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

Your solution attempt honestly works just fine as well (and it's also the first thing I thought of when I saw the question!). Indeed from ${x=\frac{1}{t}}$ you get

$${\int_{0}^{\infty}\frac{x^2 + ax + 1}{1+x^4}\arctan\left(\frac{1}{x}\right)dx=\int_{0}^{\infty}\frac{\left(\frac{1}{t}\right)^2 + \frac{a}{t} + 1}{1 + \left(\frac{1}{t}\right)^4}\arctan(t)\frac{1}{t^2}dt}$$

$${=\int_{0}^{\infty}\frac{\left(\frac{1}{t}\right)^4 + \frac{a}{t^3} + \frac{1}{t^2}}{1 + \left(\frac{1}{t}\right)^4}\arctan(t)dt=\int_{0}^{\infty}\frac{1 + at + t^2}{t^4 + 1}\arctan(t)dt}$$

And so

$${2I = \int_{0}^{\infty}\frac{x^2 + ax + 1}{1 + x^4}\left(\arctan(x) + \arctan\left(\frac{1}{x}\right)\right)dx=\frac{\pi}{2}\int_{0}^{\infty}\frac{x^2 + ax + 1}{1 + x^4}dx}$$

So

$${I = \frac{\pi}{4}\int_{0}^{\infty}\frac{x^2 + ax + 1}{x^4 + 1}dx}$$

Now, we don't actually need to use L'hopitals rule. Just notice that

$${\frac{I}{a} = \frac{1}{a}\int_{0}^{\infty}\frac{\frac{\pi}{4}x^2}{x^4 + 1}dx + \frac{\pi}{4}\int_{0}^{\infty}\frac{x}{x^4 + 1}dx + \frac{1}{a}\int_{0}^{\infty}\frac{\frac{\pi}{4}}{x^4 + 1}dx}$$

(We can split the integrals up like this because the integrals all converge separately). As ${a\rightarrow \infty}$, the left-most and right-most terms will obviously go to zero, and the only one that survives is the middle term. Hence we get that the limit is

$${=\frac{\pi}{4}\int_{0}^{\infty}\frac{x}{1+x^4}dx}$$

Now just make the substitution ${u=x^2}$:

$${\Rightarrow \frac{\pi}{8}\int_{0}^{\infty}\frac{1}{1+u^2}du}$$

And finally we are at a standard integral. We know

$${\int_{0}^{\infty}\frac{1}{1+x^2}dx=\frac{\pi}{2}}$$

And so finally we get the limit is

$${\Rightarrow \lim_{a\rightarrow\infty}\frac{I}{a}=\frac{\pi^2}{16}}$$

Answer 2

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 $\frac{\pi^2}{16}$