Year: 2018
Paper: 3
Question Number: 8
Course: UFM Pure
Section: Taylor series
The total entry was a record number, an increase of over 6% on 2017. Only question 1 was attempted by more than 90%, although question 2 was attempted by very nearly 90%. Every question was attempted by a significant number of candidates with even the two least popular questions being attempted by 9%. More than 92% restricted themselves to attempting no more than 7 questions with very few indeed attempting more than 8. As has been normal in the past, apart from a handful of very strong candidates, those attempting six questions scored better than those attempting more than six.
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1516.0
Banger Comparisons: 1
In this question, you should ignore issues of convergence.
\begin{questionparts}
\item
Let
\[
I = \int_0^1 \frac{\f(x^{-1}) } {1+x} \, \d x
\,,
\]
where $\f(x)$ is a function for which the integral exists.
Show that
\[
I = \sum_{n=1}^\infty \int_n^{n+1} \frac{\f(y) } {y(1+y)}\, \d y
\]
and deduce that, if $\f(x) = \f(x+1)$ for all $x$, then
\[
I= \int_0^1 \frac{\f(x)} {1+x} \, \d x
\,.
\]
\item
The {\em fractional part}, $\{x\}$, of a real number
$x$ is defined to be $x-\lfloor x\rfloor$ where
$\lfloor x \rfloor$ is
the largest integer less than or equal to $x$.
For example $\{3.2\} = 0.2$ and $\{3\}=0\,$.
Use the result of part (i) to evaluate
\[
\displaystyle \int _0^1 \frac { \{x^{-1}\}}{1+x}\, \d x \text{ and }
\displaystyle \int _0^1 \frac { \{2x^{-1}\}}{1+x}\, \d x \,.
\]
\item (Bonus) Use the same method to evaluate
\[
\int_0^1 \frac {x\{x^{-1}\}}{1-x^2} \, \d x \,.
\]
\item (Bonus - harder) Use the same method to evaluate
\[
\int_0^1 \frac {x^2\{x^{-1}\}}{1-x^2} \, \d x \,.
\]
\end{questionparts}\begin{questionparts}
\item \begin{align*}
&& I &= \int_0^1 \frac{f(x^{-1})}{1+x} \d x \\
u = x^{-1}, \d u = -x^{-2} \d x: &&&= \int_{\infty}^1 \frac{f(u)}{1+\frac{1}{u}} \frac{-1}{u^2} \d u \\
&&&= \int_1^{\infty} \frac{f(u)}{u(1+u)} \d u \\
&&&= \sum_{n=1}^{\infty} \int_n^{n+1} \frac{f(u)}{u(u+1)} \d u \\
\\
\text{if} f(x) = f(x+1)\, \forall x && &=\sum_{n=1}^{\infty} \int_{0}^1 \frac{f(x+n)}{(x+n)(x+n+1)} \d x \\
&&&= \sum_{n=1}^\infty \int_0^1 \frac{f(x)}{(x+n)(x+n+1)} \d x \\
&&&= \int_0^1 f(x) \l \sum_{n=1}^{\infty} \frac{1}{(x+n)(x+n+1)}\r \d x \\
&&&= \int_0^1 f(x) \l \sum_{n=1}^{\infty} \l \frac{1}{x+n} - \frac{1}{x+n+1} \r\r \d x \\
&&&= \int_0^1 f(x) \l \frac{1}{x+1} \r \d x \\
&&&= \int_0^1\frac{f(x)}{x+1} \d x \\
\end{align*}
\item Since the fractional part is periodic with period $1$, we can say \begin{align*}
&& \int_0^1 \frac{\{ x^{-1} \} }{1+x} \d x &= \int_0^1 \frac{\{ x\}}{x+1} \d x \\
&&&= \int_0^1 \frac{x}{x+1} \d x \\
&&&= \int_0^1 1-\frac{1}{x+1} \d x \\
&&&= [x - \ln (1+x) ]_0^1 \\
&&&= 1 - \ln 2
\end{align*}
\begin{align*}
&& \int_0^1 \frac{\{ 2x^{-1}\}}{1+x} \d x &= \int_0^1 \frac{\{2x\}}{x+1} \d x \\
&&&= \int_0^{1/2} \frac{2x}{x+1} \d x +\int_{1/2}^{1} \frac{2x-1}{x+1} \d x \\
&&&= 2\int_0^1 \frac{x}{x+1} \d x + \int_{1/2}^1 \frac{-1}{x+1} \d x \\
&&&= 2 - 2\ln 2 - \l \ln 2 - \ln \tfrac32 \r \\
&&&= 2 - 4 \ln 2 + \ln 3 \\
&&&= 2 + \ln \tfrac {3}{16}
\end{align*}
\item \begin{align*}
&& \int_0^1 \frac{x \{ x^{-1} \} }{1-x^2} \d x &= \frac12 \l \int_0^1 \frac{ \{ x^{-1} \}}{1-x} - \frac{\{x^{-1} \}}{1+x} \d x\r
\end{align*}
Consider for $f$ periodic with period $1$
\begin{align*}
\int_0^1 \frac{f(x^{-1})}{1-x} \d x &= \int_1^{\infty} \frac{f(u)}{u(u-1)} \d u \\
&= \sum_{n=1}^{\infty}\int_{n}^{n+1} \frac{f(u)}{u(u-1)} \d u \\
&= \sum_{n=1}^{\infty}\int_{0}^{1} \frac{f(u)}{(u+n)(u+n-1)} \d u \\
&= \int_{0}^{1} \sum_{n=1}^{\infty}\frac{f(u)}{(u+n)(u+n-1)} \d u \\
&= \int_{0}^{1} f(u) \sum_{n=1}^{\infty}\l\frac{1}{u+n-1} - \frac{1}{u+n} \r\d u \\
&= \int_0^1 \frac{f(u)}{u} \d u
\end{align*}
So we have
\begin{align*}
&& \int_0^1 \frac{x \{ x^{-1} \} }{1-x^2} \d x &= \frac12 \l \int_0^1 \frac{ \{ x^{-1} \}}{1-x} - \frac{\{x^{-1} \}}{1+x} \d x \r \\
&&&= \frac12 \int_0^1 \frac{\{ x \}}{x} \d x - \frac12 (1 - \ln 2) \\
&&&= \frac12 - \frac12 + \frac12 \ln 2 \\
&&&= \frac12 \ln 2
\end{align*}
\end{questionparts}
The fifth most popular question, this was attempted by five eighths of the candidates, and was the fourth most successfully attempted with a mean score of a little over 9/20. The first result in (i) was generally very well done, but success in the second result was usually restricted to those that spotted the possibility of using partial fractions. The first integral in (ii) was usually done correctly. However the second integral attracted a range of mistakes such as failing to see that or justify how the first part could be used and not simplifying their answers. The nature of the periodicity and the need to integrate different expressions over the differing ranges were frequent stumbling blocks.