Science:Math Exam Resources/Courses/MATH101/April 2012/Question 03 (d)

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[hide] Question 03 (d)
Determine, with explanation, whether the improper integral $\int _{-\infty }^{\infty }{\frac {x}{x^{2}+1}}\,dx$ converges or diverges.

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[show] Hint 1
Note that this improper integral has both upper and lower limits of integration being improper. To write it as a standard Type I improper integral, one first has to split it into two integrals. Then, if both integrals converge, the original integral converges. But if at least one of the integrals diverges, then the original integral diverges.

[show] Hint 2
Since this is a an improper integral, you can not argue that the integral should be zero (and hence converge) because the integrand is an odd function. Such an argument is only valid for proper integrals.

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[show] Solution
The integral is equal to the sum of the two standard Type I integrals: $\int _{-\infty }^{0}{\frac {x}{x^{2}+1}}dx+\int _{0}^{\infty }{\frac {x}{x^{2}+1}}dx$ . By the substitution u=x²+1, du = 2dx, we see that ${\begin{aligned}\int {\frac {x}{x^{2}+1}}dx&={\frac {1}{2}}\int {\frac {1}{u}}du\\&={\frac {1}{2}}\ln \|u\|+C\\&={\frac {1}{2}}\ln \|x^{2}+1\|+C\end{aligned}}$ So, we calculate ${\begin{aligned}\int _{0}^{\infty }{\frac {x}{x^{2}+1}}dx&=\lim _{a\to \infty }{\frac {1}{2}}\ln(x^{2}+1){\big \|}_{0}^{a}\\&=\lim _{a\to \infty }({\frac {1}{2}}\ln(a^{2}+1)-\ln(1))\\&=\infty \end{aligned}}$ Since this part diverges the whole integral diverges also.