Year: 2009
Paper: 3
Question Number: 8
Course: UFM Additional Further Pure
Section: Reduction Formulae
The vast majority of candidates (in excess of 95%) attempted at least five questions, and nearly a quarter attempted more than six questions, though very few doing so achieved high scores (about 2%). Most attempting more than six questions were submitting fragmentary answers, which, as the rubric informed candidates, earned little credit.
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1516.0
Banger Comparisons: 1
Let $m$ be a positive integer and let $n$ be a non-negative integer.
\begin{questionparts}
\item
Use the result $\displaystyle \lim_{t\to\infty}\e^{-mt} t^n=0$ to show that
\[
\lim_{x\to0} x^m (\ln x)^n =0\,.
\]
By writing $x^x$ as $\e^{x\ln x}$ show that
\[
\lim _{x\to0} x^x=1\,.
\]
\item Let $\displaystyle I_{n} = \int_0^1 x^m (\ln x)^n \d x\,$.
Show that
\[
I_{n+1} = - \frac {n+1}{m+1} I_{n}
\]
and hence evaluate $I_{n}$.
\item Show that
\[
\int_0^1 x^x \d x = 1 -\left(\tfrac12\right)^2 +\left(\tfrac13\right)^3
-\left(\tfrac14\right)^4 + \cdots \,.
\]
\end{questionparts}\begin{questionparts}
\item $\,$ \begin{align*}
&& \lim_{x \to 0} x^m(\ln x)^n &= \lim_{t \to \infty} (e^{-t})^m (\ln e^{-t})^n \\
&&&= \lim_{t \to \infty} e^{-mt} (-t)^n \\
&&&= (-1)^n \lim_{t \to \infty} e^{-mt} t^n = 0 \\
\\
&& \lim_{x \to 0} x^x &= \lim_{x \to 0} e^{x \ln x} \\
&&&= \exp \left (\lim_{x \to 0} x \ln x \right) \\
&&&= \exp \left (0 \right) = 1
\end{align*}
\item $\,$ \begin{align*}
&& I_{n} &= \int_0^1 x^m (\ln x)^n \d x \\
&& I_{n+1} &= \int_0^1 x^m (\ln x)^{n+1} \d x \\
&&&= \left [\frac{x^{m+1}}{m+1} (\ln x)^{n+1} \right]_0^1 - \frac{1}{m+1} \int_0^1 x^{m+1} (n+1) (\ln x)^n \cdot x^{-1} \d x \\
&&&= 0 - \frac{1}{m+1} \lim_{x \to 0} \left (x^{m+1} (\ln x)^{n+1} \right) - \frac{n+1}{m+1} \int_0^1 x^m (\ln x)^n \d x \\
&&&= - \frac{n+1}{m+1} I_n \\
\\
&& I_0 &= \int_0^1 x^m \d x = \frac{1}{m+1} \\
&& I_1 &= -\frac{1}{(m+1)^2} \\
&& I_2 &= \frac{2}{(m+1)^3} \\
&& I_n &= (-1)^n \frac{n!}{(m+1)^{n+1}}
\end{align*}
\item $\,$ \begin{align*}
&& \int_0^1 x^x \d x &= \int_0^1 e^{x \ln x} \d x \\
&&&= \int_0^1 \sum_{k=0}^{\infty} \frac{x^k(\ln x)^k}{k!} \d x \\
&&&= \sum_{k=0}^{\infty} \int_0^1 \frac{x^k(\ln x)^k}{k!} \d x\\
&&&= \sum_{k=0}^{\infty} (-1)^k \frac{k!}{k!(k+1)^{k+1}}\\
&&&= \sum_{k=0}^{\infty} \frac{(-1)^k}{(k+1)^{k+1}}\\
&&&= 1 - \frac{1}{2^2} + \frac{1}{3^3} - \frac{1}{4^4} + \cdots
\end{align*}
\end{questionparts}
Roughly the same number attempted this as question 7, with slightly less success. Usually, a candidate did not properly obtain the first three results, and so would end up having apparently finished the whole question but in fact scoring only two thirds marks. The problem was often that the limiting process was not fully understood. In part (ii), there was often odd splitting going on to attempt the integration by parts and this part often went wrong.