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Sequence and limit of integral

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Grigorios Kostakos
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Sequence and limit of integral

#1

Post by Grigorios Kostakos » Mon Apr 17, 2017 1:59 pm

  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 \[\displaystyle\mathop{\lim}\limits_{n\to+\infty}\int_{\frac{1}{n}}^{1}g_n(x)\,dx\,.\]
Grigorios Kostakos
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Grigorios Kostakos
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Re: Sequence and limit of integral

#2

Post by Grigorios Kostakos » Mon Nov 27, 2017 10:42 am

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*}
    0\leqslant g_n(x)\leqslant M\quad &\Rightarrow\quad0\leqslant\int_{0}^{\frac{1}{n}} g_n(x)\,dx\leqslant M\int_{0}^{\frac{1}{n}}dx\\
    & \Rightarrow\quad0\leqslant\mathop{\lim}\limits_{n\to+\infty}\int_{0}^{\frac{1}{n}} g_n(x)\,dx\leqslant M\cancelto{0}{\mathop{\lim}\limits_{n\to+\infty}\int_{0}^{\frac{1}{n}}dx}=0\\
    & \Rightarrow\quad \mathop{\lim}\limits_{n\to+\infty}\int_{0}^{\frac{1}{n}} g_n(x)\,dx=0\quad (1)
    \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*}
Grigorios Kostakos
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