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Math Help - Improper integral, tarctan(t)/(1+t^2)^2, from 0 to infinity, convergent or divergent?

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    Improper integral, tarctan(t)/(1+t^2)^2, from 0 to infinity, convergent or divergent?

    Hello,

    I just had a quiz and I did not know how to do this question even though it looked so familiar.

    Is the improper integral, from 0 to infinity,
    tarctan(t)/(1+t^2)^2
    convergent or divergent?

    I tried using comparison theorem but it didn't work out.
    And then I just tried solving the integral by substitution as in u=arctan(t) and du=1/1+t^2 dt but it didn't cancel out!

    Please help! Thanks in advance!
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by electricsparks View Post
    Hello,

    I just had a quiz and I did not know how to do this question even though it looked so familiar.

    Is the improper integral, from 0 to infinity,
    tarctan(t)/(1+t^2)^2
    convergent or divergent?

    I tried using comparison theorem but it didn't work out.
    And then I just tried solving the integral by substitution as in u=arctan(t) and du=1/1+t^2 dt but it didn't cancel out!

    Please help! Thanks in advance!
    comparison didn't work? really?

    try again: \int_0^\infty \frac {t \arctan t}{(1 + t^2)^2}~dt \le \frac {\pi}2 \int_0^\infty \frac t{(1 + t^2)^2}~dt
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  3. #3
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    Of course, it's a convergent integral. (Even an absolutely one.)

    The integral converges as long as \int_1^\infty\frac{t\arctan t}{(1+t^2)^2}\,dt<\infty, which in efect, it's convergent, since for all t\ge1 is \frac{t\arctan (t)}{\left( 1+t^{2} \right)^{2}}<\frac{\pi t}{2t^{4}}=\frac{\pi }{2t^{3}}. Thus our integral is bounded by a convergent integral, hence the original integral converges.

    (Beaten, ahahah.)
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  4. #4
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    Why don't we compute the exact value of the integral ?


    I =  \int_0^{\infty} \frac{x \tan^{-1}(x) }{ ( 1 + x^2 )^2 } ~dx

    Sub.  x = \frac{1}{t}  , dx = - \frac{dt}{t^2}


     I = \int_0^{\infty} \frac{t \tan^{-1}(\frac{1}{t})}{ ( t^2 + 1)^2 }~dt

    but \tan^{-1}(\frac{1}{t}) = \frac{\pi}{2} - \tan^{-1}(t)


    so we have  I = \frac{\pi}{2}\int_0^\infty \frac t{(1 + t^2)^2}~dt - I<br />

     2I = \frac{\pi}{2} \frac{1}{2} \implies I = \frac{\pi}{8}
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  5. #5
    Moo
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    simplependulum : theoretically, before using integration techniques (substitution, integration by parts, ...), you have to prove that it converges first !
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  6. #6
    Math Engineering Student
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    I really don't care about computing its value, I'm just doing what the problem asks.

    I can put \int_0^\infty\frac{\sin x}x\,dx and ask to show its convergence, and I know someone's gonna appear and batantly say \int_0^\infty\frac{\sin x}x\,dx=\frac\pi2, which is far of the answer that I was looking for.

    I know how to compute its value, but in this case, we're talking about convergence, besides, we don't know if the OP has covered computations on improper integrals.
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