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What is the value of the Mellin transform ?

  1. Feb 3, 2010 #1
    my question is what is the value of

    [tex] \int_{0}^{\infty}dtt^{s-1} [/tex]

    if s >1 the integral is divergent as x -->oo

    if s<1 the integral would become divergent as x-->0

    then what is the value of this integral, or the value (regularized i presume) of

    [tex] \int_{0}^{\infty}dtt^{s-1}cos(at) [/tex] fro every real value of 'a' thanks.
  2. jcsd
  3. Feb 3, 2010 #2


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    It has meaning only for 0<s<1. For s = 1/2, the value is √(π/2a). Since cos is even, a >0 can be assumed.
  4. Feb 4, 2010 #3
    thanks.. but could you use analytic continuation for values bigger than Re (s) >1 ??

    for example if we consider in a regularized sense that [tex] \int_{0}^{\infty}t^{s-1}dt=0 [/tex] for every 's' then

    [tex] \int_{0}^{\infty}frac(1/t)t^{s-1}dt= -\zeta(s)/s [/tex] for every s except s=1
  5. Feb 4, 2010 #4
    It is divergent on the limit [itex]x\to\infty[/itex] when [itex]s\geq 0[/itex], and on the limit [itex]x\to 0^+[/itex] when [itex]s\leq 0[/itex]. The whole integral is thus divergent for any [itex]s[/itex].

    What do you mean by this "regularization". You have given value 0 to a divergent integral! :surprised
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