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how do Gelfand and Shilov prove that the finite part

$$ F.P \int_{0}^{\infty}x^{m}dx=0 $$

however i get for non zero lower limit the recurrence of integrals in terms of the Riemann zeta function

$$ \begin{array}{l} \int\nolimits_{a}^{\infty }x^{m-s} dx =\frac{m-s}{2} \int\nolimits_{a}^{\infty }x^{m-1-s} dx +\zeta (s-m)-\sum\limits_{i=1}^{a}i^{m-s} +a^{m-s} \\ -\sum\limits_{r=1}^{\infty }\frac{B_{2r} \Gamma (m-s+1)}{(2r)!\Gamma (m-2r+2-s)} (m-2r+1-s)\int\nolimits_{a}^{\infty }x^{m-2r-s} dx \end{array} $$

which i believe it is correct . i have checked it and if we do not use regularization you get the usual results of calculus $ \int_{0}^{N}x^{m}dx = \frac{N^{m+1}}{m+1} $ if a replace the zeta fucntion by its finite sum

$ \zeta (-m) _{finite}= \sum_{i=0}^{N}i^{m} $

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What is $\;s\;$ ? –  DonAntonio Jun 26 '13 at 11:16
sorry i forgot 'm' is a real number and $ s \to 0 $ , here the parameter 's' plays the role of a zeta regulator so the integral is convergent for big 's' and after the calculations we take the limit s --->0 –  Jose Garcia Jun 26 '13 at 11:17
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