Clarification needed for first proof of this question :$\int_{-\infty}^{+\infty} f(x)dx = \int_{-\infty}^{+\infty} f\left(x - \frac{1}{x}\right)dx?$

Question

I have tried many time to know why in the first given answer of this question used Bounds $-\infty \to \infty$ in the second line however the variable change which has been used is clear dosn't give $-\infty$ and $+\infty$ , Probably a wrong typo of Author

Answer 1

Your question is not at all clear. But, I think you are asking.

When go from here: $\int_{-\infty}^{\infty}f\left(x-x^{-1}\right)dx=\int_{0}^{\infty}f\left(x-x^{-1}\right)dx+\int_{-\infty}^{0}f\left(x-x^{-1}\right)dx$

to here: $\int_{-\infty}^{\infty}f(2\sinh\theta)\,e^{\theta}d\theta+\int_{-\infty}^{\infty}f(2\sinh\theta)\,e^{-\theta}d\theta$

How do we justify the change to the limits of integration?

In the left hand integral above, we make the substitution $x = e^\theta$

This substitution acts on the limits of integration. And we must find values of $\theta$ that correspond to the $x$ values associated with $0, \infty.$ Alternatively, we can invert the substitution. $\theta = \ln x,\ \lim_\limits{x\to 0^+} \ln x = -\infty,\ \lim_\limits{x\to \infty} \ln x = \infty.$

And for the right hand intgral we do something very similar, except our substituion is $x = e^{-\theta}$