## anonymous one year ago Related to my previous question: Not as difficult, but good practice nonetheless: $\int_1^\infty \arctan\frac{1}{x^2}\,dx$

1. dan815

im a series noob, i can only look at this stuff geometrically and try to get something out

2. anonymous

No series needed, but be my guest :)

3. dan815

|dw:1435243608410:dw|

4. dan815

im gonna assume the rate at which the angles are decreasing is fast enougnh and this does infact converge

5. anonymous

Right, my previous question addressed the summation. It definitely converges because the integral converges.

6. dan815

how do you actually work out the integral of arctan

7. anonymous

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

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

1..

10. ganeshie8

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

Gotta love complex analysis :)