# a basic definite integration and its result used in evaluating limit

Question :

$$\mathbf\Omega(n)=\displaystyle\int _{0}^{2\pi}\log(n^2-2n\cos t+1)dt\ ,\ n\geq1$$

then , find :

$$\displaystyle \lim_{n \to \infty} \left(1+\dfrac{\mathbf \Omega(n)}{4\pi}\right)^{\log(n+1)}$$

my attempt:

$$\mathbf\Omega(n)=\displaystyle\int _{0}^{2\pi}\log(n-e^{it})dt+\displaystyle\int _{0}^{2\pi}\log(n-e^{-it})dt\$$

$$\mathbf\Omega(n)=2\displaystyle\int _{0}^{2\pi} \log\ n \ dt+\displaystyle \int_{0}^{2\pi}\log\left(1-\dfrac{e^{it}}{n}\right)dt+\displaystyle \int_{0}^{2\pi}\log\left(1-\dfrac{e^{-it}}{n}\right)dt$$

$$\mathbf\Omega(n)=2\pi \log\ n-\displaystyle \int_{0}^{2\pi}\displaystyle \sum_{k=1}^{\infty}\ \dfrac{e^{ikt}}{k\ n^k}dt-\displaystyle \int_{0}^{2\pi}\displaystyle \sum_{k=1}^{\infty}\ \dfrac{e^{-ikt}}{k\ n^k}dt$$

changing order of summation and integration

$$\mathbf\Omega(n)=2\pi \log\ n -\displaystyle \sum_{k=1}^{\infty}\dfrac{1}{kn^k}\displaystyle \int_{0}^{2\pi}e^{ikt}dt\ - \displaystyle \sum_{k=1}^{\infty}\dfrac{1}{kn^k}\displaystyle \int_{0}^{2\pi}e^{-ikt}dt$$

both integrals becomes zero becuase they are integration over a period of sinusoid

$$\mathbf\Omega(n)=2\pi \log\ n$$

therefore,

$$\displaystyle \lim_{n \to \infty} \left(1+\dfrac{\mathbf \Omega(n)}{4\pi}\right)^{\log(n+1)}=\displaystyle \lim_{n \to \infty} \left(1+\dfrac{\log n}{2}\right)^{\log(n+1)}$$

after this step, i don't know how to proceed to find final answer. my integral might be wrong so, please help me re-evaluate if possible and also help me find limit

You don't really need to calculate $$\Omega(n)$$.
Note that $$\Omega(n)\geq c\log n$$ for some $$c>0$$ and all sufficiently large $$n$$ (e.g., from bounding the integrand $$\log(n^2-2n\log t+1)$$ by $$2\log(n-1)$$), so $$(1+\Omega(n)/(4\pi))^{\log(n+1)}$$ is of the form $$\infty^\infty$$ so must $$\to\infty$$ as $$n\to\infty$$.