Integration of $\sin(\frac1x)$

Question

I just started to study integrals yesterday so I am not so strong in integrating functions in this moment, however today I met the integral of $\sin(\frac1x)$ and I just can't find its primitive function after pages and pages of calculus.. Is it integrable? If yes, how do you find the primitive? If not, why? Please help me, thank you very much!!

Answer 1

Hint:

Note that: $$\int\sin\left(\frac1x\right)\mathrm dx=\int\color{blue}{\sin\left(\frac1x\right)}\cdot \color{darkmagenta}1\,\mathrm dx\quad\text{and}\quad\dfrac{\mathrm d}{\mathrm dx}\sin\left(\frac1x\right)=-\dfrac{\cos(1/x)}{x^2}.$$ Using integration by parts we find that: $$\eqalign{\int\sin\left(\frac1x\right)\mathrm dx &=x\sin\left(\frac1x\right)-\int\left[-\dfrac{\cos(1/x)}{x^2}\cdot x\right]\mathrm dx \\ &=x\sin\left(\frac1x\right)+\int\dfrac{\cos(1/x)}{x}\mathrm dx. }$$ Now all what remains is to prove that: $$\int\dfrac{\cos(1/x)}{x}\mathrm dx=-\operatorname{Ci}\left(\frac1x\right),$$ where $\operatorname{Ci}$ is the cosine integral. This can be achieved by exploiting the properties of the $\operatorname{Ci}$.