# Improper Integral and Comparison Theorem

• Dec 14th 2009, 06:57 PM
Improper Integral and Comparison Theorem
I think this is my third post regarding the Comparison theorem. I can rarely tell if my reasoning is correct when using this theorem. Please LMK if I'm wrong. I want determine if the following is convergent:

$\int_0^{\infty}\frac{Arctan(x)}{2+e^x}dx$

$x\geq 0$

$0

$\frac{Arctan(x)}{2+e^x}<\frac{\pi}{4+2e^x}<\frac{\ pi}{4}$

Since $\lim_{t->\infty}\int_0^t\frac{\pi}{4}dx$ diverges, the integral $\int_0^{\infty}\frac{Arctan(x)}{2+e^x}dx$ diverges by the Comparison Theorem.

(Thinking)
• Dec 14th 2009, 08:20 PM
Drexel28
Quote:

I think this is my third post regarding the Comparison theorem. I can rarely tell if my reasoning is correct when using this theorem. Please LMK if I'm wrong. I want determine if the following is convergent:

$\int_0^{\infty}\frac{Arctan(x)}{2+e^x}dx$

$x\geq 0$

$0

$\frac{Arctan(x)}{2+e^x}<\frac{\pi}{4+2e^x}<\frac{\ pi}{4}$

Since $\lim_{t->\infty}\int_0^t\frac{\pi}{4}dx$ diverges, the integral $\int_0^{\infty}\frac{Arctan(x)}{2+e^x}dx$ diverges by the Comparison Theorem.

(Thinking)

You are a tad off. Note that $\frac{-\pi}{2}\le\arctan(x)\le\frac{\pi}{2}\implies\frac{-\pi}{e^x}\le\frac{-\pi}{2\left(2+e^x\right)}\le\frac{\arctan(x)}{2+e^ x}\le\frac{\pi}{2\left(2+e^x\right)}\le\frac{\pi}{ e^x}$. Taking the integral of the terms leads to the result that the integral converges.