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2024 Paper 2 Q2
D: 1500.0 B: 1500.0
generalised binomial theorem integration partial fractions series expansion term-by-term integration logarithm trigonometric identities

In this question, you need not consider issues of convergence.

  1. Find the binomial series expansion of \((8 + x^3)^{-1}\), valid for \(|x| < 2\). Hence show that, for each integer \(m \geqslant 0\), \[ \int_0^1 \frac{x^m}{8 + x^3}\,\mathrm{d}x = \sum_{k=0}^{\infty} \left( \frac{(-1)^k}{2^{3(k+1)}} \cdot \frac{1}{3k + m + 1} \right). \]
  2. Show that \[ \sum_{k=0}^{\infty} \frac{(-1)^k}{2^{3(k+1)}} \left( \frac{1}{3k+3} - \frac{2}{3k+2} + \frac{4}{3k+1} \right) = \int_0^1 \frac{1}{x+2}\,\mathrm{d}x\,, \] and evaluate the integral.
  3. Show that \[ \sum_{k=0}^{\infty} \frac{(-1)^k}{2^{3(k+1)}} \left( \frac{72(2k+1)}{(3k+1)(3k+2)} \right) = \pi\sqrt{a} - \ln b\,, \] where \(a\) and \(b\) are integers which you should determine.

Solution:

  1. Note that \(\,\) \begin{align*} && (8+x^3)^{-1} &= \tfrac18(1 + \tfrac18x^3)^{-1} \\ &&&= \tfrac18( 1 - \left (\tfrac{x}{2} \right)^3 + \left (\tfrac{x}{2} \right)^6 -\left (\tfrac{x}{2} \right)^9 + \cdots ) \end{align*} So \begin{align*} && \int_0^1 \frac{x^m}{8+x^3} \d x &= \int_0^1 x^m \sum_{k=0}^{\infty} \frac{1}{2^3} \left ( - \frac{x}{2} \right)^{3k} \d x \\ &&&= \sum_{k=0}^{\infty} (-1)^k \frac{1}{2^{3(k+1)}} \int_0^1 x^{m+3k} \d x \\ &&&= \sum_{k=0}^{\infty} \frac{(-1)^k}{2^{3(k+1)}} \frac{1}{3k+m+1} \\ \end{align*}
  2. Notice that \begin{align*} && S &= \sum_{k=0}^{\infty} \frac{(-1)^k}{2^{3(k+1)}} \left( \frac{1}{3k+3} - \frac{2}{3k+2} + \frac{4}{3k+1} \right) \\ &&&= \int_0^1 \frac{x^2}{8+x^3} \d x - 2\int_0^1 \frac{x^1}{8+x^3}+4\int_0^1 \frac{x^0}{8+x^3} \d x \\ &&&= \int_0^1 \frac{x^2-2x+4}{x^3+8} \d x \\ &&&= \int_0^1 \frac{1}{x+2} \d x = \left[ \ln(x+2)\right]_0^1 \\ &&&= \ln 3 - \ln 2 = \ln \tfrac32 \end{align*}
  3. Firstly, note that \begin{align*} && \frac{2k+1}{(3k+1)(3k+2)} &= \frac13 \left ( \frac{1}{3k+1} +\frac{1}{3k+2} \right) \end{align*} so \begin{align*} && S_2 &= \sum_{k=0}^{\infty} \frac{(-1)^k}{2^{3(k+1)}} \left( \frac{72(2k+1)}{(3k+1)(3k+2)} \right) \\ &&&= \sum_{k=0}^{\infty} \frac{(-1)^k}{2^{3(k+1)}} 24 \left ( \frac{1}{3k+1} +\frac{1}{3k+2} \right) \\ &&&= 24 \left [ \int_0^1 \frac{x^0}{8+x^3} \d x + \int_0^1 \frac{x^1}{8+x^3} \d x \right] \\ &&&= 24 \left [ \int_0^1 \frac{1+x}{(x+2)(x^2-2x+4)} \d x \right] \\ &&&= 24 \left [ \int_0^1 \frac{x+8}{12(x^2-2x+4)}-\frac{1}{12(x+2)} \d x \right] \\ &&&= 2\left [ \int_0^1 \frac{x+8}{x^2-2x+4}-\frac{1}{x+2} \d x \right] \\ &&&= 2\left [ \int_0^1 \frac{x-1}{x^2-2x+4}+ \frac{9}{(x-1)^2+3}-\frac{1}{x+2} \d x \right] \\ &&&=2 \left [ \int_0^1 \frac12\ln(x^2-2x+4)+3\sqrt{3}\arctan \frac{x-1}{\sqrt{3}} -\ln(x+2) \right]_0^1 \\ &&&=2 \left ( \frac12 \ln3+3\sqrt3 \arctan 0 - \ln 3 \right) - 2\left (\frac12 \ln 4-3\sqrt3 \arctan \frac{1}{\sqrt3} - \ln 2 \right) \\ &&&=\sqrt3 \pi - \ln3 \end{align*}

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2018 Paper 2 Q5
D: 1600.0 B: 1505.3
Taylor series binomial expansion integration logarithmic series exponential series term-by-term integration substitution infinite series definite integral Frullani integral

In this question, you should ignore issues of convergence.

  1. Write down the binomial expansion, for \(\vert x \vert<1\,\), of \(\;\dfrac{1}{1+x}\,\) and deduce that %. By considering %$ %\displaystyle \int \frac 1 {1+x} \, \d x %\,, %$ %show that \[ \displaystyle \ln (1+x) = -\sum_{n=1}^\infty \frac {(-x)^n}n \, \] for \(\vert x \vert <1 \,\).
  2. Write down the series expansion in powers of \(x\) of \(\displaystyle \e^{-ax}\,\). Use this expansion to show that \[ \int_0^\infty \frac {\left(1- \e^{-ax}\right)\e^{-x}}x \,\d x = \ln(1+a) \ \ \ \ \ \ \ (\vert a \vert <1)\,. \]
  3. Deduce the value of \[ \int_0^1 \frac{x^p - x^q}{\ln x} \, \d x \ \ \ \ \ \ (\vert p\vert <1, \ \vert q\vert <1) \,. \]

Solution:

  1. \begin{align*} && \frac1{1+x} &= 1 - x+ x^2 - x^3+ \cdots \\ \Rightarrow && \int_0^x \frac{1}{1+t} \d t &= \int_0^x \sum_{n=0}^{\infty} (-t)^n \d t \\ &&&= \left [\sum_{n=0}^{\infty} -\frac{(-t)^{n+1}}{n+1} \right]_0^x \\ \Rightarrow &&\ln(1+x)&=- \sum_{n=1}^\infty \frac{(-x)^n}{n} \end{align*}
  2. \begin{align*} && e^{-ax} &= \sum_{n=0}^\infty \frac{(-a)^n}{n!} x^n \\ \Rightarrow && \int_0^{\infty} \frac{1}{x} \left (1-e^{-ax} \right)e^{-x} \d x &= \int_0^{\infty} \frac{1}{x} \left (-\sum_{n=1}^\infty \frac{(-a)^n}{n!}x^n \right)e^{-x} \d x \\ &&&= -\int_0^{\infty} \sum_{n=1}^\infty \frac{(-a)^n}{n!} x^{n-1} e^{-x} \d x \\ &&&= -\sum_{n=1}^\infty \frac{(-a)^n}{n!} \int_0^{\infty} x^{n-1} e^{-x} \d x \\ &&&= -\sum_{n=1}^\infty \frac{(-a)^n}{n!} (n-1)! \\ &&&= -\sum_{n=1}^\infty \frac{(-a)^n}{n} \\ &&&= \ln (1+a) \end{align*}
  3. \begin{align*} && \int_0^1 \frac{x^p - x^q}{\ln x} \, \d x &= \int_0^1 \frac{x^p(1 - x^{q-p})}{\ln x} \, \d x \\ e^{-u} = x, \d x = -e^{-u} \d u: &&&=\int_{u=\infty}^{0} \frac{e^{-pu}-e^{-qu}}{-u} (-e^{-u})\d u \\ &&&= \int_0^\infty \frac{e^{-u}(e^{-qu}-e^{-pu})}{u} \d u \\ &&&= \int_0^\infty \frac{e^{-(1+q)u}(1-e^{-(p-q)u})}{u} \d u \\ v = (1+q)u, \d v = (1+q) \d u: &&&=\int_0^{\infty} \frac{e^{-v}(1-e^{-\left(\frac{p-q}{1+q}\right)v}}{v}\d v \\ &&&= \ln \left(1 + \frac{p-q}{1+q} \right) \\ &&&= \ln \left ( \frac{1+p}{1+q} \right) \end{align*}

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