Year: 2015
Paper: 3
Question Number: 1
Course: UFM Additional Further Pure
Section: Reduction Formulae
A very similar number of candidates to 2014 once again ensured that all questions received a decent number of attempts, with seven questions being very popular rather than five being so in 2014, but the most popular questions were attempted by percentages in the 80s rather than 90s. All but one question was answered perfectly at least once, the one exception receiving a number of very close to perfect solutions. About 70% attempted at least six questions, and in those cases where more than six were attempted, the extra attempts were usually fairly superficial.
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
\begin{questionparts}
\item Let
\[
I_n= \int_0^\infty \frac 1 {(1+u^2)^n}\, \d u \,,
\]
where $n$ is a positive integer. Show that
\[
I_n - I_{n+1} = \frac 1 {2n} I_n
\]
and deduce that
\[
I_{n+1} = \frac{(2n)!\, \pi}{2^{2n+1}(n!)^2} \,.
\]
\item Let
\[
J = \int_0^\infty \f\big( (x- x^{-1})^2\big ) \, \d x \,,
\]
where $\f$ is any function for which the integral exists. Show that
\[
J = \int_0^\infty x^{-2} \f\big( (x- x^{-1})^2\big) \, \d x \, = \frac12 \int_0^\infty
(1 + x^{-2})
\f\big( (x- x^{-1})^2\big )
\, \d x \,
= \int_0^\infty \f\big(u^2\big) \,\d u \,.
\]
\item Hence evaluate
\[
\int_0^\infty \frac {x^{2n-2}}{(x^4-x^2+1)^n} \, \d x \,,
\]
where $n$ is a positive integer.
\end{questionparts}\begin{align*}
I_n - I_{n+1} &= \int_0^\infty \frac 1 {(1+u^2)^n}\, \d u - \int_0^\infty \frac 1 {(1+u^2)^{n+1}}\, \d u \\
&= \int_0^\infty \l \frac 1 {(1+u^2)^n}- \frac 1 {(1+u^2)^{n+1}} \r\, \d u \\
&= \int_0^\infty \frac {u^2} {(1+u^2)^{n+1}} \, \d u \\
&= \left [ u \frac{u}{(1+u^2)^{n+1}} \right]_0^{\infty} - \frac{-1}{2n}\int_0^{\infty} \frac{1}{(1+u^2)^n} \d u \tag{$IBP: u = u, v' = \frac{u}{(1+u^2)^{n+1}}$}\\
&= \frac{1}{2n} I_n
\end{align*}
$\displaystyle I_1 = \int_0^{\infty} \frac{1}{1+u^2} \d u = \left [ \tan^{-1} u \right]_0^\infty = \frac{\pi}{2}$ as expected.
We also have, $I_{n+1} = \frac{2n(2n-1)}{2n \cdot 2n} I_n $, by rearranging the recurrence relation. Therefore, when we multiply out the top we will have $2n!$ and the bottom we will have two factors of $n!$ and two factors of $2^n$ combined with the original $\frac{\pi}{2}$ we get
\[ I_{n+1} = \frac{(2n)! \pi}{2^{2n+1} (n!)^2} \]
\begin{align*}
J = \int_0^\infty f\big( (x- x^{-1})^2\big ) \, \d x &= \int_{u = \infty}^{u = 0} f((u^{-1}-u)^2)(-u^{-2} )\d u \tag{$u = x^{-1}, \d u = -x^{-2} \d x$} \\
&= \int^{u = \infty}_{u = 0} f((u^{-1}-u)^2)u^{-2} \d u \\
&= \int^{\infty}_{0} u^{-2}f((u-u^{-1})^2) \d u \\
\end{align*}
Therefore adding the two forms for $J$ we have
\begin{align*}
2 J &= \int_0^\infty f\big( (x- x^{-1})^2\big ) \, \d x + \int_0^\infty x^{-2} f\big( (x- x^{-1})^2\big ) \, \d x \\
&= \int_0^\infty (1+x^{-2}) f\big( (x- x^{-1})^2\big ) \, \d x
\end{align*}
And letting $u = x - x^{-1}$, we have $\d u = (1 + x^{-2}) \d x$, and $u$ runs from $-\infty$ to $\infty$ so we have:
\begin{align*}
\int_0^\infty (1+x^{-2}) f\big( (x- x^{-1})^2\big ) \, \d x &= \int_{-\infty}^\infty f(u^2) \, \d u \\
&=2 \int_{0}^\infty f(u^2) \, \d u
\end{align*}
Since both of these are $2J$ we have the result we are after.
Finally,
\begin{align*}
\int_0^\infty \frac {x^{2n-2}}{(x^4-x^2+1)^n} \, \d x &= \int_0^{\infty} \frac{x^{2n-2}}{x^{2n}(x^2-1+x^{-2})^n} \d x \\
&= \int_0^{\infty} \frac{x^{-2}}{((x-x^{-1})^2+1)^n} \d x \\
&= \int_0^{\infty} \frac{1}{(x^2+1)^n} \d x \tag{Where $f(x) = (1+x^2)^{-n}$ in $J$ integral} \\
&= I_n = \frac{(2n-2)! \pi}{2^{2n-1} ((n-1)!)^2}
\end{align*}
This was the most popular question, being attempted by 85% of candidates, it was however only moderately successful although a number achieved full marks. Quite often, candidates ignored the helpful approach suggested by the LHS of the first required result, though, of course, it was possible to start from the first defined integral and achieve the same result. Many needlessly lost marks through omitting fairly straightforward steps such as the final evaluation in the last part of the question and failing to substantiate the simplified form of the result of part (i). Some got very carried away with tan or sinh substitutions in part (i), usually unsuccessfully and leading to monstrous amounts of algebraic working. A few failed to change the limits of integration in part (ii).