Editing Mellin transform (section)

==Examples==

===Cahen–Mellin integral===
{{anchor|Cahen–Mellin}}<!-- This Anchor tag serves to provide a permanent target for incoming section links. Please do not remove it, nor modify it, except to add another appropriate anchor.  If you modify the section title, please anchor the old title. It is always best to anchor an old section header that has been changed so that links to it will not be broken. See [[Template:Anchor]] for details. This template is {{subst:Anchor comment}} -->The Mellin transform of the function <math> f(x) = e^{-x} </math> is
<math display="block">\Gamma(s) = \int_0^\infty x^{s-1}e^{-x} dx </math>
where <math>\Gamma(s)</math> is the [[gamma function]].  <math>\Gamma(s)</math> is a [[meromorphic function]] with simple [[zeros and poles|poles]] at <math>z = 0, -1, -2, \dots</math>.<ref>{{cite book |first1=E.T. |last1=Whittaker |author-link1=E. T. Whittaker|first2=G.N. |last2=Watson|author-link2=G. N. Watson |title=[[A Course of Modern Analysis]] |year=1996 |publisher=Cambridge University Press}}</ref>  Therefore, <math>\Gamma(s)</math> is analytic for <math>\Re(s)>0</math>.  Thus, letting <math>c>0</math> and <math>z^{-s}</math> on the [[principal branch]], the inverse transform gives
<math display="block"> e^{-z}= \frac{1}{2\pi i} \int_{c-i\infty}^{c+i\infty} \Gamma(s) z^{-s} \; ds .</math>

This integral is known as the Cahen–Mellin integral.<ref>{{cite journal |first1=G. H. |last1=Hardy|author-link1=G. H. Hardy |first2=J. E. |last2=Littlewood|author-link2=J. E. Littlewood |title=Contributions to the Theory of the Riemann Zeta-Function and the Theory of the Distribution of Primes |journal=[[Acta Mathematica]] |volume=41 |issue=1 |year=1916 |pages=119–196 |doi=10.1007/BF02422942 |url=https://zenodo.org/record/2294397 |doi-access=free }} ''(See notes therein for further references to Cahen's and Mellin's work, including Cahen's thesis.)''</ref>

===Polynomial functions===
Since <math display="inline">\int_0^\infty x^a dx</math> is not convergent for any value of <math>a\in\mathbb{R}</math>, the Mellin transform is not defined for polynomial functions defined on the whole positive real axis.  However, by defining it to be zero on different sections of the real axis, it is possible to take the Mellin transform.  For example, if
<math display="block"> f(x) = \begin{cases} x^a  & x < 1, \\ 0 & x > 1, \end{cases}</math>
then
<math display="block">
\mathcal M f (s)= \int_0^1 x^{s-1}x^adx = \int_0^1 x^{s+a-1}dx = \frac 1 {s+a}.
</math>

Thus <math>\mathcal M f (s)</math> has a simple pole at <math>s=-a</math> and is thus defined for <math>\Re (s)>-a</math>.  Similarly, if
<math display="block">f(x)=\begin{cases} 0  & x < 1, \\ x^b & x > 1, \end{cases}</math>
then
<math display="block">
\mathcal M f (s)= \int_1^\infty x^{s-1}x^bdx = \int_1^\infty x^{s+b-1}dx = - \frac 1 {s+b}.
</math>
Thus <math>\mathcal M f (s)</math> has a simple pole at <math>s=-b</math> and is thus defined for <math>\Re (s)<-b</math>.

===Exponential functions===
For <math>p > 0 </math>, let <math>f(x)=e^{-px}</math>.  Then
<math display="block">
\mathcal M f (s) = \int_0^\infty x^{s} e^{-px}\frac{dx}{x} = \int_0^\infty \left(\frac{u}{p} \right)^{s}e^{-u} \frac{du}{u} = \frac{1}{p^s}\int_0^\infty u^{s}e^{-u} \frac{du}{u} = \frac{1}{p^{s}}\Gamma(s).
</math>

===Zeta function===
It is possible to use the Mellin transform to produce one of the fundamental formulas for the [[Riemann zeta function]], <math>\zeta(s)</math>.  Let <math display="inline">f(x)=\frac{1}{e^x-1}</math>.  Then
<math display="block">
\begin{alignat}{3}
\mathcal M f (s) &= \int_0^\infty x^{s-1}\frac{1}{e^x-1}dx &&= \int_0^\infty x^{s-1}\frac{e^{-x}}{1-e^{-x}}dx \\
&= \int_0^\infty x^{s-1}\sum_{n=1}^\infty e^{-nx}dx &&= \sum_{n=1}^\infty \int_0^\infty x^{s}e^{-nx}\frac{dx}{x} \\
&= \sum_{n=1}^\infty \frac{1}{n^s}\Gamma(s)=\Gamma(s)\zeta(s) .
\end{alignat}
</math>
Thus,
<math display="block"> \zeta(s) = \frac{1}{\Gamma(s)}\int_0^\infty x^{s-1}\frac{1}{e^x-1} dx. </math>

===Generalized Gaussian===
For <math>p > 0</math>, let <math>f(x)=e^{-x^p}</math> (i.e. <math>f</math> is a [[Generalized normal distribution|generalized Gaussian distribution]] without the scaling factor.)  Then
<math display="block">
\begin{alignat}{3}
\mathcal M f (s) &= \int_0^\infty x^{s-1}e^{-x^p}dx &&= \int_0^\infty x^{p-1}x^{s-p}e^{-x^p}dx \\
&= \int_0^\infty x^{p-1}(x^p)^{s/p-1}e^{-x^p}dx &&= \frac{1}{p}\int_0^\infty u^{s/p-1}e^{-u}du \\
&= \frac{\Gamma(s/p)}{p} .
\end{alignat}</math>
In particular, setting <math>s=1</math> recovers the following form of the gamma function
<math display="block">
\Gamma\left(1+\frac{1}{p}\right) = \int_0^\infty e^{-x^p}dx.
</math>

===Power series and Dirichlet series===

Generally, assuming the necessary convergence, we can connect Dirichlet series and [[power series]]
<math display="block">F(s)=\sum\limits_{n=1}^{\infty}\frac {a_n}{n^s}, \quad f(z)=\sum\limits_{n=1}^{\infty}a_nz^n</math>
by this formal identity involving the Mellin transform:<ref>{{cite journal |first1=Aurel |last1=Wintner|author-link1=Aurel Wintner |title=On Riemann's Reduction of Dirichlet Series to Power Series |journal=[[American Journal of Mathematics]] |volume=69 |issue=4 |year=1947 |pages=769–789 |doi=10.2307/2371798|doi-access=free |jstor=2371798 }}</ref>
<math display="block">\Gamma(s)F(s)=\int_{0}^{\infty}x^{s-1}f(e^{-x})dx</math>