How to evaluate this integral $\int_{1/3}^{3} \frac{\sin(\frac1x -x)dx}{x}$? 
Evaluate
  $$\int_{1/3}^{3} \frac{\sin(\frac1x -x)dx}{x}.$$

I am unable to think of any way to solve this.
 A: Hint. Note that by letting $t=1/x$, we get
$$\int_{1/3}^{3} \frac{\sin(\frac1x -x)dx}{x}=\int_{3}^{1/3} 
\frac{\sin(t-\frac1t)(-dt/t^2)}{1/t}=-\int_{1/3}^{3} 
\frac{\sin(\frac1t-t)dt}{t}.$$
A: Enforce the substitution $x=\frac{-t+\sqrt{t^2+4}}{2}$, which is equivalent to $t=\frac{1}{x}-x$.  Then, we have $dx=\frac12\left(-1+\frac{t}{\sqrt{t^2+4}}\right)\,dt$ and 
$$\begin{align}
\int_{1/3}^3 \frac{\sin\left(\frac1x-x\right)}{x}\,dx&=\int_{8/3}^{-8/3}\frac{\sin(t)}{\frac{-t+\sqrt{t^2+4}}{2}}\,\left(-\frac12+\frac{t}{2\sqrt{t^2+4}}\right)\,dt\\\\
&=\int_{-8/3}^{8/3}\underbrace{\frac{\sin(t)}{\sqrt{t^2+4}}}_{\text{an odd function}}\\\\
&=0
\end{align}$$
A: $$I=\int_{\frac{1}{3}}^3 \dfrac{\sin \left( \dfrac{1}{x}-x\right)}{x}dx=\underbrace{\int_{\frac{1}{3}}^1\dfrac{\sin \left( \dfrac{1}{x}-x\right)}{x}dx}_J+\underbrace{\int_1^3\dfrac{\sin \left( \dfrac{1}{x}-x\right)}{x}dx}_H$$
$$J=\int_{\frac{1}{3}}^1\dfrac{\sin \left( \dfrac{1}{x}-x\right)}{x}dx\quad t=\dfrac{1}{x}\rightarrow x=\dfrac{1}{t}\rightarrow dx=-\dfrac{1}{t^2}\quad \text{and}\quad \left\{ \begin{array}{lcc}
             x=\dfrac{1}{3}\rightarrow t=3 \\
             \\ x=1\rightarrow t=1 
             \end{array}
   \right.$$
$$J=\int_3^1\dfrac{\sin \left( t-\dfrac{1}{t}\right)}{\dfrac{1}{t}}\left( -\dfrac{1}{t^2}\right)dt=-\int_1^3\dfrac{\sin \left[ -\left( \dfrac{1}{t}-t\right) \right]}{-t}dt=-\int_1^3 \dfrac{-\sin \left( \dfrac{1}{t}-t\right)}{-t}dt=$$
$$=-\int_1^3\dfrac{\sin \left( \dfrac{1}{t}-t\right)}{t}dt=-\int_1^3\dfrac{\sin \left( \dfrac{1}{x}-x\right)}{x}dx=-H\rightarrow I=-H+H=\boxed{0}$$
A: Let $\displaystyle I =\int^{3}_{\frac{1}{3}}\frac{1}{x}\cdot \sin\bigg(\frac{1}{x}-x\bigg)dx$
put $x=e^t,$ Then $dx = e^t dt$ and changing limits
So $\displaystyle I = \int^{\ln(3)}_{-\ln(3)}\underbrace{\sin (e^{-t}-e^t)}_{\bf{odd\; function}}dt =0.$
