Question

Given,


`\displaystyle\rm\Omega(n)=\sum_(k=2)^(n-1) \int_(0)^{(\pi)/(4)} (\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\rm\lim _(n \rightarrow \infty) (\Omega(n))/(n)`​

Answer 1

Let
`s_a = \sin(ax)` and
`c_a = \cos(ax)`, so the integrand is

`\frac{{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

`= \frac{{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)}`

`= \frac{\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)}`

`= \frac{\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)}`

`= 2 * \frac{{s_2}^k}{(1-c_2)^(k+1) - {s_2}^(k+1) + (1+c_2)^(k+1)}`

After simplifying, substitute y = 2x. Then

`\displaystyle \int_0^(\frac\pi4) (\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) (\sin^k(2x))/((1-\cos(2x))^(k+1) - \sin^(k+1)(2x) + (1 + \cos(2x))^(k+1)) \, dx\\\\\\ = \int_0^(\frac\pi2) (\sin^k(y))/((1-\cos(y))^(k+1) - \sin^(k+1)(y) + (1+\cos(y))^(k+1)) \, dy`

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 \int_0^1 (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 \frac1{k+1} \int_0^1 (dz)/(z^2-z+1) = (2\pi)/(3\sqrt3 (k+1))`

Then Ω(n) is simply

`\displaystyle \Omega(n) = \sum_(k=2)^(n-1) (2\pi)/(3\sqrt3(k+1)) = (2\pi)/(3\sqrt3) \sum_(k=3)^n \frac1k = (2\pi)/(3\sqrt3) \left(H_n - \frac32\right)`

where Hₙ denotes the n-th harmonic number,

`H_n = \displaystyle \sum_(k=1)^n \frac1k`

It's known that

`\displaystyle \lim_(n\to\infty) (H_n - \ln(n)) = \gamma`

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

`\displaystyle \lim_(n\to\infty) \frac{\Omega(n)}n = (2\pi)/(3\sqrt3) \lim_(n\to\infty) ((H_n - \ln(n)) + \ln(n) - \frac32)/(n) = \boxed{(2\gamma\pi)/(3\sqrt3)} \approx 0.687969`