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$$\text{Does }\int _0^{\infty }\:\left(\frac{\arctan\left(x\right)}{\:2+e^{3x}}\right)dx \text { converge/diverge?}$$

I have to use the convergence/divergence theorem to check if that integral converges or diverges.

I notice that:

$$\frac{\arctan(x)}{2+e^{3x}} \leq \frac{\arctan(x)}{e^{3x}} \leq \frac{\arctan(x)}{e^{x}}$$

I also notice that $$-\pi/2 \leq \arctan(x) \leq \pi/2$$.

So I can say that

$$\frac{\arctan(x)}{2+e^{3x}} \leq \frac{\arctan(x)}{e^{3x}} \leq \frac{\arctan(x)}{e^{x}} \ \leq \frac{\pi/2}{e^{x}}$$

So here, would I just prove that $$\int^{\infty}_{0}\frac{\pi/2}{e^x}dx$$ converges?

Could I also take it one more step and say that

$$\frac{\pi/2}{e^x} \leq {\pi/2}$$ since we want to maximize the numerator while making the denominator smaller, and we know that $\forall x \in [0,\infty), 1\leq e^x \leq \infty$?

So in the end, can I just prove that $\int^{\infty}_{0} \pi/2 dx$ converges?

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There are a couple of issues in the OP that need to be addressed. First, we should bound the integrand as

$$\left|\frac{\arctan(x)}{2+e^{3x}}\right|\le \frac{\pi/2}{2+e^{3x}}$$

Note the absolute value sign prevents the possibility of the integrand from running to far negative. In this problem, that isn't an issue inasmuch as the integrand is non-negative on the domain of integration.


The first question from the OP is:

"So I can say that

$$\frac{\arctan(x)}{2+e^{3x}} \leq \frac{\arctan(x)}{e^{3x}} \leq \frac{\arctan(x)}{e^{x}} \ \leq \frac{\pi/2}{e^{x}}"$$

">"So here, would I just prove that $$\int^{\infty}_{0}\frac{\pi/2}{e^x}dx$$ converges?"

Yes, the inequalities are correct and one can proceed to show

$$\begin{align} \left|\int_0^\infty \frac{\arctan(x)}{2+e^{3x}}\,dx\right|&\le \frac{\pi}{2}\int_0^\infty e^{-x}\,dx\\\\ &=1\\\\ &<\infty \end{align}$$

And we are done!


The second question from the OP is:

"Could I also take it one more step and say that

$$\frac{\pi/2}{e^x} \leq {\pi/2}$$ since we want to maximize the numerator while making the denominator smaller, and we know that $\forall x \in [0,\infty), 1\leq e^x \leq \infty$?"

"So in the end, can I just prove that $\int^{\infty}_{0} \pi/2 dx$ converges?"

No. While it is true that $\frac{\pi/2}{e^x}\le \frac{\pi}{2}$ for $x\in[0,\infty)$, this upper bound provides no insight into the convergence or lack thereof of the integral of interest. That is to say, it shows

$$\left|\int_0^L\frac{\arctan(x)}{2+e^{3x}}\,dx\right| \le \int_0^L \frac{\pi}{2}\,dx\to \infty\,\,\text{as}\,\,L\to \infty \tag 1$$

So, from $(1)$ we could not deduce that the integral of interest converges.

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You are very close when you write

"So here, would I just prove that $$\int^{\infty}_{0}\frac{\pi/2}{e^x}dx$$ converges?"

And the answer is yes.

Since $(e^{-x})' = -e^{-x}$, $\int e^{-x} dx = -e^{-x}$, so $\int_0^{\infty} e^{-x} dx = (-e^{-x})\big|_0^{\infty} =1$, so it converges.

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