Improper integral, infinite limits of integration

[nick.martinez]

∞
∫ x/(x^2+1) dx
-∞

I basicaly evaluated the integral and Ln (x^2+1) as the antiderivative and when taking the limits I get ∞-∞

(ln |1| -ln|b+1|) + (ln|n+1|- ln|1|)

lim b-> neg. infinity lim n-> infinity

does this function converge or diverge? this was a question on the quiz and everyone in class after said the function converges to zero. I'm a littled puzzled by this.

 

[pwsnafu]

##\int_{-\infty}^\infty \frac{x}{x^2+1} \, dx = \lim_{a \to -\infty} \int_{a}^{0} \frac{x}{x^2+1} \, dx + \lim_{b\to\infty} \int_{0}^{b} \frac{x}{x^2+1} \, dx##

The limits are separate, and you need convergence in both. Your friends are thinking of the Cauchy principle value, but unless the problem explicitly asks for it, you should not give that as the answer.

 

[nick.martinez]

so does it converge or diverge?

 

[nick.martinez]

Also I looked on harvard, and university of portlands website, and it says that the function diverges.

http://www.math.harvard.edu/~keerthi/files/1b_fall_2011/worksheet11_solns.pdf
end of pg. 1 and beginning of page 2

 

[nick.martinez]

##\int_{-\infty}^\infty \frac{x}{x^2+1} \, dx = \lim_{a \to -\infty} \int_{a}^{0} \frac{x}{x^2+1} \, dx + \lim_{b\to\infty} \int_{0}^{b} \frac{x}{x^2+1} \, dx##

The limits are separate, and you need convergence in both. Your friends are thinking of the Cauchy principle value, but unless the problem explicitly asks for it, you should not give that as the answer.

Also, when looking at the antiderivative of Ln|x^2+1| i see that it is an even function does this mean you can subtract ∞-∞ since the function is approaching infinity at the same rate? If not, when is this method applied, if it is at all? ha? I really want to understand this problem. Apparently my teacher loves surprising us on tests and quizzes.

 

[jbunniii]

so does it converge or diverge?

For large positive values of ##x##, the integrand is approximately ##1/x##. What can you conclude?

 

[nick.martinez]

it would have to diverge because taking the limit as Ln|x| as x-> infinity diverges, correct?

 

[jbunniii]

it would have to diverge because taking the limit as Ln|x| as x-> infinity diverges, correct?

Yes, that's right.