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

thanks in advance

up vote 2 down vote accepted

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 $n\to\infty$.

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