0^{\frac\pi2} \frac{\sin^k(y)}{(1-\cos(y))^{k+1} - \sin^{k+1}(y) + (1+\cos(y))^{k+1}} \, dy\end{gathered}$

Now substitute t = tan(y/2). Under this change of variable, we have

dt = 1/2 sec²(y/2) dy ⇒ dy = 2/(1 + t²) dt

sin(y) = 2 sin(y/2) cos(y/2) = 2t/(1 + t²)

cos(y) = cos²(y/2) - sin²(y/2) = (1 - t²)/(1 + t²)

Making these replacements and simplifying the integrand reduces it significantly to

$\displaystyle \rm\int_0^1 \frac{t^k}{t^{2k+2} - t^{k+1} + 1} \,dt$

Substitute once more with z = tᵏ⁺¹ and dz = (k + 1) tᵏ dt to reduce it to the trivial

$\displaystyle \rm\frac1{k+1} \int_0^1 \frac{dz}{z^2-z+1} = \frac{2\pi}{3\sqrt3 (k+1)}$

Then Ω(n) is simply

$\displaystyle \rm \text{\O}mega(n) = \sum_{k=2}^{n-1} \frac{2\pi}{3\sqrt3(k+1)} = \frac{2\pi}{3\sqrt3} \sum_{k=3}^n \frac1k = \frac{2\pi}{3\sqrt3} \left(H_n - \frac32\right)$

where Hₙ denotes the n-th harmonic number,

$\rm H_n = \displaystyle \rm \sum_{k=1}^n \frac1k$

It's known that

$\displaystyle \rm\lim_{n\to\infty} (H_n - \ln(n))$=γ

where γ ≈ 0.577216 (the Euler-Mascheroni constant). Then

$\displaystyle \rm \lim_{n\to\infty} \frac{\text{\O}mega(n)}n = \frac{2\pi}{3\sqrt3} \lim_{n\to\infty} \frac{(H_n - \ln(n)) + \ln(n) - \frac32}{n} = \boxed{\frac{2\gamma\pi}{3\sqrt3}} \approx 0.687969$

Answer 2

Step-by-step explanation:

Now substitute t = tan(y/2). Under this change of variable, we have

dt = 1/2 sec²(y/2) dy ⇒ dy = 2/(1 + t²) dt

sin(y) = 2 sin(y/2) cos(y/2) = 2t/(1 + t²)

cos(y) = cos²(y/2) - sin²(y/2) = (1 - t²)/(1 + t²)

Making these replacements and simplifying the integrand reduces it significantly to