## 175 – Quiz 5

Here is quiz 5.

Problem 1 asks for the partial fractions decomposition of ${\displaystyle\frac1{x^4-x^2}.}$ To find this, first we factor the denominator:

${x^4-x^2=x^2(x^2-1)=x^2(x-1)(x+1).}$ This means that there must be constants ${A,B,C,D}$ such that

$\displaystyle \frac1{x^4-x^2}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x-1}+\frac{D}{x+1}.$

To find these constants, we add the fractions on the right hand side, and obtain

$\displaystyle \frac1{x^4-x^2}=\frac{Ax(x^2-1)+B(x^2-1)+Cx^2(x+1)+Dx^2(x-1)}{x^4-x^2}.$

This means that

$\displaystyle 1=Ax(x^2-1)+B(x^2-1)+Cx^2(x+1)+Dx^2(x-1).$

Setting ${x=0}$ gives us

$\displaystyle 1=-B\mbox{ or }B=-1.$

Setting ${x=1}$ gives us

$\displaystyle 1=2C\mbox{ or }C=1/2.$

Setting ${x=-1}$ gives us

$\displaystyle 1=-2D\mbox{ or }D=-1/2.$

Setting ${x=2}$ and using the values of ${B,C,D}$ just found, gives us

$\displaystyle 1=6A-3+6-2\mbox{ or }1=6A+1\mbox{ or }A=0.$

Putting this together, the partial fractions decomposition is

$\displaystyle \frac1{x^4-x^2}=-\frac1{x^2}+\frac{1/2}{x-1}-\frac{1/2}{x+1}.$

Problem 2 asks to determine whether ${\displaystyle\int_1^2\frac{dx}{x^4-x^2}}$ converges. Note that the expression we are integrating is defined in ${(1,2],}$ but not at ${x=1,}$ so this is an improper integral of Type II and to evaluate it we use the definition:

$\displaystyle \int_1^2\frac{dx}{x^4-x^2}=\lim_{a\rightarrow1^+}\int_a^2\frac{dx}{x^4-x^2}.$

To evaluate this last expression, we use the result from Problem 1:

$\displaystyle \begin{array}{rl} \displaystyle \lim_{a\rightarrow1^+}\int_a^2\frac{dx}{x^4-x^2}&=\displaystyle\lim_{a\rightarrow1^+}\int_a^2-\frac1{x^2}+\frac{1/2}{x-1}-\frac{1/2}{x+1}\,dx\\ &=\displaystyle\lim_{a\rightarrow1^+}\left[\left.\frac1x+\frac12\ln(x-1)-\frac12\ln(x+1)\right]\right|_a^2\\ &=\displaystyle\lim_{a\rightarrow1^+}\left[\frac12+\frac12\ln(1)-\frac12\ln(3)\right]\\ &\hspace{1cm}\displaystyle-\left[\frac1a+\frac12\ln(a-1)-\frac12\ln(a+1)\right]. \end{array}$

This expression diverges because ${\displaystyle \lim_{a\rightarrow1^+}\frac1a-\frac12\ln(a+1)=1-\frac12\ln2,}$ but

$\displaystyle \lim_{a\rightarrow1^+}\frac12\ln(a-1)=-\infty.$

Problem 3 asks to determine whether ${\displaystyle\int_2^\infty\frac{dx}{x^4-x^2}}$ converges. Note that the expression we are integrating is defined in ${[2,\infty)}$ but, of course, the interval of integration is infinite, so this is an improper integral of Type I and to evaluate it we use the definition:

$\displaystyle \int_2^\infty\frac{dx}{x^4-x^2}=\lim_{N\rightarrow\infty}\int_2^N\frac{dx}{x^4-x^2}.$

To evaluate this last expression, we proceed as in Problem 2:

$\displaystyle \begin{array}{rl} \displaystyle \lim_{N\rightarrow\infty}\int_2^N\frac{dx}{x^4-x^2}&=\displaystyle\lim_{N\rightarrow\infty}\left[\left.\frac1x+\frac12\ln(x-1)-\frac12\ln(x+1)\right]\right|_2^N\\ &=\displaystyle\lim_{N\rightarrow\infty}\left[\frac1N+\frac12\ln(N-1)-\frac12\ln(N+1)\right]\\ &\hspace{1cm}\displaystyle-\left[\frac12+\frac12\ln(1)-\frac12\ln(3)\right]. \end{array}$

To evaluate the expression within the first set of parentheses, we use that

$\displaystyle \frac12\ln(N-1)-\frac12\ln(N+1)$ $\displaystyle =\ln\sqrt{N-1}+\ln\left(\frac1{\sqrt{N+1}}\right)$ $\displaystyle=\ln\left(\sqrt{\frac{N-1}{N+1}}\right).$

The limit of this expression as ${N\rightarrow\infty}$ is ${\ln\sqrt1=0,}$ because ${\displaystyle\lim_{N\rightarrow\infty}\frac{N-1}{N+1}=\lim_{N\rightarrow\infty}\frac{1}{1}=1,}$ using l’Hôpital’s rule. This means that

$\displaystyle \int_2^\infty\frac{dx}{x^4-x^2}=-\frac12+\frac12\ln3.$

(In particular, the integral converges.)

Typeset using LaTeX2WP. Here is a printable version of this post.