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

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
Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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: $\color {blue}c=2$