Is the Mellin transform useful to solve differential equations? 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
 A: 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,
$$\begin{eqnarray*}
f(t) &=& \sum_{n=0}^\infty \mathrm{Res}_{s=-n} t^{-s} \Gamma(s) \\
&=& \sum_{n=0}^\infty t^{n}\frac{(-1)^n}{n!} \\
&=& e^{-t}.
\end{eqnarray*}$$
