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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 $\approx .408108504052.$

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    @Jonathan Valid point. Will change the wording - thanks.2012-10-02

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HINT

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

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    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?2012-10-02