This improper integral $\int_{1}^{+\infty}x\sin{x}\sin{x^4}dx$ is absolutely convergent? Discuss the improper integral
$$\int_{1}^{+\infty}x\sin{x}\sin{x^4}dx$$ absolute convergence?
My idea: since
$$\sin{x}=x-\dfrac{x^3}{6}+\cdots$$
$$\sin{x^4}=x^4-\dfrac{x^{12}}{6}+\cdots$$
so
$$x\sin{x}\sin{x^4}=x^6-\cdots$$
then I can't sure this integral is absolutely convergent. Thank you
 A: This is a partial answer, using Lebesgue integral.
Call $\mu$ the Lebesgue measure on $\mathbb{R}$. For all $n \in \mathbb{N}$ call
$$A_n = \{ x \in [1, + \infty) : |x \sin x \sin x^4| \geq \frac{1}{n} \}$$
Then, it sufficies to show that $\exists n : \mu(A_n) = + \infty$.
A: Approximating $\sin t$ for small $|t|$ is of no help here.
Denote by $A_n$ the interval of length ${\pi\over2}$ with midpoint $(2n+1){\pi\over2}$. Then
$$|x\sin x|\geq n\pi{\sqrt{2}\over2}\qquad(x\in A_n)\ .$$
It follows that
$$\int_{A_n}\left|x\sin x\sin(x^4)\right|\>dx\geq n\pi{\sqrt{2}\over2}\int_{A_n}|\sin(x^4)|\>dx\ .\tag{1}$$
Now when $n\gg1$ the function $g:\>x\mapsto \sin(x^4)$ is oscillating with high frequency on $A_n$. Between two successive zeros of $g$ the mean value of $|g|$ is about ${2\over \pi}$ for all half-waves, so that we obtain approximatively
$$\int_{A_n}|\sin(x^4)|\>dx\doteq{2\over\pi}\>|A_n|=1\qquad(n\gg1)\ .$$
Together with $(1)$ this shows that the integral in question is definitively not absolutely convergent.
A: This is also a partial answer.
$|\sin x|>1/2$ for more than half of $[0,N]$ as soon as $N>\pi/3$.
$|\sin x^4|>1/2$ for more than half of $[0,N]$ after a while, but I'm not sure when.
I would expect that $|\sin x\sin x^4|>1/4$ for more than a quarter of $[0,N]$ eventually.
A: No, this integral does not converge absolutely.
First take a look at:
$$\int_{1}^{+\infty}|x\sin{x}|dx$$
It does not converge for sure, does it?
Now you multiply the function by $|\sin(x^4)|$. This $|\sin(x^4)|$ fluctuates very-very frequently when $x$ is large. Within each period of $|\sin(x^4)|$ the $|x\sin{x}|$ is almost constant. When you multiply almost constant function by $|\sin(x)|$ it's integral gets multiplied by 2/$\pi$. Well, approximately.
So, the original integral diverges almost as quickly as $\int_{1}^{+\infty}|x\sin{x}|dx$. "Almost" means slower by a factor of 2/$\pi$.
I guess this answer is partial since it is not an actual proof. But I hope you got the idea.
