2
$\begingroup$

How to evaluate $$\int_{0}^{\infty }\frac{x\arctan x}{1+x^{4}}\, dx$$ Define $$I\left ( a \right )=\int_{0}^{\infty }\frac{x\arctan \left ( ax \right )}{1+x^{4}}\, \mathrm{d}x\Rightarrow I'\left ( a \right )=\int_{0}^{\infty }\frac{x^{2}}{\left ( 1+a^{2}x^{2}\right )\left (1+x^{4} \right )}\, \mathrm{d}x$$ but it seems too complicated,is there another way for it?

  • 0
    this is a simple integral if you are familiar with the residue theorem – 2017-01-10

2 Answers 2

7

You can do (here we do $x\mapsto 1/x$ and use the fact that $\arctan(1/x)=\pi/2-\arctan x$) $$ \begin{aligned} \int_0^{+\infty}\frac{x\arctan x}{1+x^4}\,dx&=\int_0^{+\infty}\frac{x(\pi/2-\arctan x)}{1+x^4}\,dx\\ &=\frac{\pi}{4}\int_0^{+\infty}\frac{x}{1+x^4}\,dx\\ &=\frac{\pi}{4}\bigl[\frac{1}{2}\arctan(x^2)\bigr]_0^{+\infty}=\frac{\pi^2}{16}. \end{aligned} $$

  • 3
    symmetry based solutions rock! (+1) – 2017-01-10
  • 0
    sry,didn't see your answer just now – 2017-01-10
  • 0
    @Renascence_5. Don't worry, I know how it feels. You get a +1 from me since I agree with your solution :) – 2017-01-10
4

I think this is the best way!

Let $x\rightarrow \dfrac{1}{x}$, we get $$\int_{0}^{\infty }\frac{x\arctan x}{1+x^{4}}\, \mathrm{d}x=\int_{0}^{\infty }\frac{x\arctan\left ( \dfrac{1}{x} \right )}{1+x^4}\, \mathrm{d}x$$ notice that $$\arctan x +\arctan\left ( \dfrac{1}{x} \right ) =\frac{\pi}{2}$$hence \begin{align*} \int_{0}^{\infty }\frac{x\arctan x}{1+x^{4}}\, \mathrm{d}x&=\frac{1}{2}\int_{0}^{\infty }\frac{x\left (\arctan x +\arctan\left ( \dfrac{1}{x} \right ) \right )}{1+x^{4}}\, \mathrm{d}x\\ &=\frac{\pi }{4}\int_{0}^{\infty }\frac{x}{1+x^{4}}\, \mathrm{d}x \end{align*} then you can take it from here.