Question

Let $f(x)$ be an even function such that $\int\limits_0^{\infty} f(x) d x=\frac{\pi}{2}$, then the value of the definite integral $\int\limits_0^{\infty} f \left( x -\frac{1}{ x }\right) dx$ is equal to

• A. $\pi$
• B. $\frac{\pi}{2}$
• C. $\frac{3 \pi}{2}$
• D. $2 \pi$

Answer 1

Answer: (B)

$I=\int\limits_0^{\infty} f \left( x -\frac{1}{ x }\right) dx$......(1)
put $x=\frac{1}{t}$
$I=\int\limits_0^{\infty} f\left(\frac{1}{t}-t\right) \frac{d t}{t^2}$
$I=\int\limits_0^{\infty} f\left(x-\frac{1}{x}\right) \frac{d x}{x^2}$.....(2)
$(1)+(2)$
$2 I=\int\limits_0^{\infty} f\left(x-\frac{1}{x}\right)\left(1+\frac{1}{x^2}\right) d x$
put $x-\frac{1}{x}=t$
$=\int\limits_{-\infty}^{\infty} f ( t ) dt =2 \int\limits_0^{\infty} f ( x ) dt =\pi $
$\therefore I=\frac{\pi}{2}$