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Math Help - Integral convergance

  1. #1
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    Integral convergance

    I am suppose to evaluate the integral of 1/(x^2 + 1) from 1 to infinity.

    So I evaluated it from 1 to b and then took the antiderivative and got the inverse tan of b - inverse tan of 1. I think this integral converges at -pi/2 but the answer shows it converges at pi/4. Can somone tell me what I am doing wrong? I know that as be gets very large the inverse tan of it goes to zero but I can not figure out how to get pi/4!!!!! Thanks to all of you out there who help!
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Frostking View Post
    I am suppose to evaluate the integral of 1/(x^2 + 1) from 1 to infinity.

    So I evaluated it from 1 to b and then took the antiderivative and got the inverse tan of b - inverse tan of 1. I think this integral converges at -pi/2 but the answer shows it converges at pi/4. Can somone tell me what I am doing wrong? I know that as be gets very large the inverse tan of it goes to zero but I can not figure out how to get pi/4!!!!! Thanks to all of you out there who help!
    You are correct. The antiderivative is arctangent.

    However, by the FTC, we have {\color{blue}\lim_{b\to\infty}\tan^{-1}\left(b\right)}-\tan^{-1}\left(1\right)

    The limit (in blue) yields the value of \frac{\pi}{2}

    Thus, we are left with \frac{\pi}{2}-\tan^{-1}\left(1\right)=\frac{\pi}{2}-\frac{\pi}{4}=\color{red}\boxed{\frac{\pi}{4}}

    Does this make sense?

    --Chris
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  3. #3
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    Integral convergence at pi/4

    So, can I think of it as arctan equal to zero degrees which would be at pi/2 and then subtracting the pi/4 to get pi/4??? or is this too simplistic a view. My trig class was back about 30 years ago so I am rusty to say the least. Thanks very much for your time and willingness to help!
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