Does $\int_{0}^{\infty}\frac{\cos(\ln x)}{\sqrt{x^{4}+1}}\text dx$ converges?

Question

Given $$\int_{0}^{\infty}\frac{\cos(\ln x)}{\sqrt{x^{4}+1}}\text dx$$

I need to check if the integral absolutely converges, Conditionally converges or diverges.

what I did was to use Linearity, given $$\int_{1}^{\infty}\frac{\cos(\ln x)}{\sqrt{x^{4}+1}}\text dx$$

I've tried to make u-substitution as: $x^2 = t$ and got $$\frac{1}{2}\int_{1}^{\infty}\frac{\cos(\frac{1}{2}\ln t))}{\sqrt{t^{3}+t}}\text dx$$

than I tried to make another u-substitution as: $\ln t = k$ so that: $$\frac{1}{2}\int_{1}^{\infty}\frac{\cos(\frac{k}{2})\exp^{k}}{\sqrt{\exp^{3k}+\exp^{k}}}\text dx$$

but on that point I got stuck as another substitution would not help and also Integration by parts will not work.

Answer 1

You are looking for $$\text{Re}\int_{0}^{+\infty}\frac{x^i}{\sqrt{x^4+1}}\,\text dx = \frac{1}{4\sqrt{\pi}}\,\text{Re}\left[\Gamma\left(\tfrac{1+i}{4}\right)\Gamma\left(\tfrac{1-i}{4}\right)\right]=\frac{1}{4\sqrt{\pi}}\left\|\Gamma\left(\tfrac{1+i}{4}\right)\right\|^2$$ by Euler's Beta function. It approximately equals $\frac{43}{50}$.

Answer 2

The numerator is absolutely bounded by $1$, while the denominator is asymptotic to $x^2$.