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 Post subject: Sequence and limit of integralPosted: Mon Apr 17, 2017 1:59 pm
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Joined: Mon Nov 09, 2015 1:36 am
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Location: Ioannina, Greece
1. Examine whether the sequence of functions $g_n:[0, 1] \longrightarrow \mathbb{R}$ defined as $g_n(x)=\begin{cases}\dfrac{x^{\frac{1}{n}}\log(1+x)}{x\,(1+x^{\frac{2}{n}})^{\frac{3}{2}}}\,,& x\in(0,1]\\ 0\,,& x=0\end{cases}\,,\quad n\in\mathbb{N}\,,$ converges uniformly on $[0, 1]$ or not.

2. Evaluate $\displaystyle\mathop{\lim}\limits_{n\to+\infty}\int_{\frac{1}{n}}^{1}g_n(x)\,dx\,.$

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Grigorios Kostakos

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 Post subject: Re: Sequence and limit of integralPosted: Mon Nov 27, 2017 10:42 am
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Joined: Mon Nov 09, 2015 1:36 am
Posts: 460
Location: Ioannina, Greece
We give a solution:

1. The sequence of functions $g_n:[0, 1] \longrightarrow \mathbb{R}$ defined as $g_n(x)=\begin{cases}\dfrac{x^{\frac{1}{n}}\log(1+x)}{x\,(1+x^{\frac{2}{n}})^{\frac{3}{2}}}\,,& x\in(0,1]\\ 0\,,& x=0\end{cases}\,,\quad n\in\mathbb{N}\,,$ converges pointwise to the function
$g(x)=\begin{cases}\dfrac{\sqrt{2}\log(1+x)}{4x}\,,& x\in(0,1]\\ 0\,,& x=0\end{cases}\,.$
The functions $g_n(x)$ are continuous since $\mathop{\lim}\limits_{x\to0^{+}}g_n(x)=0=g_n(0)$. Becauce $\mathop{\lim}\limits_{x\to0^{+}}g(x)=\frac{\sqrt{2}}{4}\neq g(0)$, $g$ it's not continuous on $[0,1]$. Therefore the sequence $g_n$ does not converges uniformly to $g$.

2. The sequence $g_n$ is uniformly bounded. More specifically, there exists $M>0$ such that for $n\in\mathbb{N}\,,\; x\in[0,1]$ :
\begin{align*}
\end{align*}
By the bounded convergence theorem, we have that $\mathop{\lim}\limits_{n\to+\infty}\int_{0}^{1} g_n(x)\,dx=\int_{0}^{1} g(x)\,dx\stackrel{(*)}{=}\frac{\pi^2\sqrt{2}}{48}\quad (2)\,.$
So, we have that \begin{align*}
\mathop{\lim}\limits_{n\to+\infty}\int_{\frac{1}{n}}^{1}g_n(x)\,dx&=\mathop{\lim}\limits_{n\to+\infty}\bigg(\int_{0}^{1}g_n(x)\,dx-\int_{0}^{\frac{1}{n}} g_n(x)\,dx\bigg)\\
&=\mathop{\lim}\limits_{n\to+\infty}\int_{0}^{1}g_n(x)\,dx-\mathop{\lim}\limits_{n\to+\infty}\int_{0}^{\frac{1}{n}} g_n(x)\,dx\\
&\stackrel{(1)\,(2)}{=\!=\!=}\frac{\pi^2\sqrt{2}}{48}\,.
\end{align*}

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Grigorios Kostakos

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