<s>[INST] About an interesting improper integral $\int_{0}^{\infty} \frac{x dx}{1+x^6 \sin^2 x}$ Is there a non-negative $C^\infty$ function $f(x)$ which is not bounded in $[a, \infty)$, but whose improper integral $\int_{a}^{\infty} f(x) dx$ exists?
Yes!
$\frac{x}{1+x^6 \sin^2 x}$ is not bounded in $[0, \infty)$ and not negative, but 
$$\int_{0}^{\infty} \frac{x dx}{1+x^6 \sin^2 x}$$
exists.
I found this example in a calculus book written by Teiji Takagi.
His proof of the above fact is the following:  
$$\int_{n \pi}^{(n+1) \pi} \frac{x dx}{1+x^6 \sin^2 x} < (n+1) \pi \int_{0}^{\pi} \frac{x dx}{1+(n \pi)^6 \sin^2 x} < \frac{1}{n^2}$$
I cannot derive the following inequaltiy:  
$$(n+1) \pi \int_{0}^{\pi} \frac{x dx}{1+(n \pi)^6 \sin^2 x} < \frac{1}{n^2}$$
Please derive it.
 [/INST] Using the fact that
\begin{align}
\frac{2}{\pi}x \leq \sin x \ \ \text{ for } \ \ 0\leq x\leq \frac{\pi}{2}
\end{align}
and
\begin{align}
1-\frac{2}{\pi}\left(x-\frac{\pi}{2}\right) \leq \sin x \ \ \text{ for } \ \ \frac{\pi}{2}\leq x \leq \pi.
\end{align}
Hence we have 
\begin{align}
\int^\pi_0 \frac{xdx}{1+(n\pi)^6\sin^2 x} =&\ \int^{\pi/2}_0 \frac{xdx}{1+(n\pi)^6\sin^2 x}+\int^\pi_{\pi/2} \frac{xdx}{1+(n\pi)^6\sin^2 x}\\
\leq& \int^{\pi/2}_0 \frac{xdx}{1+4n^6\pi^4 x^2}+\int^\pi_{\pi/2} \frac{xdx}{1+(n\pi)^6(1-\frac{2}{\pi}(x-\frac{\pi}{2}))^2}\\
=&\ \frac{\tan^{-1}(n\pi)^3}{2\pi n^3} \leq \frac{1}{4n^3}.
\end{align}
Hence everything else follows immediately. 
</s>