The Mellin transform of a function $f$ is another function $\varphi$ such that
$$ f(x)=\frac{1}{2 \pi i} \int_{c-i \infty}^{c+i \infty} x^{-s} \varphi(s) d s . $$For a value $s$, the output $\varphi(s)$ represents "how important" is the power $x^{-s}$ to construct, by "continuous linear combination" (i.e., by integration) the function $f$. This coefficients can be calculated with
$$ \varphi(s)=\int_0^{\infty} x^{s-1} f(x) d x . $$For example, $\varphi(-1)$ should be the first derivative of $f$ evaluated at 0. But
$$ \varphi(-1)=\int_0^{\infty} x^{-2} f(x) d x $$is not $f'(0)$, or it is? Observe the similarity with Cauchy integral formula...
It is related to Fourier transform and therefore to Laplace transform.
It is somewhat related to Riemann zeta function.
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Author of the notes: Antonio J. Pan-Collantes
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