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Thread: Improper Integral and Comparison Theorem

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    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:

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

    $\displaystyle x\geq 0$

    $\displaystyle 0<Arctan(x)<\frac{\pi}{2}$

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

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

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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by adkinsjr View Post
    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:

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

    $\displaystyle x\geq 0$

    $\displaystyle 0<Arctan(x)<\frac{\pi}{2}$

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

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

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