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If $f$ is a locally integrable function then its Mellin transform $\mathcal{M}[f]$ is defined by $$ \mathcal{M}[f] (s) = \int_0^{\infty} x^{s - 1} f (x) dx . $$ This integral usually converges in a strip $\alpha < Re \; s < \beta$ and defines an analytic function. For our purposes we can assume that $\mathcal{M}[f]$ converges in the right half-plane.

Let us denote $F (s) =\mathcal{M}[f] (s)$. Provided that the corresponding Mellin transforms exist, the basic general theory tells us that, for instance, $$ \mathcal{M} \left[ \frac{d}{d x} f (x) \right] = - (s - 1) F (s - 1),\quad \mathcal{M} \left[ x^{\mu} f (x) \right] = F (s + \mu) . $$ This allows us to translate a differential equation for $f (x)$ into a functional equation for its Mellin transform $F (s)$.

Example: For instance, the function $f (x) = e^{- x}$ satisfies the differential equation $$ f' (x) + f (x) = 0 $$ which translates to the functional equation $$ - (s - 1) F (s - 1) + F (s) = 0 $$ for its Mellin transform. Of course, the Mellin transform of $e^{- x}$ is nothing but the gamma function $\Gamma (s)$ which is well-known for satisfying exactly this functional equation.

Now, let us assume that we are given a function $f (x)$ and its Mellin transform $F (s)$. Further, suppose that we know that $F (s)$, just as the gamma function, can be analytically extended to the whole complex plane with poles at certain nonpositive integers. We also know that $F (s)$ satisfies a functional equation which we would like to translate back into a differential equation for $f (x)$. Formally, we obtain, say, a third order differential equation with polynomial coefficients. Can we conclude that $f (x)$ solves this DE?

The issue is that in our case the derivatives of $f (x)$ develop singularities in the domain and are no longer integrable. So Mellin transforms can't be defined in the usual way for them (and so we can't just use Mellin inversion).

What I am looking for is conditions under which we can still conclude that the functional equation for $F (s)$ translates into a differential equation for $f (x)$. Preferably, these should be conditions on $F (s)$ and not on $f (x)$. If it helps, we can assume $f (x)$ to be compactly supported.

Any help or references are greatly appreciated!

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2 Answers 2

As long as the singularities are not "too bad", the answer will be yes, $f(x)$ will represent a solution of the differential equation. The very same way that $$ \sum_{x=0}^{\infty} (-1)^{x}n!x^{n+1} $$ represents a solution of $x^2y'+y=x$. One might object that the series diverges, but resummation theory says this is irrelevant, that sum nevertheless represents a unique function (in a sector) which is a solution of the differential equation.

Balser's book, "From divergent series to analytic differential equations", would be a good place to start. Since the theory fundamentally uses Mellin transforms, you should be able to find what you want (perhaps indirectly) there.

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+1 for the ${}_2 F_0$ series example. :D –  J. M. Aug 19 '10 at 16:01
    
Thank you! I will definitely check out Balser's book. –  Armin Straub Aug 21 '10 at 8:41
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Dear Jacques Carette -- I've tried following up on this and haven't been able to find anything related to Mellin transforms. Can you be more precise about where in his book they are discussed? (Also, which book? It seems you've combined the title of two of his books.) Thanks in advance. –  JBorger Aug 25 '10 at 6:50
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By defining the Mellin transform for distributions as for instance done in Transform Analysis of Generalized Functions by O. Misra, J. Lavoine it follows that the functional equation for $F(s)$ translates into a differential equation of which $f(x)$ is a weak solution.

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