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anonymous
 one year ago
Related to my previous question: Not as difficult, but good practice nonetheless:
\[\int_1^\infty \arctan\frac{1}{x^2}\,dx\]
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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dan815
 one year ago
Best ResponseYou've already chosen the best response.0im a series noob, i can only look at this stuff geometrically and try to get something out

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0No series needed, but be my guest :)

dan815
 one year ago
Best ResponseYou've already chosen the best response.0im gonna assume the rate at which the angles are decreasing is fast enougnh and this does infact converge

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Right, my previous question addressed the summation. It definitely converges because the integral converges.

dan815
 one year ago
Best ResponseYou've already chosen the best response.0how do you actually work out the integral of arctan

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0One 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0As for the integral of \(\arctan\), we have \[\int\arctan x\,dx=x\arctan x\int\frac{x}{1+x^2}\,dx\\ \begin{matrix} u=\arctan x&&&dv=dx\\ du=\dfrac{dx}{1+x^2}&&&v=x \end{matrix}\]

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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Gotta love complex analysis :)
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