Trigonometric integral with the function- sin

Question

I need some help with this integral please:

$\displaystyle\int x\sin\frac{1}{x}dx$

Answer 1

$\int x\sin\dfrac{1}{x}dx$

$=\int x\sum\limits_{n=0}^\infty\dfrac{(-1)^nx^{-2n-1}}{(2n+1)!}dx$

$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^nx^{-2n}}{(2n+1)!}dx$

$=\sum\limits_{n=0}^\infty\dfrac{(-1)^nx^{1-2n}}{(2n+1)!(1-2n)}+C$

$=\sum\limits_{n=0}^\infty\dfrac{(-1)^{n+1}}{(2n+1)!(2n-1)x^{2n-1}}+C$

Answer 2

Hint: Let $~t=\dfrac1x$ , then, after a bit of integration by parts, use the trigonometric integral $\text{Si}(x)$.