How to show that $ \int_1^{+\infty} \frac{\sin \frac{1}{x}}{x^a} \mathrm{d}x$ is not integrable?

Question

I wonder how to show that $$ \int_1^{+\infty} \frac{\sin \frac{1}{x}}{x^a} \mathrm{d}x$$ is not Lebesgue-integrable, where $a<1$.

Thanks in advance!

Answer 1

Hint: $$ \begin{align} \int_1^\infty\frac{\sin\left(\frac1x\right)}{x^a}\,\mathrm{d}x &=\int_0^1\sin(x)\,x^{a-2}\,\mathrm{d}x\\ &=\int_0^1\frac{\sin(x)}x\,x^{a-1}\,\mathrm{d}x \end{align} $$