Question

If $2f\left(x\right)+f(-x)=\frac{1}{x}\sin (x-\frac{1}{x})$, then the value of $I={\int }_{1/e}^{e}f\left(x\right)dx$ is equal to

Answer 1

$2f\left(x\right)+f(-x)=\frac{1}{x}sin(x-\frac{1}{x})$ ..... (i)
Replace $x$ by $-x$
$2f(-x)+f\left(x\right)=\frac{1}{x}sin(x-\frac{1}{x})$ ..... (ii)
$2\left(i\right)-\left(ii\right)$ gives
$3f\left(x\right)=\frac{1}{x}sin(x-\frac{1}{x})$
$\Rightarrow f\left(x\right)=\frac{1}{3x}sin(x-\frac{1}{x})$
$I={\int }_{1/e}^{e}f\left(x\right)dx={\int }_{1/e}^e\frac{1}{3x}sin(x-\frac{1}{x})dx$
put $x=\frac{1}{t}\Rightarrow dx=\frac{-1}{t^2}dt$
$I={\int }_e^{1/e}\frac{1}{3}tsin(\frac{1}{t}-t)\left(\frac{-1}{t^2}dt\right)$
$={\int }_e^{1/e}\frac{1}{3t}sin(t-\frac{1}{t})dt$
$=-{\int }_{1/e}^e\frac{1}{3x}sin(x-\frac{1}{x})dx\Rightarrow I=-I\Rightarrow I=0$