Question

Let $f(x)$ be an even function such that $\underset{0}{\overset{\infty }{\int }}f(x)dx=\frac{\pi }{2}$, then the value of the definite integral $\underset{0}{\overset{\infty }{\int }}f\left(x-\frac{1}{x}\right)dx$ is equal to

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

Answer 1

Answer: (B)

$I=\underset{0}{\overset{\infty }{\int }}f\left(x-\frac{1}{x}\right)dx$......(1)
put $x=\frac{1}{t}$
$I=\underset{0}{\overset{\infty }{\int }}f\left(\frac{1}{t}-t\right)\frac{dt}{t^2}$
$I=\underset{0}{\overset{\infty }{\int }}f\left(x-\frac{1}{x}\right)\frac{dx}{x^2}$.....(2)
$(1)+(2)$
$2I=\underset{0}{\overset{\infty }{\int }}f\left(x-\frac{1}{x}\right)\left(1+\frac{1}{x^2}\right)dx$
put $x-\frac{1}{x}=t$
$=\underset{-\infty }{\overset{\infty }{\int }}f(t)dt=2\underset{0}{\overset{\infty }{\int }}f(x)dt=\pi$
$\therefore I=\frac{\pi }{2}$