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If $\int_{-\infty}^{\infty}f(x)dx$$=1$ then $\int_{-\infty}^{\infty}f(x-\frac{1}{x})dx$ equals $?$

1.My try: I tried assuming a function that enclose an area of 1 as it takes values from - $\infty$ to $\infty$ and then replacing $x$ to $x$-$\frac{1}{x}$.But I was unable to reach the answer. Please tell me some good approach to solve this problem.

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Let $y:=x-\frac{1}{x}$ for $x\neq 0$. Then, there exists a $2$-to-$1$ correspondence between $x$ and $y$ (namely, $x$ and $-\frac{1}{x}$ are mapped to the same $y$). That is, $$\int_{x=-\infty}^{x=+\infty}\,f\left(x-\frac{1}{x}\right)\,\text{d}\left(x-\frac{1}{x}\right)=2\,\int_{-\infty}^{+\infty}\,f(x)\,\text{d}x=2\,.$$ Now, with $t:=-\frac{1}{x}$, we have $$\int_{x=-\infty}^{x=+\infty}\,f\left(x-\frac{1}{x}\right)\,\text{d}\left(-\frac{1}{x}\right)=\int_{-\infty}^{+\infty}\,f\left(t-\frac{1}{t}\right)\,\text{d}t\,.$$ Since $$\int_{x=-\infty}^{x=+\infty}\,f\left(x-\frac{1}{x}\right)\,\text{d}\left(x-\frac{1}{x}\right)=\int_{-\infty}^{+\infty}\,f\left(x-\frac{1}{x}\right)\,\text{d}x+\int_{x=-\infty}^{x=+\infty}\,f\left(x-\frac{1}{x}\right)\,\text{d}\left(-\frac{1}{x}\right)\,,$$ we get $$\int_{x=-\infty}^{x=+\infty}\,f\left(x-\frac{1}{x}\right)\,\text{d}\left(x-\frac{1}{x}\right)=2\,\int_{-\infty}^{+\infty}\,f\left(x-\frac{1}{x}\right)\,\text{d}x\,.$$ Thus, $$\int_{-\infty}^{+\infty}\,f\left(x-\frac1x\right)\,\text{d}x=1\,.$$

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    I didn't get the " 2-to-1 correspondence between x and y" thing . Can you please explain? – 2017-01-19
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    In other words, note that $$\int_{x=0}^{x=+\infty}\,f\left(x-\frac1x\right)\,\text{d}\left(x-\frac1x\right)=\int_{x=-\infty}^{x=0}\,f\left(x-\frac1x\right)\,\text{d}\left(x-\frac1x\right)=\int_{-\infty}^{\infty}\,f(x)\,\text{d}x\,.$$ – 2017-01-19