Prove that $\int^{4\pi}_0(3x^2\sin\frac{1}{x}-x\cos\frac{1}{x})\,dx=\frac{32\sqrt2}{\pi^3}$

Question

Prove that $$\int^{4\pi}_0(3x^2\sin\frac{1}{x}-x\cos\frac{1}{x})\,dx=\frac{32\sqrt2}{\pi^3}.$$

This question is regarding improper integrals of the second kind. I'm stuck at it. Letting $1/x=u$, I have,

$$\int^\frac{1}{4\pi}_\infty(-3\sin u+u\cos u)\, du = [4\cos u+u\sin u]\Bigm|^\frac{1}{4\pi}_\infty$$ and after this I'm stuck.

Comments and hints appreciated!

Answer 1

Your integral is $[x^3\sin\frac{1}{x}]_0^{4\pi}=64\pi^3\sin\frac{1}{4\pi}-\lim_{x\to 0}x^3\sin\frac{1}{x}$. Since $|\sin\frac{1}{x}|\le 1$, $\lim_{x\to 0}x^3\sin\frac{1}{x}=0$. The final result is $64\pi^3\sin\frac{1}{4\pi}$.