The Mellin transform is defined as: $$F(\mu)=\int_0^\infty f(x)x^{\mu-1}dx$$ The derivative of the Mellin transform is: $$F'(\mu)=-(\mu-1)F(\mu-1)$$ Applying this property, for example to the Bessel equation: $$x^2 y''+xy'+(x^2-\nu^2)y$$ we can transform it in the complex difference equation: $$Y(\mu + 2) = ±(\mu − \nu)(\mu + \nu)Y(\mu)$$ where $Y(\mu)$ is the Mellin transform of $y(x)$.

Is this method useful in general to solve ODE? Thanks

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    $\begingroup$ The Mellin transform is mainly used in (analytic) number theory. One property it has is that it is well-behaved under multiplication by a scalar (the measure being used is $\frac{dx}{x}$). But I've never seen it in the context of ODEs. $\endgroup$
    – M Turgeon
    Jul 6 '12 at 14:45
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    $\begingroup$ Did you see this? $\endgroup$
    – Artem
    Jul 6 '12 at 16:36

Here we give a simple example of the use of the Mellin transform to solve a differential equation.

Consider the differential equation $$\begin{equation*} f'(t) + f(t) = 0.\tag{1} \end{equation*}$$ On taking the Mellin transform we find that the corresponding complex difference equation is $-(s-1)\phi(s-1) + \phi(s) = 0$, or $$\phi(s+1) = s \phi(s).$$ This is the recurrence relation for the gamma function, so $$\phi(s) = \Gamma(s)$$ up to an overall numerical factor.

The solution to (1) must be $$\begin{equation*} f(t) = \frac{1}{2\pi i} \int_{c-i\infty}^{c+i\infty} ds\, t^{-s} \Gamma(s),\tag{2} \end{equation*}$$ where $c$ picks out some suitable contour avoiding the poles of $\Gamma$. Recall that $\Gamma$ is meromorphic with simple poles at $0, -1, -2, \ldots$. Existence of the inverse transform (2) can be shown by appealing to the asymptotic behavior of $\Gamma$, $$|\Gamma(x+i y)| \sim \sqrt{2\pi} |y|^{x-1/2} e^{-|y|\pi/2},$$ in the limit $|y|\to \infty$.

We choose $c>0$. With this choice of contour the integral (2) is the Cahen-Mellin integral. Pushing the contour to the left, we pick up all the poles of the gamma function. To calculate the residue, recall that near $s=-n$ $$\Gamma(s) = \frac{(-1)^n}{n!}\frac{1}{s+n} + O(1).$$ This is a straightforward consequence of the recurrence relation for $\Gamma$. Thus, $$\begin{eqnarray*} f(t) &=& \sum_{n=0}^\infty \mathrm{Res}_{s=-n} t^{-s} \Gamma(s) \\ &=& \sum_{n=0}^\infty t^{n}\frac{(-1)^n}{n!} \\ &=& e^{-t}. \end{eqnarray*}$$


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