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

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

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

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

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

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

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    $\begingroup$ You can notice that: $$\left|\frac{\cos(\ln(x))}{\sqrt{1+x^4}} \right| \leq \frac{1}{\sqrt{1+x^4}}$$ and $$\int_0^\infty \frac{1}{\sqrt{1+x^4}} dx$$ is absolutely convergent. $\endgroup$ – Delta-u Apr 23 '18 at 13:32
  • $\begingroup$ This integral can computed symbolically. $\endgroup$ – Dr. Sonnhard Graubner Apr 23 '18 at 13:55
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You are looking for $$\text{Re}\int_{0}^{+\infty}\frac{x^i}{\sqrt{x^4+1}}\,dx = \frac{1}{4\sqrt{\pi}}\,\text{Re}\left[\Gamma\left(\tfrac{1+i}{4}\right)\Gamma\left(\tfrac{1-i}{4}\right)\right]=\frac{1}{4\sqrt{\pi}}\left\|\Gamma\left(\tfrac{1+i}{4}\right)\right\|^2$$ by Euler's Beta function. It approximately equals $\frac{43}{50}$.

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    $\begingroup$ Seems a little overkill, if the question is just checking for convergence $\endgroup$ – Dylan Apr 23 '18 at 15:01
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    $\begingroup$ @Dylan: I know, but on the other hand convergence is trivial, and I thought it was interesting to point out something non-trivial. $\endgroup$ – Jack D'Aurizio Apr 23 '18 at 15:05
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The numerator is absolutely bounded by $1$, while the denominator is asymptotic to $x^2$.

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