Science:Math Exam Resources/Courses/MATH103/April 2013/Question 02 (c)

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[hide] Question 02 (c)
Integration: Short Answer Problems - Evaluate the following integrals: state if a definite integral does not exist; use limits for improper integrals. (full marks for correct answer; work must be shown for partial marks) $\displaystyle I_{c}=\int _{0}^{2}{\frac {1}{x^{2}-1}}\,dx$

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[show] Hint 1
Watch out! It's a trap! This integral is actually improper! First start out by splitting up at the singularity in the interval.

[show] Hint 2
Try integration by partial fractions.

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[show] Solution 1
First, note that the integral above is improper with a singularity at $\displaystyle x=1$ . Splitting up the integral there, we see that ${\begin{aligned}\int _{0}^{2}{\frac {1}{x^{2}-1}}\ dx&=\int _{0}^{1}{\frac {1}{x^{2}-1}}\ dx+\int _{1}^{2}{\frac {1}{x^{2}-1}}\ dx\\&=\lim _{b\to 1^{-}}\int _{0}^{b}{\frac {1}{x^{2}-1}}\ dx+\lim _{a\to 1^{+}}\int _{a}^{2}{\frac {1}{x^{2}-1}}\ dx\\\end{aligned}}$ Now, let $\displaystyle {\frac {1}{x^{2}-1}}={\frac {A}{x-1}}+{\frac {B}{x+1}}$ by the partial fraction decomposition. This gives via cross multiplication $\displaystyle 1=A(x+1)+B(x-1)$ METHOD 1: Plugging in $\displaystyle x=-1$ shows that $\displaystyle B=-{\frac {1}{2}}$ and plugging in $\displaystyle x=1$ shows that $\displaystyle A={\frac {1}{2}}$ . METHOD 2: Expanding the right yields $\displaystyle 1=A-B+(A+B)x$ Since the coefficients of the constant term are equal and the coefficients of the x term must be equal, we have $\displaystyle A-B=1$ and $\displaystyle A+B=0$ Adding the two equations gives $\displaystyle 2A=1$ so $\displaystyle A={\frac {1}{2}}$ as above and subtracting the two equations gives $\displaystyle -2B=1$ and so $\displaystyle B={\frac {-1}{2}}$ . Now that we have the coefficients, we note that $\displaystyle {\frac {1}{x^{2}-1}}={\frac {\tfrac {1}{2}}{x-1}}+{\frac {\tfrac {-1}{2}}{x+1}}$ and go back to the integral to see that ${\begin{aligned}\int _{0}^{2}{\frac {1}{x^{2}-1}}\ dx&=\int _{0}^{1}{\frac {1}{x^{2}-1}}\ dx+\int _{1}^{2}{\frac {1}{x^{2}-1}}\ dx\\&=\lim _{b\to 1^{-}}\int _{0}^{b}{\frac {1}{x^{2}-1}}\ dx+\lim _{a\to 1^{+}}\int _{a}^{2}{\frac {1}{x^{2}-1}}\ dx\\&=\lim _{b\to 1^{-}}\int _{0}^{b}\left({\frac {\tfrac {1}{2}}{x-1}}+{\frac {\tfrac {-1}{2}}{x+1}}\right)\ dx+\lim _{a\to 1^{+}}\int _{a}^{2}\left({\frac {\tfrac {1}{2}}{x-1}}+{\frac {\tfrac {-1}{2}}{x+1}}\right)\ dx\\&=\lim _{b\to 1^{-}}{\tfrac {1}{2}}\ln \|x-1\|-{\tfrac {1}{2}}\ln \|x+1\|{\big \|}_{0}^{b}+\lim _{a\to 1^{+}}{\tfrac {1}{2}}\ln \|x-1\|-{\tfrac {1}{2}}\ln \|x+1\|{\big \|}_{a}^{2}\\&=\lim _{b\to 1^{-}}{\tfrac {1}{2}}\ln \|b-1\|-{\tfrac {1}{2}}\ln \|b+1\|-({\tfrac {1}{2}}\ln \|1\|-{\tfrac {1}{2}}\ln \|1\|)\\&\quad +\lim _{a\to 1^{+}}{\tfrac {1}{2}}\ln \|1\|-{\tfrac {1}{2}}\ln \|3\|-({\tfrac {1}{2}}\ln \|a-1\|-{\tfrac {1}{2}}\ln \|a+1\|)\\\end{aligned}}$ Since $\displaystyle \lim _{b\to 1^{-}}{\tfrac {1}{2}}\ln \|b-1\|$ diverges, we have that the entire integral diverges and thus the integral does not exist.

[show] Solution 2
First, note that the integral above is improper with a singularity at $\displaystyle x=1$ . Splitting up the integral there, we see that ${\begin{aligned}\int _{0}^{2}{\frac {1}{x^{2}-1}}\ dx&=\int _{0}^{1}{\frac {1}{x^{2}-1}}\ dx+\int _{1}^{2}{\frac {1}{x^{2}-1}}\ dx\\&=\lim _{b\to 1^{-}}\int _{0}^{b}{\frac {1}{x^{2}-1}}\ dx+\lim _{a\to 1^{+}}\int _{a}^{2}{\frac {1}{x^{2}-1}}\ dx\\\end{aligned}}$ Now, let $\displaystyle {\frac {1}{x^{2}-1}}={\frac {A}{x-1}}+{\frac {B}{x+1}}$ by the partial fraction decomposition. We note that ${\begin{aligned}\int _{0}^{1}{\frac {1}{x^{2}-1}}\ dx&=\int _{0}^{1}{\frac {A}{x-1}}\ dx+\int _{0}^{1}{\frac {B}{x+1}}\ dx\end{aligned}}$ and ${\begin{aligned}\int _{0}^{1}{\frac {A}{x-1}}\ dx\end{aligned}}$ doesn't seem to converge since it behaves like $\int _{-1}^{0}{\frac {1}{x}}\ dx$ , so the whole integral does not converge. (In particular, we don't even need to worry about what $\displaystyle A$ and $\displaystyle B$ are, since they are non-zero constants!) Let's prove this rigorously. ${\begin{aligned}\int _{0}^{1}{\frac {A}{x-1}}\ dx&=A\cdot \lim _{b\to 1^{-}}\int _{0}^{b}{\frac {1}{x-1}}\ dx\\&=A\cdot \lim _{b\to 1^{-}}\ln \|x-1\|_{0}^{b}\\&=A\cdot \left(\lim _{b\to 1^{-}}\ln \|b-1\|-\ln \|0-1\|\right)\\\end{aligned}}$ Since $\displaystyle \lim _{b\to 1^{-}}\ln \|b-1\|$ diverges, we have that the integral diverges and thus the entire integral does not exist.