Let [tex]s_a = \sin(ax)[/tex] and [tex]c_a = \cos(ax)[/tex], so the integrand is
[tex]\dfrac{{s_1}^{2k+1} {c_1}^k + {s_1}^k {c_1}^{2k+1}}{{s_1}^{3k+3} + {c_1}^{3k+3}}[/tex]
Factorize everything and recall the identities
2 s₁ c₁ = s₂
c₁² = (1 + c₂)/2
s₁² = (1 - c₂)/2
[tex]= \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)}[/tex]
[tex]= \dfrac{\left(s_1c_1\right)^k}{\left({s_1}^2\right)^{k+1} - \left(s_1 c_1)^{k+1} + \left({c_1}^2\right)^{k+1}}[/tex]
[tex]= \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}}[/tex]
[tex]= 2 \times \dfrac{{s_2}^k}{(1-c_2)^{k+1} - {s_2}^{k+1} + (1+c_2)^{k+1}}[/tex]
After simplifying, substitute y = 2x. Then
[tex]\displaystyle \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 \\\\\\ = 2 \int_0^{\frac\pi4} \frac{\sin^k(2x)}{(1-\cos(2x))^{k+1} - \sin^{k+1}(2x) + (1 + \cos(2x))^{k+1}} \, dx\\\\\\ = \int_0^{\frac\pi2} \frac{\sin^k(y)}{(1-\cos(y))^{k+1} - \sin^{k+1}(y) + (1+\cos(y))^{k+1}} \, dy[/tex]
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
[tex]\displaystyle \int_0^1 \frac{t^k}{t^{2k+2} - t^{k+1} + 1} \, dt[/tex]
Substitute once more with z = tᵏ⁺¹ and dz = (k + 1) tᵏ dt to reduce it to the trivial
[tex]\displaystyle \frac1{k+1} \int_0^1 \frac{dz}{z^2-z+1} = \frac{2\pi}{3\sqrt3 (k+1)}[/tex]
Then Ω(n) is simply
[tex]\displaystyle \Omega(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)[/tex]
where Hₙ denotes the n-th harmonic number,
[tex]H_n = \displaystyle \sum_{k=1}^n \frac1k[/tex]
It's known that
[tex]\displaystyle \lim_{n\to\infty} (H_n - \ln(n)) = \gamma[/tex]
where γ ≈ 0.577216 (the Euler-Mascheroni constant). Then
[tex]\displaystyle \lim_{n\to\infty} \frac{\Omega(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[/tex]
