Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Only using the Comparison test, I am trying to see if the following integral converges: $$\int_0^{\infty} \frac{\arctan x} {2+e^{x}} \ dx$$

I first noted that $\arctan x \lt (2+e^{x}) \ \forall x \in \mathbb{R}$ which allows me to say that

$$\int_0^{\infty} \frac{\arctan x} {2+e^{x}} \ dx \lt \infty$$

I'm not sure where to progress from here though.

Mathematica reports the integral converging to $\approx .408108504052.$

share|cite|improve this question
You can't evaluate an integral with a comparison test. You can only use it to try to determine if something converges or not. – Jonathan Oct 2 '12 at 4:42
@Jonathan Valid point. Will change the wording - thanks. – Joe Oct 2 '12 at 4:43
up vote 2 down vote accepted


$\vert\arctan(x)\vert \in \left[ 0, \pi/2\right]$ and $2+e^x > e^x$. Can you now finish it off?

share|cite|improve this answer
Right now I have: $$\int_0^{\infty} \frac {\arctan x} {2+e^x} \ dx \lt \int_0^{\infty} \frac {\pi}{2(2+e^x)} \ dx \lt \int_0^{\infty}\frac{\pi}{2(e^x)} = \frac{\pi}{2}$$ I believe that's all I can say about this integral - that it converges to $\lt \frac{\pi}{2}$. Is that correct? – Joe Oct 2 '12 at 4:53

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.