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On Wikipedia articles, I often see $$\frac{dx}{x}$$ where the $\frac{1}{x}$ can often be absorbed into powers of $x$ in the integrand.

For example, the Mellin transform is expressed as $$\int_0^\infty x^s f(x) \frac{dx}{x}$$ instead of $$\int_0^\infty x^{s-1} f(x) \, dx.$$

What is the reason behind?

Michael Hardy
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Szeto
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    Oftentimes it just looks nicer. They mean the exact same thing. – B. Mehta Feb 15 '18 at 01:15
  • If you want to get really deep into it, $\int f,\dfrac{\mathrm{d}x}{x}$ is a Lebesgue-Stieltjes (or Riemann-Stieltjes) integral corresponding to the map $x \mapsto \log(x)$. But, unless you want to go deeper into the mathematical field of analysis, then you are probably safe regarding it as "just a notation." – Xander Henderson Feb 15 '18 at 01:34

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It is about the Haar measure on the multiplicative group $\left((0,\infty),\dfrac{dx}{x}\right)$. And when you are dealing with convolution in this group, one writes that $f\ast g(x)=\displaystyle\int_{0}^{\infty}g(y^{-1}x)f(y)\dfrac{dy}{y}$.

user284331
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  • Could you please elaborate? I know very little about convolution and Haar measure. – Szeto Feb 15 '18 at 01:48
  • What I wrote is the definition of Haar measure with respect the multiplicative group with respect to the measure $\dfrac{dx}{x}$. In general, you have something like $f\ast g(x)=\displaystyle\int_{X}g(y^{-1}x)f(y)d\mu(y)$. – user284331 Feb 15 '18 at 01:50