1989 Paper 1 Q2

Year: 1989
Paper: 1
Question Number: 2

Course: UFM Pure
Section: Sequences and series, recurrence and convergence

Difficulty: 1484.0 Banger: 1500.0

integration integration by parts Riemann sum sequences convergence logarithm area approximation factorial

Problem

For \(x>0\) find \(\int x\ln x\,\mathrm{d}x\). By approximating the area corresponding to \(\int_{0}^{1}x\ln(1/x)\, \d x\) by \(n\) rectangles of equal width and with their top right-hand vertices on the curve \(y=x\ln(1/x)\), show that, as \(n\rightarrow\infty\), \[ \frac{1}{2}\left(1+\frac{1}{n}\right)\ln n-\frac{1}{n^{2}}\left[\ln\left(\frac{n!}{0!}\right)+\ln\left(\frac{n!}{1!}\right)+\ln\left(\frac{n!}{2!}\right)+\cdots+\ln\left(\frac{n!}{(n-1)!}\right)\right]\rightarrow\frac{1}{4}. \] {[}You may assume that \(x\ln x\rightarrow0\) as \(x\rightarrow0\).{]}

Solution

Integrating by parts we obtain: \begin{align*} \int x \ln x \, \d x &= [\frac12 x^2 \ln x] - \int \frac12x^2 \cdot \frac1x \d x \\ &= \frac12 x^2 \ln x - \frac14 x^2 + C \end{align*}

We should have: \begin{align*} \int_0^1 x \ln \frac{1}{x} \d x &= \lim_{n \to \infty} \sum_{i=1}^n \frac{1}{n} \frac{i}{n} \ln \left ( \frac{n}{i} \right) \\ \left [ -\frac12 x^2 \ln x + \frac14 x^2 \right]_0^1 &= \lim_{n \to \infty} \sum_{i=1}^n \frac{1}{n} \frac{i}{n} \ln \left ( \frac{n}{i} \right) \\ \frac{1}{4} &=\lim_{n \to \infty} \frac{1}{n^2} \sum_{i=1}^n \l i \ln n - i \ln i \r \\ &= \lim_{n \to \infty} \frac{1}{n^2}\l \frac{n(n+1)}{2} \ln n - \sum_{i=1}^n i \ln i \r \\ &= \lim_{n \to \infty} \l \frac{1}{2}(1+\frac{1}n) \ln n - \frac{1}{n^2}\sum_{i=1}^n i \ln i \r \\ &= \lim_{n \to \infty} \l \frac{1}{2}(1+\frac{1}n) \ln n - \frac{1}{n^2}\sum_{i=1}^n \sum_{k=1}^i \ln i \r \\ &= \lim_{n \to \infty} \l \frac{1}{2}(1+\frac{1}n) \ln n - \frac{1}{n^2}\sum_{k=0}^{n-1} \sum_{i=0}^k \ln (n-i) \r \\ &= \lim_{n \to \infty} \l \frac{1}{2}(1+\frac{1}n) \ln n - \frac{1}{n^2}\sum_{k=0}^{n-1} \ln \frac{n!}{(n-k)!}\r \\ \end{align*}

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Problem source

For $x>0$ find $\int x\ln x\,\mathrm{d}x$. 
By approximating the area corresponding to $\int_{0}^{1}x\ln(1/x)\, \d x$ by $n$ rectangles of equal width and with their top right-hand vertices on the curve $y=x\ln(1/x)$, show that, as $n\rightarrow\infty$,
\[
\frac{1}{2}\left(1+\frac{1}{n}\right)\ln n-\frac{1}{n^{2}}\left[\ln\left(\frac{n!}{0!}\right)+\ln\left(\frac{n!}{1!}\right)+\ln\left(\frac{n!}{2!}\right)+\cdots+\ln\left(\frac{n!}{(n-1)!}\right)\right]\rightarrow\frac{1}{4}.
\]
{[}You may assume that $x\ln x\rightarrow0$ as $x\rightarrow0$.{]}

Solution source

Integrating by parts we obtain: 
\begin{align*}
\int x \ln x \, \d x &= [\frac12 x^2 \ln x] - \int \frac12x^2 \cdot \frac1x \d x \\
&= \frac12 x^2 \ln x - \frac14 x^2 + C
\end{align*}

\begin{center}
    \begin{tikzpicture}[scale=4]
       \draw[domain = 0.01:1, samples=100, variable = \x]  plot ({\x},{-2*\x * ln(\x)});

       \draw[->] (0,0) -- (1.1, 0);
       \draw[->] (0,0) -- (0, 1.1);

       \node at (1.1,0) [right] {$x$};
       \node at (0,1.1) [above] {$y$};

       \node at (0.9, 0.6) {$y = -x \ln x$};

       \foreach \i in {1,...,10} {
            \draw ({(\i-1)/10}, {0}) -- ({(\i-1)/10}, {-2*\i/10 * ln(\i/10)}) -- ({\i/10}, {-2*\i/10 * ln(\i/10)}) -- ({(\i/10}, {0});
       }
    \end{tikzpicture}
\end{center}

We should have:

\begin{align*}
\int_0^1 x \ln \frac{1}{x} \d x &= \lim_{n \to \infty} \sum_{i=1}^n \frac{1}{n} \frac{i}{n} \ln \left ( \frac{n}{i} \right) \\
\left [ -\frac12 x^2 \ln x + \frac14 x^2  \right]_0^1 &= \lim_{n \to \infty} \sum_{i=1}^n \frac{1}{n} \frac{i}{n} \ln \left ( \frac{n}{i} \right) \\ 
\frac{1}{4} &=\lim_{n \to \infty}  \frac{1}{n^2} \sum_{i=1}^n \l i \ln n - i \ln i \r \\
&= \lim_{n \to \infty}  \frac{1}{n^2}\l \frac{n(n+1)}{2} \ln n -  \sum_{i=1}^n i \ln i \r \\
&=  \lim_{n \to \infty}  \l \frac{1}{2}(1+\frac{1}n) \ln n -  \frac{1}{n^2}\sum_{i=1}^n i \ln i \r \\
&=  \lim_{n \to \infty}  \l \frac{1}{2}(1+\frac{1}n) \ln n -  \frac{1}{n^2}\sum_{i=1}^n \sum_{k=1}^i \ln i \r \\
&=  \lim_{n \to \infty}  \l \frac{1}{2}(1+\frac{1}n) \ln n -  \frac{1}{n^2}\sum_{k=0}^{n-1} \sum_{i=0}^k \ln (n-i) \r \\
&=  \lim_{n \to \infty}  \l \frac{1}{2}(1+\frac{1}n) \ln n -  \frac{1}{n^2}\sum_{k=0}^{n-1} \ln \frac{n!}{(n-k)!}\r \\
\end{align*}

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