Science:Math Exam Resources/Courses/MATH101/April 2018/Question 10 (ii)

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MATH101 April 2018

• Q1 (i) • Q1 (ii) • Q1 (iii) • Q1 (iv) • Q2 (i) • Q2 (ii) • Q2 (iii) • Q3 (i) • Q3 (ii) • Q4 (i) • Q4 (ii) • Q5 (i) • Q5 (ii) • Q6 (i) • Q6 (ii) • Q7 (i) • Q7 (ii) • Q8 (i) • Q8 (ii) • Q9 (i) • Q9 (ii) • Q10 (i) • Q10 (ii) • Q10 (iii) • Q11 (i) • Q11 (ii) •

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Question 10 (ii)
Let ${\textstyle c>0}$ and consider the improper integral ${\textstyle I}$ defined by $I\equiv \int _{0}^{1}{\frac {\left[cx-\log(1+2x)\right]}{x^{2}}}\,dx\,.$ For which value of ${\textstyle c}$ does the integral converge? Explain clearly. Do not evaluate the integral for this part.

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Hint 1
Use the Maclaurin series found in part (i).

Hint 2
If a power series $\sum _{n=0}^{\infty }A_{n}(x-a)^{n}$ has the radius of convergence $R$ , then for any $\|t-a\|<R$ , we can interchange the summation and the integral; $\int _{a}^{t}\sum _{n=0}^{\infty }A_{n}(x-a)^{n}dx=\sum _{n=0}^{\infty }\int _{a}^{t}A_{n}(x-a)^{n}dx=\sum _{n=0}^{\infty }{\frac {A_{n}}{n+1}}(t-a)^{n+1}.$

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Solution
We first split the interval of the given integral into two parts. $\int _{0}^{1}{\frac {\left[cx-\log(1+2x)\right]}{x^{2}}}dx=\int _{0}^{\frac {1}{4}}{\frac {\left[cx-\log(1+2x)\right]}{x^{2}}}dx+\int _{\frac {1}{4}}^{1}{\frac {\left[cx-\log(1+2x)\right]}{x^{2}}}dx.$ Since the Maclaurin series of $\log(1+2x)$ has the radius of convergence ${\frac {1}{2}}$ and $\|{\frac {1}{4}}-0\|<{\frac {1}{2}}$ , we can plug the series into the first integral and consider the integral of each term: ${\begin{aligned}\int _{0}^{\frac {1}{4}}{\frac {\left[cx-\log(1+2x)\right]}{x^{2}}}dx=&\int _{0}^{\frac {1}{4}}{\frac {\left[cx-\sum _{n=0}^{\infty }(-1)^{n}{\frac {(2x)^{n+1}}{n+1}}\right]}{x^{2}}}dx\\=&\int _{0}^{\frac {1}{4}}{\frac {cx-2x}{x^{2}}}dx-\int _{0}^{\frac {1}{4}}{\frac {\sum _{n=1}^{\infty }(-1)^{n}{\frac {(2x)^{n+1}}{n+1}}}{x^{2}}}dx\\=&(c-2)\int _{0}^{\frac {1}{4}}{\frac {1}{x}}dx-\sum _{n=1}^{\infty }{\frac {(-1)^{n}2^{n-1}}{n+1}}\int _{0}^{\frac {1}{4}}(x)^{n-1}dx.\end{aligned}}$ Since we have $\int _{0}^{\frac {1}{4}}{\frac {1}{x}}dx=\lim _{t\to 0^{+}}\int _{t}^{\frac {1}{4}}{\frac {1}{x}}dx=\lim _{t\to 0^{+}}(\ln x{\bigg \|}_{t}^{\frac {1}{4}})=\lim _{t\to 0^{+}}(\ln {\frac {1}{4}}-\ln t)=+\infty ,$ the integral $(c-2)\int _{0}^{\frac {1}{4}}{\frac {1}{x}}dx$ diverges for $c\neq 2$ . On the other hand, $\int _{0}^{\frac {1}{4}}x^{n-1}dx={\frac {1}{n}}x^{n}{\bigg \|}_{0}^{\frac {1}{4}}={\frac {1}{n}}2^{-2n},$ so that the integral $\sum _{n=1}^{\infty }{\frac {(-1)^{n}2^{n-1}}{n+1}}\int _{0}^{\frac {1}{4}}(x)^{n-1}dx={\frac {1}{2}}\sum _{n=1}^{\infty }{\frac {(-1)^{n}2^{n}}{n+1}}{\frac {1}{n}}2^{-2n}={\frac {1}{2}}\sum _{n=1}^{\infty }{\frac {(-1)^{n}}{n(n+1)}}2^{-n}$ converges. This follows from the comparison test; $\sum _{n=1}^{\infty }{\frac {(-1)^{n}}{n(n+1)}}2^{-n}\leq \sum _{n=1}^{\infty }\left\|{\frac {(-1)^{n}}{n(n+1)}}2^{-n}\right\|\leq \sum _{n=1}^{\infty }2^{-n}<+\infty .$ Therefore, the integral in the interval $[0,1/4]$ converges only when $c=2$ . Finally, the integral in the remaining interval $[1/4,1]$ converges for any $c$ . This is because $\left\|{\frac {\left[cx-\log(1+2x)\right]}{x^{2}}}\right\|\leq {\frac {\|c\|\|x\|+\|\log(1+2x)\|}{x^{2}}}\leq {\frac {1}{(1/4)^{2}}}(\|c\|+\|\log 3\|)<+\infty$ and hence ${\begin{aligned}\int _{\frac {1}{4}}^{1}{\frac {\left[cx-\log(1+2x)\right]}{x^{2}}}dx&\leq \int _{\frac {1}{4}}^{1}\left\|{\frac {\left[cx-\log(1+2x)\right]}{x^{2}}}\right\|dx\\&\leq \int _{\frac {1}{4}}^{1}{\frac {1}{(1/4)^{2}}}(\|c\|+\|\log 3\|)dx={\frac {1}{(1/4)^{2}}}(\|c\|+\|\log 3\|)(1-1/4)<+\infty .\end{aligned}}$ Therefore, the given integral converges when $c=2$ and diverges otherwise. Answer: $\color {blue}c=2$