Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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} $

share|improve this question
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.