Mellin inversion theorem

If ${\displaystyle \varphi (s)}$ is analytic in the strip ${\displaystyle a<\Re (s)<b}$, and if it tends to zero uniformly as ${\displaystyle \Im (s)\to \pm \infty }$ for any real value c between a and b, with its integral along such a line converging absolutely, then if

${\displaystyle f(x)=\{{\mathcal {M}}^{-1}\varphi \}={\frac {1}{2\pi i}}\int _{c-i\infty }^{c+i\infty }x^{-s}\varphi (s)\,ds}$

we have that

${\displaystyle \varphi (s)=\{{\mathcal {M}}f\}=\int _{0}^{\infty }x^{s}f(x)\,{\frac {dx}{x}}.}$

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

${\displaystyle \varphi (s)=\int _{0}^{\infty }x^{s}f(x)\,{\frac {dx}{x}}}$

is absolutely convergent when ${\displaystyle a<\Re (s)<b}$. Then f is recoverable via the inverse Mellin transform from its Mellin transform ${\displaystyle \varphi }$.

We may strengthen the boundedness condition on ${\displaystyle \varphi (s)}$ if f(x) is continuous. If ${\displaystyle \varphi (s)}$ is analytic in the strip ${\displaystyle a<\Re (s)<b}$, and if ${\displaystyle |\varphi (s)|<K|s|^{-2}}$, 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 ${\displaystyle \varphi }$ for at least ${\displaystyle a<\Re (s)<b}$.

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 ${\displaystyle \varphi }$ to simply make it of polynomial growth in any closed strip contained in the open strip ${\displaystyle a<\Re (s)<b}$.

We may also define a Banach space version of this theorem. If we call by ${\displaystyle L_{\nu ,p}(R^{+})}$ the weighted Lp space of complex valued functions f on the positive reals such that

${\displaystyle \|f\|=\left(\int _{0}^{\infty }|x^{\nu }f(x)|^{p}\,{\frac {dx}{x}}\right)^{1/p}<\infty }$

where ν and p are fixed real numbers with p>1, then if f(x) is in ${\displaystyle L_{\nu ,p}(R^{+})}$ with ${\displaystyle 1<p\leq 2}$, then ${\displaystyle \varphi (s)}$ belongs to ${\displaystyle L_{\nu ,q}(R^{+})}$ with ${\displaystyle q=p/(p-1)}$ and

${\displaystyle f(x)={\frac {1}{2\pi i}}\int _{\nu -i\infty }^{\nu +i\infty }x^{-s}\varphi (s)\,ds.}$

Here functions, identical everywhere except on a set of measure zero, are identified.

Since the two-sided Laplace transform can be defined as

${\displaystyle \left\{{\mathcal {B}}f\right\}(s)=\left\{{\mathcal {M}}f(-\ln x)\right\}(s)}$

these theorems can be immediately applied to it also.

• Mellin transform
• Nachbin's theorem

• Flajolet, P.; Gourdon, X.; Dumas, P. (1995). "Mellin transforms and asymptotics: Harmonic sums" (PDF). Theoretical Computer Science. 144 (1–2): 3–58. doi:10.1016/0304-3975(95)00002-E.
• McLachlan, N. W. (1953). Complex Variable Theory and Transform Calculus. Cambridge University Press.
• Polyanin, A. D.; Manzhirov, A. V. (1998). Handbook of Integral Equations. Boca Raton: CRC Press. ISBN 0-8493-2876-4.
• Titchmarsh, E. C. (1948). Introduction to the Theory of Fourier Integrals (Second ed.). Oxford University Press.
• Yakubovich, S. B. (1996). Index Transforms. World Scientific. ISBN 981-02-2216-5.
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• Tables of Integral Transforms at EqWorld: The World of Mathematical Equations.