# Thread: Comparison test for improper integrals problem

1. ## Comparison test for improper integrals problem

I have a problem using the comparison test for integrals and I'm stuck. The question asks whether the integral of (arctanx)/(2+e^x) dx from 0 to infinity diverges or converges using the comparison test for improper integrals.

My reasoning is that you can ignore the arctanx since the max value of that is pi/2, so it's pretty much a constant so you take that out. Then compare 1/(2+e^x) to 1/(e^x) to see if that diverges. However, how do you find the integral of 1/(e^x) in order to see if it converges or diverges. Should I even compare the original with 1/(e^x)? Help please? Thanks.

2. Since $\displaystyle \arctan x<\frac{\pi }{2}$ for all $\displaystyle x\in\mathbb R$ then $\displaystyle \frac{\arctan x}{2+e^{x}}<\frac{\pi }{2\left( 2+e^{x} \right)}<\frac{\pi }{e^{x}}=\pi e^{-x},$ and your integral converges since $\displaystyle \int_0^\infty e^{-x}\,dx$ does.

3. Thanks Krizalid, but how do you know that the integral of e^(-x) from 0 to infinity converges? Is the integral of that simply -e^(-x)? That's what I get from u-substitution but an online integrator gives me an expression with cosh and sinh.

4. $\displaystyle \int_{0}^{\infty }{e^{-x}\,dx}=\underset{\alpha\to \infty }{\mathop{\lim }}\,\int_{0}^{\alpha }{e^{-x}\,dx}=\underset{\alpha\to \infty }{\mathop{\lim }}\,\left( 1-e^{-\alpha } \right)=1.$