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 $\approx .408108504052.$

  • $\begingroup$ You can't evaluate an integral with a comparison test. You can only use it to try to determine if something converges or not. $\endgroup$ – Jonathan Oct 2 '12 at 4:42
  • $\begingroup$ @Jonathan Valid point. Will change the wording - thanks. $\endgroup$ – Joe Oct 2 '12 at 4:43


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

  • 1
    $\begingroup$ 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? $\endgroup$ – Joe Oct 2 '12 at 4:53

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