Integration can leave the convergence range · source reference

Printed question

(i) Using standard series from the List of Formulae (MF26), expand $\ln(k+2x)^n$ in ascending powers of $x$ as far as the term in $x^3$, where $k$ and $n$ are real constants, $k>0$, $n<0$ and $x>-\frac k2$. [2]

(ii) State the range of values of $x$ for which the expansion in part (i) is valid. [1]

(iii) Find the exact value of $\int_0^1\ln(1+2x)^{-2}\,dx$. [3]

(iv) Use your series from part (i) to estimate $\int_0^1\ln(1+2x)^{-2}\,dx$. [2]

(v) Calculate the percentage error of the estimate in part (iv), and comment on the accuracy of your approximation using your answer from part (ii). [2]

[2020/NYJC/Prelim/I/9]

Printed working

$\ln(k+2x)^n=n\ln(k+2x)=n\ln[k(1+2x/k)]=n[\ln k+\ln(1+2x/k)]$.

$=n[\ln k+2x/k-\frac12(2x/k)^2+\frac13(2x/k)^3-\cdots]=n[\ln k+2x/k-2x^2/k^2+8x^3/(3k^3)-\cdots]$.

$-1<2x/k\le1\Rightarrow-k/2<x\le k/2$.

By parts, $u=\ln(1+2x)$, $dv/dx=1$, $du/dx=2/(1+2x)$, $v=x$.

$$\begin{aligned}\int_0^1\ln(1+2x)^{-2}dx&=-2\int_0^1\ln(1+2x)dx\\&=-2\{[x\ln(1+2x)]_0^1-\int_0^1\tfrac{2x}{1+2x}dx\}\\&=-2\{\ln3-\int_0^1[1-1/(1+2x)]dx\}\\&=-2\{\ln3-[x-\tfrac12\ln(1+2x)]_0^1\}\\&=-2(\tfrac32\ln3-1)=2-3\ln3.\end{aligned}$$

$n=-2,k=1$ estimate $-2\int_0^1(2x-2x^2+8x^3/3)dx=-2[x^2-2x^3/3+2x^4/3]_0^1=-2$.

Percentage error $|-2-(2-3\ln3)|/|2-3\ln3|\times100\approx54.3\%$. The series is valid for $-1/2<x\le1/2$ while the integral covers $0\le x\le1$; the use outside that interval is not valid and results in large error.

Teacher note 1

Need to factorise out $k$ to make the constant term 1. Apply rule of logarithms $\ln(ab)=\ln a+\ln b$. From MF27, use $\ln(1+x)=x-x^2/2+x^3/3-\cdots$, where $-1<x\le1$, replacing $x$ by $2x/k$.

Teacher note 2

Use integration by parts: $\int u\frac{dv}{dx}dx=uv-\int v\frac{du}{dx}dx$.

Teacher note 3

Percentage error $=\frac{|\text{approximate value}-\text{actual value}|}{|\text{actual value}|}\times100$.

Mathematical review

(i) $n\ln(k+2x)=n\ln k+(2n/k)x-(2n/k^2)x^2+(8n/(3k^3))x^3+\cdots$.

(ii) The logarithm series is valid for $-1<2x/k\le1$, hence $-k/2<x\le k/2$; the positive endpoint converges conditionally.

(iii) Integration by parts gives $-2[x\ln(1+2x)-x+\tfrac12\ln(1+2x)]_0^1=2-3\ln3$.

(iv) With $n=-2,k=1$, integrate $-4x+4x^2-16x^3/3$: the result is $-2$.

(v) $100|-2-(2-3\ln3)|/|2-3\ln3|\approx54.34\%$. The series represents the function only up to $x=1/2$, while the integral reaches $x=1$; the estimate is unreliable.

Provenance

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