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

Answer

b
Put $x=y-\frac{1}{y}$

$\Rightarrow 1 \Rightarrow \int_{-\infty}^{\infty} f\left(y-\frac{1}{y}\right)\left(1+\frac{1}{y^{2}}\right) d y$

$=\int_{-\infty}^{0} f\left(y-\frac{1}{y}\right) d y+\int_{-\infty}^{0} f\left(y-\frac{1}{y}\right) \frac{d y}{y^{2}}$

Putting $z=-\frac{1}{y}=\int_{-\infty}^{0} f\left(y-\frac{1}{y}\right) d y+\int_{0}^{\infty} f\left(z-\frac{1}{z}\right) d z=1$

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