Find $\lim_{a\to \infty}\frac{1}{a}\int_0^{\infty}\frac{x^2+ax+1}{1+x^4}\cdot\arctan(\frac{1}{x})dx$

Question

Find $$ \lim_{a\to \infty} \frac{1}{a} \int_0^{\infty}\frac{x^2+ax+1}{1+x^4} \arctan\left(\frac{1}{x}\right)dx $$

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I tried to find $$ \int_0^{\infty} \frac{x^2+ax+1}{1+x^4}\arctan\left(\frac{1}{x}\right) dx $$ Let $$ I(a) = \int_0^\infty \frac{x^2+ax+1}{1+x^4}\arctan\left(\frac{1}{x}\right) dx $$ Let $$ \begin{split} t &= \arctan(1/x) \\ \frac{dt}{dx} &= \frac{-1}{1+x^2} \\ dt &= \frac{-1}{1+x^2}dx \end{split} $$

I am stuck here, there seems no way to further solving.

Answer 1

One may observe that, as $a\to \infty$, $$ \begin{align} &\frac1a \int_0^{\infty}\frac{x^2+ax+1}{1+x^4} \arctan\left(1/x\right)\:dx- \color{red}{\int_0^{\infty}\!\!\frac{x}{1+x^4} \arctan\left(1/x\right)\:dx} \\\\&=\frac1a\int_0^{\infty}\frac{x^2+ax+1}{1+x^4} \arctan\left(1/x\right)\:dx- \frac1a\int_0^{\infty}\frac{ax}{1+x^4} \arctan\left(1/x\right)\:dx \\\\&=\frac1a\int_0^{\infty}\frac{x^2+1}{1+x^4} \arctan\left(1/x\right)\:dx \: \longrightarrow \: 0,\tag1 \end{align} $$ since the latter integral is convergent.

On the other hand, we have $$ \int_0^{\infty}\frac{x}{1+x^4} \:\arctan\left(1/x\right)\:dx=\int_0^{\infty}\frac{x}{1+x^4}\: \arctan\left(x\right)\:dx\quad (x \to 1/x) $$ using, $\displaystyle \arctan\left(x\right)+\arctan\left(1/x\right)=\frac{\pi}2$ ($x>0$), one gets $$ \begin{align} &\int_0^{\infty}\!\!\frac{x}{1+x^4}\: \arctan\left(x\right) \:dx+\int_0^{\infty}\!\!\frac{x}{1+x^4}\:\arctan\left(1/x\right)\:dx \\\\& =\frac{\pi}2\int_0^{\infty}\!\!\frac{x\:dx}{1+x^4} \\\\&=\frac{\pi}4\int_0^{\infty}\!\!\frac{du}{1+u^2} \\\\& =\frac{\pi^2}8, \end{align} $$ thus $$\displaystyle \color{red}{\int_0^{\infty}\!\frac{x}{1+x^4}\:\arctan\left(1/x\right)\:dx}=\frac{\pi^2}{16}$$ giving, from $(1)$,

$$ \lim_{a\to \infty} \frac{1}{a} \int_0^{\infty}\frac{x^2+ax+1}{1+x^4}\: \arctan\left(\frac{1}{x}\right)dx=\frac{\pi^2}{16}.\tag2 $$