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In a proof of Frullani's Theorem I'm trying to understand there's a step where $$\int_x^y \frac{f(at)}{t}dt - \int_x^y \frac{f(bt)}{t}dt = \int_{ax}^{ay} \frac{f(u)}{u/a} \frac{du}{a} - \int_{bx}^{by} \frac{f(u)}{u/b} \frac{du}{b}$$ It seems that the substitutions $u=at$ and $u=bt$ are made. It is given that $a<b$.
I don't understand how this works since it doesn't seem like $at=bt$.
I can't post there because I don't have 50 point :/
Thanks in advance!

Lee David Chung Lin
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Brian Lilley
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    The $t$ and $u$ in the integrals are dummy variables. Namely, writing $\int_x^y \frac{f(at)}{t}dt $ then changing to $u=at$ is the same as writing $\int_x^y \frac{f(a s)}sds$ then changing to $r=as$ or writing $\int_x^y \frac{f(aw)}{w}dw $ then changing to $z=aw$ or with any other symbols. – Lee David Chung Lin Dec 03 '21 at 02:11
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    It is just a sloppy convention. What is truly done there is that the substitutions $u_1 = at$ and $u_2 = bt$ are applied to the integrals, and then the dummy variables $u_1$ and $u_2$ inside integral symbols are silently renamed as $u$: \begin{align}\int_{x}^{y}\frac{f(at)}{t},\mathrm{d}t-\int_{x}^{y}\frac{f(bt)}{t},\mathrm{d}t&=\int_{ax}^{ay}\frac{f(u_1)}{u_1},\mathrm{d}u_1-\int_{bx}^{by}\frac{f(u_2)}{u_2},\mathrm{d}u_2\&=\int_{ax}^{ay}\frac{f(u)}{u},\mathrm{d}u-\int_{bx}^{by}\frac{f(u)}{u},\mathrm{d}u\end{align} – Sangchul Lee Dec 03 '21 at 02:20
  • Thanks that makes sense! Hopefully I can recognize this in the future. – Brian Lilley Dec 03 '21 at 05:09
  • The thing to remember is that a dummy variable is only defined within its notation. Thus the $t$ in the first integration is a different variable than the $t$ in the second integration, and similarly for the $u$ on the rh side. They may have given the same name, but they are not the same variable. It is quite common to use the same dummy variable in many different notations. Normally you don't even notice it because the usages do not conflict with each other. In this case there was a conflict, but it only occurred because you were misintepreting them as being the same. – Paul Sinclair Dec 03 '21 at 18:06

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