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I'm currently trying to prove the existence of the following improper integral:

$$I := \int_0^\infty \frac{\cos(x)}{\sqrt{x}+x^3} dx$$

and to show that $|I| \leq \frac{5}{2}$.

$\begin{align}|I| &\leq \int_0^\infty \frac{|\cos(x)|}{|\sqrt{x}+x^3|} dx\\ &\leq \int_0^\infty \frac{1}{\sqrt{x}+x^3} dx\end{align}$

However, I'm pretty much stuck at this point, I could use a substitution $u = \sqrt{x}$ which leads to $$ 2 \cdot \int_0^\infty \frac{1}{1+u^5}du$$ but how can I evaluate this improper integral the easiest way to get the upper bound?

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    $\begingroup$ For $u\in[0,1]$, we have $\frac1{1+u^5}<1$, and for $u\in[1,\infty)$, we have $\frac1{1+u^5}<\frac1{u^5}$. $\endgroup$ – Simply Beautiful Art Aug 26 '17 at 21:37
  • $\begingroup$ Thank you @SimplyBeautifulArt, so I just split the integral into $\int_0^1 1 dx + \int_1^\infty \frac{1}{u^5}du$ which leads to $\frac{5}{4}$ multiplied with $2$ giving the wanted result :) $\endgroup$ – PeterMcCoy Aug 26 '17 at 21:45
  • $\begingroup$ Yes, that would do. $\endgroup$ – Simply Beautiful Art Aug 26 '17 at 21:52
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    $\begingroup$ The initial substitution wasn't necessary, of course. On $[0, 1]$, $\frac{1}{\sqrt{x}+x^3}\leq \frac{1}{\sqrt{x}}$, and on $[1, \infty)$, $\frac{1}{\sqrt{x}+x^3}\leq \frac{1}{x^3}$. Then, $$\int_0^1 \frac{\mathrm{d}x}{\sqrt{x}} = 2$$ and $$\int_1^{\infty} \frac{\mathrm{d}x}{x^3} = \frac{1}{2}$$ $\endgroup$ – Michael Lee Aug 26 '17 at 22:21
  • $\begingroup$ Sooner or later, you will learn that (for $n >1$) $$\int_0^\infty \frac{du}{1+u^n}=\frac \pi n \,\csc \left(\frac{\pi }{n}\right)$$ which, for $n=5$, gives $\frac{\pi}{5} \sqrt{2+\frac{2}{\sqrt{5}}} \approx 1.06896$. $\endgroup$ – Claude Leibovici Aug 27 '17 at 4:04

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