Question

Let $f\left(x\right)$ be an even function & $I_1={\int }_0^{\infty }f\left(x\right)dx,$
$I_2={\int }_0^{\infty }f(3x-\frac{1}{2x})dx,$ then the value of $\frac{I_1}{I_2}$ is (where $I_1\&I_2$ are finite)

Answer 1

$3x-\frac{1}{2x}=tx=\frac{1}{6}(t+\sqrt{t^2+6}),x>0dx=\frac{1}{6}(1+\frac{t}{\sqrt{t^2+6}})dt{I}_2={\int }_{-\infty }^{\infty }f(3x-\frac{1}{2x})dx=\frac{1}{6}{\int }_{-\infty }^{\infty }f\left(t\right)(1+\frac{t}{\sqrt{t^2+6}})dt=\frac{1}{6}{\int }_{-\infty }^{\infty }f\left(t\right)dt+\frac{1}{6}{\int }_{-\infty }^{\infty }\frac{tf\left(t\right)}{\sqrt{t^2+6}}dt$
where first integrand is an even function and second is an odd function.
$\therefore I_2=\frac{2}{6}{\int }_0^{\infty }f\left(t\right)dt\Rightarrow I_2=\frac{1}{3}{I}_1\Rightarrow \frac{I_1}{I_2}=3$