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Let \begin{align} f \in C^1 \text{ with } f(0) , f(\infty) \in \Bbb R \text{ and } b>a>0 \tag{1}. \end{align} Show \begin{align*} \int_0^\infty \frac{f(ax)-f(bx)}{x}dx = \left( f(0) - f(\infty) \right) \log\left( \frac{b}{a} \right). \end{align*}

There is a proof of this identity in this answer.

How do we justify the application of Fubini theorem to interchange the order of integration.
I don't see why (1) is sufficient to guarantee the finiteness of \begin{align} \int_0^\infty \int_a^b \left| f'(yx) \right| dy dx , \end{align} and \begin{align} \int_a^b \int_0^\infty \left| f'(yx) \right| dx dy . \end{align}

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That was probably my answer you're referring to, and I guess I didn't read the exact conditions the OP was stating too carefully. Assume that $f$ is monotone and everything is ok. (More general solutions were already given.) – mrf Jan 16 '13 at 21:47
I believe that the justification lies in the absolute convergence of each of the integrals in the double integral above, see T.W. Koerner, Fourier Analysis Secs. 47 & 48. (BTW I had occasion to use this result here:…) – Ron Gordon Jan 16 '13 at 21:57
I updated my answer in the other thread. It would probably have been a better idea to place a comment there. It was sheer luck that I saw this question. – mrf Jan 16 '13 at 22:47
@rlgordonma Something eludes me in the other answer and I just left a comment to that effect. – Did Jan 21 '13 at 9:18

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