title: Mellin inversion theorem
text: # Mellin inversion theorem
  
  In mathematics, the Mellin inversion formula (named after Hjalmar Mellin) tells us conditions under which the inverse Mellin transform, or equivalently the inverse two-sided Laplace transform, are defined and recover the transformed function.
  
  ## Method
  
  If $\varphi (s)$ is analytic in the strip $a<\Re (s)<b$, and if it tends to zero uniformly as $\Im (s)\to \pm \infty$ for any real value c between a and b, with its integral along such a line converging absolutely, then if
  
      $f(x)=\{{\mathcal {M}}^{-1}\varphi \}={\frac {1}{2\pi i}}\int _{c-i\infty }^{c+i\infty }x^{-s}\varphi (s)\,ds$
  
  we have that
  
      $\varphi (s)=\{{\mathcal {M}}f\}=\int _{0}^{\infty }x^{s-1}f(x)\,dx.$
  
  Conversely, suppose $f(x)$ is piecewise continuous on the positive real numbers, taking a value halfway between the limit values at any jump discontinuities, and suppose the integral
  
      $\varphi (s)=\int _{0}^{\infty }x^{s-1}f(x)\,dx$
  
  is absolutely convergent when $a<\Re (s)<b$. Then $f$ is recoverable via the inverse Mellin transform from its Mellin transform $\varphi$. These results can be obtained by relating the Mellin transform to the Fourier transform by a change of variables and then applying an appropriate version of the Fourier inversion theorem.
  
  ## Boundedness condition
  
  The boundedness condition on $\varphi (s)$ can be strengthened if $f(x)$ is continuous. If $\varphi (s)$ is analytic in the strip $a<\Re (s)<b$, and if $|\varphi (s)|<K|s|^{-2}$, where K is a positive constant, then $f(x)$ as defined by the inversion integral exists and is continuous; moreover the Mellin transform of $f$ is $\varphi$ for at least $a<\Re (s)<b$.
  
  On the other hand, if we are willing to accept an original $f$ which is a generalized function, we may relax the boundedness condition on $\varphi$ to simply make it of polynomial growth in any closed strip contained in the open strip $a<\Re (s)<b$.
  
  We may also define a Banach space version of this theorem. If we call by $L_{\nu ,p}(R^{+})$ the weighted Lp space of complex valued functions $f$ on the positive reals such that
  
      $\|f\|=\left(\int _{0}^{\infty }|x^{\nu }f(x)|^{p}\,{\frac {dx}{x}}\right)^{1/p}<\infty$
  
  where ν and p are fixed real numbers with $p>1$, then if $f(x)$ is in $L_{\nu ,p}(R^{+})$ with $1<p\leq 2$, then $\varphi (s)$ belongs to $L_{\nu ,q}(R^{+})$ with $q=p/(p-1)$ and
  
      $f(x)={\frac {1}{2\pi i}}\int _{\nu -i\infty }^{\nu +i\infty }x^{-s}\varphi (s)\,ds.$
  
  Here functions, identical everywhere except on a set of measure zero, are identified.
  
  Since the two-sided Laplace transform can be defined as
  
      $\left\{{\mathcal {B}}f\right\}(s)=\left\{{\mathcal {M}}f(-\ln x)\right\}(s)$
  
  these theorems can be immediately applied to it also.
  
  