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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
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