$\text{ Let }f\left(x\right)\text{ be an even function such that }{\int }_0^{\mathrm{\infty }} f\left(x\right)dx=\frac{\pi }{2},\text{ then the value of }$$\text{ the definite integral }{\int }_0^{\mathrm{\infty }} f\left(x-\frac{1}{x}\right)dx\text{ is equal to }$

$\begin{array}{l}\text{ Let }I={\int }_0^{\mathrm{\infty }} f\left(x-\frac{1}{x}\right)dx\dots \dots \dots \dots ..\left(1\right)\\ \text{ Put }x=\frac{1}{t}\\ \Rightarrow dx=\frac{-1}{t^2}dt\\ \mathrm{From}\left(1\right):I={\int }_{\mathrm{\infty }}^0 f\left(\frac{1}{t}-t\right)\left(-\frac{dt}{t^2}\right)\left(\text{ Use }{\int }_a^b f\left(x\right)dx=-{\int }_b^a f\left(x\right)dx\right)\\ \Rightarrow I={\int }_0^{\mathrm{\infty }} f\left(\frac{1}{t}-t\right)\frac{dt}{t^2}\\ \Rightarrow I={\int }_0^{\mathrm{\infty }} f\left(t-\frac{1}{t}\right)\frac{dt}{t^2}(\because f\text{ is even function })\\ \Rightarrow I={\int }_0^{\mathrm{\infty }} f\left(x-\frac{1}{x}\right)\frac{dx}{x^2}\dots \dots \dots \dots ...\left(2\right)\end{array}$

Adding equations (1) & (2) 

$\begin{array}{l}\Rightarrow 2I={\int }_0^{\mathrm{\infty }} f\left(x-\frac{1}{x}\right)dx+{\int }_0^{\mathrm{\infty }} f\left(x-\frac{1}{x}\right)\frac{dx}{x^2}\\ \Rightarrow 2I={\int }_0^{\mathrm{\infty }} f\left(x-\frac{1}{x}\right)\left(1+\frac{1}{x^2}\right)dx\\ \text{ Put }x-\frac{1}{x}=u\\ \Rightarrow \left(1+\frac{1}{x^2}\right)dx=du\end{array}$

$\begin{array}{l}\therefore 2I={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }} f\left(u\right)du\left(\because \text{ If }f\text{ is even }{\int }_{-a}^a f\left(x\right)=2\times {\int }_0^a f\left(x\right)dx\right)\\ =2{\int }_0^{\mathrm{\infty }} f\left(x\right)dx\\ =2\left(\frac{\pi }{2}\right)\\ \Rightarrow 2I=\pi \\ \therefore I=\frac{\pi }{2}\end{array}$