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Math Help - Improper Integral

  1. #1
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    Improper Integral

    I'm having trouble determining if this integral is convergent or not:

    \int_0^{\infty}\frac{xArctan(x)}{(1+x^2)^2}dx=\lim  _{t->\infty}\int_0^{t}\frac{xArctan(x)}{(1+x^2)^2}dx

    If I graph the integrand, it appears that

    \frac{xArctan(x)}{(1+x^2)^2}\leq \frac{Arctan(x)}{1+x^2} for x>0

    So I attempted the comparison theorem:

    \lim_{t->\infty}\int_0^{\infty}\frac{Arctan(x)}{1+x^2}dx

    u=Arctan(x)\Rightarrow du=\frac{dx}{1+x^2}

    \lim_{t->\infty}\int_0^{Arctan(t)}udu

    =\int_0^{\frac{\pi}{2}}udu

    This is divergent right? So this was useless??
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  2. #2
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    wait


    or perhaps...

    \lim_{t->\infty}\int_0^{Arctan(t)}udu

    =\lim_{t->\infty}\frac{1}{2}(Arctan(t))^2-0

    =\frac{\pi^2}{8}
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  3. #3
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    split it into two sets, one for [0,1] and the other one for [1,\infty), so for each x\ge1 is \frac{x\arctan x}{\left( 1+x^{2} \right)^{2}}<\frac{\arctan x}{x^{3}}<\frac{\pi }{2x^{3}}, and the integral converges.
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  4. #4
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    I understand everything except the point of writing:

    \int_0^{\infty}\frac{xArctan(x)}{(1+x^2)^2}dx=\int  _0^1\frac{xArctan(x)}{(1+x^2)^2}dx+\int_1^{\infty}  \frac{xArctan(x)}{(1+x^2)^2}dx

    Why split the interval?
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  5. #5
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    i could've picked any number there, but to make my life easier i took a "one."

    besides, one doesn't care about the first piece, since is a continuous function being integrated over a compact set.
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