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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

zeraoulia rafik
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  • Focus on one of the terms: $\int_{-\infty}^0 f(x-1/x)dx$. Let $x=-e^{\theta}$, which gives $\theta=\infty$ when $x=-\infty$ and $\theta=-\infty$ when $x=0$. Everything seems fine? – Alex R. Aug 08 '18 at 17:12

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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 $x = e^{-\theta}$

Doug M
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