Question

If ${\int }_{-\infty }^{\infty }f\left(x\right)dx=1$ then ${\int }_{-\infty }^{\infty }f(x-\frac{1}{x})dx$ is equal to:

• A. $0$
• B. $1$
• C. $-1$
• D. $2$

Answer 1

The correct option is D $2$

Let,

$x-\frac{1}{x}=t$

$\Rightarrow \frac{dt}{dx}=1+\frac{1}{x^2}$

$\Rightarrow \frac{dt}{dx}>0$

Thus $t$ is a monotonically increasing function and odd with a discontinuity at $x=0$

Also $t$ goes from $-\infty$ to $+\infty$ for $x\in (-\infty ,0)$ and $x\in (0,+\infty )$.

Hence t runs from $-\infty$ to $+\infty$ $TWICE$ when $x$ runs from $-\infty$ to $+\infty$ once.

$\therefore {\int }_{-\infty }^{+\infty }f\left(t\right)dx=2{\int }_{-\infty }^{+\infty }f\left(x\right)dx=2$