If $$\Omega_n=\sum \limits_{k=1}^n \left(\int \limits_{-\frac{1}{k}}^{\frac{1}{k}}(2x^{10} + 3x^8 + 1)\cos^{-1}(kx)dx\right)$$Then find $$\Omega=\lim \limits_{n\to \infty}\left(\Omega_n-\pi H_n\right)$$
1. **Change of Variable:**
   We start by changing the variable in the integral. Let \( t = kx \). Then \( dt = k \, dx \) and the limits of integration change from \( x = -\frac{1}{k} \) to \( x = \frac{1}{k} \) to \( t = -1 \) to \( t = 1 \). The integral becomes:
   \[
   \int_{-\frac{1}{k}}^{\frac{1}{k}} (2x^{10} + 3x^8 + 1) \cos^{-1}(kx) \, dx = \int_{-1}^{1} \left(2 \left(\frac{t}{k}\right)^{10} + 3 \left(\frac{t}{k}\right)^{8} + 1\right) \cos^{-1}(t) \frac{dt}{k}
   \]
   Simplifying the integrand, we get:
   \[
   \int_{-1}^{1} \left(\frac{2t^{10}}{k^{11}} + \frac{3t^8}{k^9} + \frac{1}{k}\right) \cos^{-1}(t) \, dt
   \]

2. **Evaluating the Integral:**
   We can split the integral into three separate integrals:
   \[
   \int_{-1}^{1} \left(\frac{2t^{10}}{k^{11}} + \frac{3t^8}{k^9} + \frac{1}{k}\right) \cos^{-1}(t) \, dt = \frac{2}{k^{11}} \int_{-1}^{1} t^{10} \cos^{-1}(t) \, dt + \frac{3}{k^9} \int_{-1}^{1} t^8 \cos^{-1}(t) \, dt + \frac{1}{k} \int_{-1}^{1} \cos^{-1}(t) \, dt
   \]

3. **Using Known Integrals:**
   The integrals \(\int_{-1}^{1} t^{10} \cos^{-1}(t) \, dt\), \(\int_{-1}^{1} t^8 \cos^{-1}(t) \, dt\), and \(\int_{-1}^{1} \cos^{-1}(t) \, dt\) are known and can be evaluated using integration by parts or known results:
   \[
   \int_{-1}^{1} t^{10} \cos^{-1}(t) \, dt = \frac{\pi}{11}, \quad \int_{-1}^{1} t^8 \cos^{-1}(t) \, dt = \frac{\pi}{9}, \quad \int_{-1}^{1} \cos^{-1}(t) \, dt = \pi
   \]

4. **Substituting Back:**
   Substituting these results back into the integrals, we get:
   \[
   \frac{2}{k^{11}} \cdot \frac{\pi}{11} + \frac{3}{k^9} \cdot \frac{\pi}{9} + \frac{1}{k} \cdot \pi = \frac{2\pi}{11k^{11}} + \frac{3\pi}{9k^9} + \frac{\pi}{k}
   \]

5. **Summing Over \( k \):**
   Now, we sum this expression over \( k \) from 1 to \( n \):
   \[
   \Omega_n = \sum_{k=1}^n \left( \frac{2\pi}{11k^{11}} + \frac{3\pi}{9k^9} + \frac{\pi}{k} \right)
   \]
   This can be written as:
   \[
   \Omega_n = \pi \sum_{k=1}^n \left( \frac{2}{11k^{11}} + \frac{3}{9k^9} + \frac{1}{k} \right)
   \]

6. **Separating the Harmonic Series:**
   The term \(\sum_{k=1}^n \frac{1}{k}\) is the \( n \)-th Harmonic number \( H_n \). Therefore, we can write:
   \[
   \Omega_n = \pi \left( \sum_{k=1}^n \frac{2}{11k^{11}} + \sum_{k=1}^n \frac{3}{9k^9} + H_n \right)
   \]

7. **Taking the Limit:**
   We are interested in the limit:
   \[
   \Omega = \lim_{n \to \infty} \left( \Omega_n - \pi H_n \right)
   \]
   Substituting \(\Omega_n\) from above, we get:
   \[
   \Omega = \lim_{n \to \infty} \left( \pi \left( \sum_{k=1}^n \frac{2}{11k^{11}} + \sum_{k=1}^n \frac{3}{9k^9} + H_n \right) - \pi H_n \right)
   \]
   Simplifying, we have:
   \[
   \Omega = \pi \lim_{n \to \infty} \left( \sum_{k=1}^n \frac{2}{11k^{11}} + \sum_{k=1}^n \frac{3}{9k^9} \right)
   \]

8. **Using Zeta Function:**
   The sums \(\sum_{k=1}^\infty \frac{1}{k^{11}}\) and \(\sum_{k=1}^\infty \frac{1}{k^9}\) are known to be the Riemann zeta functions \(\zeta(11)\) and \(\zeta(9)\), respectively. Therefore:
   \[
   \Omega = \pi \left( \frac{2}{11} \zeta(11) + \frac{3}{9} \zeta(9) \right)
   \]

The final answer is \(\boxed{\pi \left( \frac{2}{11} \zeta(11) + \frac{3}{9} \zeta(9) \right)}\)