Find the value of $\lim _{a \to \infty} \frac{1}{a} \int_{0}^{\infty} \frac{x^{2}+a x+1}{1+x^{4}} \cdot \tan ^{-1}\left(\frac{1}{x}\right) \,d x $

Question

Find the value of : $$
\lim _{a \rightarrow \infty} \frac{1}{a} \int_{0}^{\infty} \frac{x^{2}+a x+1}{1+x^{4}} \cdot \tan ^{-1}\left(\frac{1}{x}\right) \,d x
$$ I have tried to evaluate this integral by L'Hospitals rule, by separately diffrentiating the numerator and the denominator in the following steps: $$\int_{0}^{\infty} \frac{x^{2}+a x+1}{1+x^{4}} d x\cdot\dfrac{\pi}{2} (\text{by taking x=1/t and adding the integrands)}
$$ I end up converting the above integral in the form $(\text{by Leibnitz's integral rule})$ $$ \dfrac{\pi}{2}
\left(\int_{0}^{\infty} \frac{x}{1+x^{4}} d x\right)
$$ please provide an approach to this problem after this step.

Answer 1

We can rewrite the integral as

$$\lim_{a\to\infty} \frac{1}{a}\int_0^\infty \frac{x^2+1}{1+x^4}\tan^{-1}\left(\frac{1}{x}\right)\:dx + \int_0^\infty \frac{x}{1+x^4}\tan^{-1}\left(\frac{1}{x}\right)\:dx$$

which we are allowed to split up since both pieces are absolutely convergent. Then notice that

$$ \frac{x}{1+x^4}\tan^{-1}\left(\frac{1}{x}\right) = \frac{1}{x^2}\cdot \frac{\frac{1}{x}}{1+\frac{1}{x^4}}\tan^{-1}\left(\frac{1}{x}\right) = \frac{d}{dx}\left[-\frac{1}{4}\arctan^2\left(\frac{1}{x}\right)\right]$$

therefore the integral evaluates to $\frac{\pi^2}{16}$