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anonymous

  • one year ago

Related to my previous question: Not as difficult, but good practice nonetheless: \[\int_1^\infty \arctan\frac{1}{x^2}\,dx\]

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  1. dan815
    • one year ago
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    im a series noob, i can only look at this stuff geometrically and try to get something out

  2. anonymous
    • one year ago
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    No series needed, but be my guest :)

  3. dan815
    • one year ago
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    |dw:1435243608410:dw|

  4. dan815
    • one year ago
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    im gonna assume the rate at which the angles are decreasing is fast enougnh and this does infact converge

  5. anonymous
    • one year ago
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    Right, my previous question addressed the summation. It definitely converges because the integral converges.

  6. dan815
    • one year ago
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    how do you actually work out the integral of arctan

  7. anonymous
    • one year ago
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    One possible way is to first set \(\sqrt u=\dfrac{1}{x}\) so that \(\dfrac{du}{2\sqrt u}=-\dfrac{dx}{x^2}\), giving \[\int_1^\infty \arctan\frac{1}{x^2}\,dx=\frac{1}{2}\int_0^1\frac{\arctan u}{u^{3/2}}\,du\] then apply IBP.

  8. anonymous
    • one year ago
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    As for the integral of \(\arctan\), we have \[\int\arctan x\,dx=x\arctan x-\int\frac{x}{1+x^2}\,dx\\ \begin{matrix} u=\arctan x&&&dv=dx\\ du=\dfrac{dx}{1+x^2}&&&v=x \end{matrix}\]

  9. anonymous
    • one year ago
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    1..

  10. ganeshie8
    • one year ago
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    just for the sake of alternative \[\int_1^\infty \arctan\frac{1}{x^2}\,dx=\int_1^\infty \arg (x^2+i)\,dx = -i\int_1^\infty \log (\text{sgn}(x^2+i))\,dx \approx \href{http:///www.wolframalpha.com/input/?s=15&_=1435358463969&i=%5cint_1%5e%7b%5cinfty%7d+-i*log(sgn(x%5e2%2bi))+dx&fp=1&incTime=true}{0.95}\]

  11. anonymous
    • one year ago
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    Gotta love complex analysis :)

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