# Mellin inverse transform of $s^{k}$

is it posible to prove rigorously that the Mellin transform of the distribution

$$(xD)^{k}\delta (x-1)$$

is just $s^{k}$ ?? i have proved it by integration by parts i mean

$$s^{k}= \int_{0}^{\infty}dxx^{s-1}(xD)^{k}\delta (x-1)$$

and of course $D = \frac{d}{dx}$

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Do you know what the Mellin transform of $\delta(x-1)$? –  Mhenni Benghorbal Aug 6 '13 at 21:23
the mellin transform of $\delta (x-1)$ is just 1 but perhaps from integration by parts we get powers of 's' –  Jose Garcia Aug 6 '13 at 21:46