Show the integral is convergent and find the value it converges to.

$$\int_1^\infty \frac{\arctan x}{x^2} ~dx$$

I have found the indefinite integral to be

$$-\frac{\arctan x}{x} + \ln|x| -\frac{1}{2}\ln(x^2+1)$$

When I take the limit as t goes to $\infty$, I end up having to deal with this

$$\lim_{t\to \infty}\left(-\frac{\arctan x}{x} + \ln|x| -\frac{1}{2}\ln(x^2+1)\right)$$

I know the first term goes to 0, but how do I find the limit of the rest of it? I tried combining the two terms to see if I could use L'Hospital's rule, but I couldn't get anywhere.

I got to this and was not sure how to proceed.

$$\lim_{t\to \infty}\left(\ln \left(\frac{|t|}{\sqrt{t^2+1}}\right)\right)$$

  • 2
    $\begingroup$ Use the properties of the logarithm: $$\ln|x| - \frac{1}{2} \ln (x^2+1) = \ln \frac{|x|}{\sqrt{x^2+1}} \to \ln 1 =0$$ $\endgroup$ – Crostul Mar 12 '16 at 22:57
  • $\begingroup$ Right I was able to reach that point, but I am not sure how to prove that it is equal to zero. $\endgroup$ – Zack Mar 12 '16 at 23:06
  • $\begingroup$ use l'Hopital again? $\endgroup$ – Jens Renders Mar 12 '16 at 23:18
  • 1
    $\begingroup$ You can show convergence without checking the improper integral; $\arctan(x)/x^2$ is nonnegative for $x > 1$ and $\arctan(x) \le \pi/2$ for all $x \in \mathbb R$, so your integral is bounded by $\displaystyle{\frac{\pi}{2}\int_1^\infty \frac{dx}{x^2}} = \frac{\pi}{2} < \infty$ and thus converges. $\endgroup$ – amcerbu Mar 13 '16 at 0:35

Using the continuity of the logarithm, we have

$$\begin{align} \lim_{t\to \infty}\log\left(\frac{t}{\sqrt{t^2+1}}\right)&=\log\left(\lim_{t\to \infty}\frac{t}{\sqrt{t^2+1}}\right)\\\\ &=\log(1)\\\\ &=0 \end{align}$$

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